The Cauchy Method of Residues (Vol. 2)

The Cauchy Method of Residues Volume 2

Mathematics and Its Applications

Managing Editor:

M.HAZEWINKEL Centre for Mathematics and Computer Science, Amsterdam, The Netherlands

Volume 259

The Cauchy Method of Residues Volume 2 Theory and Applications

by

Dragoslav S. Mitrinovic and

Jovan D. Keckic Dtpartment of MaJMmaticS. Univtrsity of Btlgradt. Btlgrade. Yugoslavia

SPRINGER SCIENCE+BUSINESS MEDIA, B.V.

Library of Congress Cataloging-in-Publication Data (Revised for voI. 2) Mitrinovi~. Dragos1av S.

The Cauchy method of residues.

(v. 1: Mathematics and its app1ications. East European series) (v. 2: Mathematics and its app1icatians ; v. 259) "Volume 1 is a revised. extevded. and updated trans1ation of ••• Cauchyjev racun ostataka sa primenama published ••• by Nau~na knjiga. Be1grade. in 1978. whereas the greater part of volume 2 is based upon the second Serbian edition of the mentioned book from 1991"--Pref.. v. 2. VoI. 2 published by Springer-Science+Business Media, BV 1. Ana1ytic functions. 2. Ca1cu1us of residues. 1. Ke~kic. Jovan D. II. Series: Mathematics and its app1ications (D. Reidel Pub1ishing Company). East European series. III. Series: Mathematics and its applications (Springer-Science+Business Media, B.v.) ; v. 259. IV. Tit1e. QA331.M65713 1984 515.9 83-24697 ISBN 978-94-010-4883-5 ISBN 978-94-011-2000-5 (eBook) DOI 10.1007/978-94-011-2000-5

This book is a revised and updated translation by J.D. KeCkic of the supplement published in the second edition of D.S. Mitrinovic - 1.D. Kec"kic: Cauchyjev raeun ostataka sa primeru:una, Belgrade 1991. pp. 264-399. with new material added.

Printed on acid-free paper AU Rights Reserved

© 1993 Springer Science+Business Media Dordrecht

Originally published by Kluwer Academic Publishers in 1993 No part of the material protected by this copyright notice may be reproduced or utilized in any form or by any means. electronic or mechanical. including photocopying. recording or by any information storage and retrieval system. without written permission from the copyright owner.

Table of Contents

Preface

ix

Contents of Volume 1

x

1.

I 1 1

Introduction

1.1. Organization and References 1.2. Errata for Volume 1 1.3. Notations, Definitions and Theorems

3

2.

Evaluation of Residues

6

3.

Applications of Calculus of Residues in the Theory of Functions

3.1. 3.2. 3.3. 3.4. 3.5. 3.6. 3.7.

A Generalization of the Principle of the Argument Runge's Phenomenon Expansion into BOrmann's Series Carleman's Theorem Analytic Continuation of Cauchy Type Integrals An Asymptotic Formula Miscellaneous Applications

4.

Evaluation of Real Dermite Integrals by Means of Residues Integrals with Infinite Limits Integrals with Finite Limits CebySev's Approximation of the Integral of a Positive Function A Note on some Papers of Ostrogradski and Bouniakowski

4.1. 4.2. 4.3. 4.4.

5. Evaluation of Finite and Inf'mite Sums by Residues 5.1. Gauss' Sums 5.2. The Riemann Zeta Function 5.3. Miscellaneous Summations

18 18 19

22 23 26 26 29 36 36

47

64

67 69 69 72

74

vi

Table ofCootents

6. Applications of Calculus of Residues to Special Functions 6.1. Polygamma Functions of ArbitraIy Order 6.2. A Connection Between the Exponential and the Gamma Function 6.3. Residues of Some Functions Related to the Gamma Function 6.4. Some Integrals Involving the Gamma Function

80 80 82 84 87

Master's dissertation of J. V. Sohocki Introduction Properties of Residues Two Formulas of Lagrange Continued Fractions Legendre's Polynomials Expansion of a Function by Means of Continued Fractions

93 93 95 100 101 103 108

On the Principal and the Generalized Value of Improper Integrals

110

7. 7.1. 7.2. 7.3. 7.4. 7.5. 7.6. 8.

(DRAGAN S. DIMITROVSKI)

8.1. Substitution in Complex Integrals 8.2. The Principal Value for Higher Order Poles 8.3. The Principal Value in the Case when the Limits of Integration are Singular Points 8.4. Generalized Value of an Improper Integral with Infinite Limits 8.5. Generalized Value of an Improper Integral Between Finite Limits

110 112 114

9.

Applications of the Calculus of Residues to Numerical Evaluation of Integrals (DoBRILO D. To§lC)

133

10.

Inclusive Calculus of Residues

140

Complex Polynomials Orthogonal on the Semicircle

147

(MIODRAG S. PETKOVlC)

11.

117 128

(GRADIMIR V. MILOvANOVlC)

11.1. Introduction 11.2. Orthogonality on the Semicircle 11.3. Existence and Representation of 1rn 11.4. Recurrence Relation 11.5. Jacobi Weight 11.6. Symmetric Weights and Gegenbauer Weights 11.7. The zeros of 1rn(z)

147 149

12.

162

A Representation of Half Plane Meromorphic Functions (DRAGl§A MITROVlC)

150 155

156 157 159

Table of Contents

13.

Calculus of Residues and Distributions

vii 167

(DRAGIIA MrrROVlC)

13.1. 13.2. 13.3. 13.4. 13.S. 13.6. 13.7.

Test Functions and Distributions The SpacesD andD' The Spaces E and E' TheSpacesOQ andOQ ' A Distributional Representation of Half Plane Meromorphic Functions A Generalization of the Residue Theorem A Generalization of the Cauchy Integral Theorem for an Infinite Strip

167 167

170 171 173 176 180

Name Index

187

Subject Index

190

Preface Volume 1, i. e. the monograph The Cauchy Method of Residues - Theory and Applications published by D. Reidel Publishing Company in 1984 is the only book that covers all known applications of the calculus of residues. They range from the theory of equations, theory of numbers, matrix analysis, evaluation of real definite integrals, summation of finite and infinite series, expansions of functions into infinite series and products, ordinary and partial differential equations, mathematical and theoretical physics, to the calculus of finite differences and difference equations. The appearance of Volume 1 was acknowledged by the mathematical community. Favourable reviews and many private communications encouraged the authors to continue their work, the result being the present book, Volume 2, a sequel to Volume 1. We mention that Volume 1 is a revised, extended and updated translation of the book Nau~na knjiga, Belgrade in 1978, whereas the greater part of Volume 2 is based upon the second Serbian edition of the mentioned book from 1991.

Cauchyjev raeun ostataka sa primenama published in Serbian by

Chapter 1 is introductory while Chapters 2 - 6 are supplements to the corresponding chapters of Volume 1. They mainly contain results missed during the preparation of Volume 1 and also some new results published after 1982. Besides, certain topics which were only briefly mentioned in Volume 1 are treated here in more detail. Chapter 7 is devoted to the Master's thesis of I. V. Sohocki from 1868, a text of great historical interest, since it contains many original contributions which have always been ascribed to other mathematicians. The remaining Chapters 8 - 13 are special contributions written by various authors and are based mainly on their own research work. They include topics such as the generalized value of an improper integral (D. S. Dimitrovski), numerical evaluation of definite integrals (D. D. T~iC), inclusive calculus of residues (M. S. PetkoviC), polynomials orthogonal on a semicircle (G. V. MilovanoviC), and an interesting generalization of the residue theorem to distributions (D. MitroviC).

It seems reasonable to suppose that all those who found Volume 1 useful will also be interested in this book. Some parts can be used for both undergraduate and graduate courses, and like Volume 1 the book as a whole is suitable for postgraduate courses. It ix

x will also be of interest to researchers in complex analysis, physicists and engineers and all those who apply complex functions. The authors wish to express their appreciation to Kluwer Academic Publishers, well known for their high quality productions, and particularly to the publisher Dr. D. J. Lamer, as well as the series editor, Professor Dr. M. Hazewinkel, for their most efficient handling of the publication of this monograph. Belgrade, 27 January 1993 St. Sava's Day

D. S. Mitrinovic and J. D. KeCkic

Contents of Volume 1 1. Introduction

1

2. Definition and Evaluation of Residues

5

3. Contour Integration

24

4. Applications of the Calculus of Residues in the Theory of Functions

51

5. Evaluation of Real Definite Integrals by Means of Residues

108

6. Evaluation of Finite and Infinite Sums by Residues

211

7. Differential and Integral Equations

253 286 308 323 343

8. Applications of Calculus of Residues to Special Functions 9. Calculus of Finite Differences 10. Augustin-Louis Cauchy Notes Added in Proof

(ISBN 90-277-1623-4)

Chapter 1

Introduction 1.1. Organization and References 1.1.1. This book is divided into chapters, each chapter into sections, and many sections into subsections. The numeration of theorems, definitions, remarks, examples and formulas runs continuously throughout a subsection, or a section which does not contain subsections. There are many cross-references in the book. If a theorem referred to belongs to the same subsection only its number is given, while if it belongs to another, the numbers of the chapter, section, and subsection are given. So, for example, 3.1.5 means Chapter 3, Section 1, Subsection 5. 1.1.2. Bibliographical references are quoted after each subsection if it exists, or after each section if it does not. The titles of papers and books written in English, French, German, or Italian are quoted such as they are. All the other titles are translated into English and the original language is given in parenthesis. There are many references to our previous book D. S. Mitrinovic and J. D. Keckic: The Cauchy Method ofResidues. Theory and Applications. D. Reidel Publishing Company, Dordrecht Boston - Lancaster 1984; in further text it will shortly be referred to as Volume 1.

1.2. Errata for Volume 1 Professor P. 1. de Doelder (Eindhoven) and Professor D. D .. To~ic (Belgrade) have pointed out to the authors that there are a few misprints and errors in Volume 1. They are listed and corrected below. 1.2.1. At the bottom of p. 37 the term e-mlR should be replaced by e-mR • In other words, the last line but one on p. 37 should read

_ &7r(1 &7r - -e-mR) <-, m

m

1

2

Chapter 1

Ll.l. The term 2m in the first formula at the top of p. 117 should be replaced by 2n: Hence the formula should read

e'"

21r

../2,

- -..e J-chxd x =l+e

+<0

-eo

1r = -trI' cos2

1.1.3. A minor misprint is present in the formulation of Theorem 5 on p. 133: the word its was misprinted as it. Namely, the second line of that theorem should read

10 it is analytic in the whole plane; its singularities which do not belong to 1.2.4. In the definition of the contour C at the beginning of p. 149 the expression 21r n

was, in two places, misprinted as 2m. The correct definition of the contour C is as n

follows

C= {zIO~Rez~R, Imz =O}u{zllzl=R, u{z

O~argz~ 21r}

IO~lzl~R,

n

21r argz=-}.

n

1.1.5. The value of the second integral in 5.3.4.9 on p. 157 is incorrect. The correct result is +J"'cosax 1 d 1rcha v.p. ocosbxl+~ x= 2chb' 1.1.6. The term e'C!l+IR+It)" in the integrand in formula (7) on p. 164 should be replaced by e i C!l+IR+il)"/2. Also, an absolute value sign is missing. Hence, the first line offormula (7) on p. 164 should read

Jf(z)dz

DB

= Je-.JW(R+IR+It) -1 dt. .JW

t/(R+IR+It)"/2

0

1.1.7. The proof of 6.7.6 onp. 246 holds provided thatxz *ki. 1.1.8. The integrand in the formula (2) on p. 287 has a spurious factor should be replaced by (Z2 + l); i. e. the formula (2) should read

d

1

J

+'"

-logF(z)=K+logz---2 y dy. dz 2z 0 (z2 + y2)(e2q -1)

(Z2

+ y2)2 which

3

Introduction

1.2.9. There is a misprint in Reference 1 on p. 349. The pages of the quoted article are: 157-160.

1.3. Notations, Definitions and Theorems 1.3.1. Since standard notations are used throughout the book, it is believed unnecessary to define them all. Hence, we list only a few of them.

N

the set of all positive integers

No

the set of all nonnegative integers

Z

the set of all integers

R

the set of all real numbers

R+

the set of all positive numbers

R~

the set of all nonnegative real numbers

C

the set of all complex numbers

dgP

degree of the polynomial P

intr

for a closed contour r, int denotes the finite (internal) region bounded byr for a closed contour r, ext denotes the region bounded by which contains the point 00

ext

r

OG v. p.

r

r

r

the boundary of the region G

J

the principal value of the integral

For technical reasons the integral of denoted in two ways, namely:

tf(z)dz r

and

f

along (positively oriented) closed contour

r

is

1

f(z) dz.

r

1.3.2. The reader is assumed to be familiar with complex analytic functions, their derivatives and integrals, with Laurent's expansions and singularities. Volume 1 begins with the definition of the residue. In order to make this book selfcontained we repeat some basic definitions and theorems relevant to residues. THEOREM 1 (Cauchy)./f f is a regular function in a simply connected region G, then

tf(z)dz = 0, r

where r(c G) is a closed contour. DEFINITION l. Suppose that the point z = a is the only singularity of the analytic function

4

Chapter 1

f in the region G = {zl!z-al
= LA,,(z-a)"

I

in the region {z 0
(2)

Resf(z)=A_1• r-"

REMARK 2. Fonnula (2) is often taken as a definition of the residue offat Q. DEFINITION 2.lf f

is regular in a neighbourhood of the point 00 which can be an isolated singularity of J, and if r is a circle so large that all other singularities are contained in int r, then we define Resf(z) = _1_. ff(z)dz. 2='" 2 trl r-

REMARK 3. In Definition 2 the integral is taken along negatively oriented contour rsince in that case the neighbourhood of 00, i. e. the region extr, remains left from the contour. THEOREM 2.

We have

Resf(z) =-Resi- f(!). r=0 Z

r='"

THEOREM 3.lfthe point

Z

a (:,t 00) is a regular point or a removable singularity of f, then

Resf(z) =O. r="

REMARK 4. The above theorem does not hold for the point fez)

=1 + 1/ z

is regular at

00,

00.

Indeed, the function

dermed by

but Resf(z) =-1. Again, the point 00 is a removable singularity of ZOG

..

the function g dermed by g(Z)=(Zl_Zl +1)/ Zl, but Resg(z) = 1. THEOREM 4.lfthe point

f

'" a (:,t 00) is a pole of order k of the function f, then

5

Introduction

Res/(z) = Z=8

1 d k- 1 lim--.::l(z-at I(z). (k-l)1 H 8 dz

5 (Cauchy's residue theorem). Let 1 be an analytic junction in the region G, bounded by a closed contour r, and let 1 be continuous on r. Furthermore. let aI' ... ,an be isolated singularities 01 1 contained in G. Then THEOREM

f/(z)dz =2ni:t~es/(z).

r

1-=1

Z-G"

THEOREM 6. If an analytic lunction 1 has only isolated singularities in the extended plane. then the total sum olresidues ollis equal to zero.

Chapter 2

Evaluation of Residues 2.1. Suppose that g and h are regular functions at z = a and that z =a is a zero of order m of the function h, i.e. (1)

L a,,(z-a)",

+~

g (z)=

,,=0

L b,,(z-a)"

+~

h(z)=(z-a)m

,,=0

In Volume 1, Example 8, pp. 14-15 it was shown that

(2)

Res!(z) =_1

I hIho I

hom!

Z-Q

0 ho hI

h2

i hm-

I

hm- z

... 0 0 0 hI

ao al

°2

Qm-I

This last formula can be written in such a way that the coefficients from (1) do not figure in (2). Indeed, we have (3)

Ok

g(k) (a)

=--, k!

},(m+k) (a) hk = - - - - - .

(m+k)!

ao,

ai' ... bo, b l ,

...

(k =0, 1, ... )

and those values can be substituted in (2). This form of (2) can be found in [1], but with an error. The table on the next page is also given in [1]. REFERENCE

1. J. E. Marsden: Basic Complex Analysis. San Francisco 1973. 2.2. Suppose that f is an analytic function and define the function g by g(z) = !(z") where n E N. If Zo is a pole of g, then 2m

(E,,=-~

n

k=I, ... ,n-!)

are also poles of g and (1)

Res g(z) =.( Res g(z). 6=Z.

z=z"

6

7


Type of sin2UIaritv

Condition

fez)

Resf(z) 6=.

g(z) h(z) fez)

g and h have a zero at z = a of the same order

Removable

0

lim (z-a) fez)

Simple pole

lim(z-a) fez)

g(z)

g(a) ~o, h(a) hi (a) ~ 0

h(z) g(z) h(z)

Z-+
~o

=0

g(z)

~o

g(a) ~O, g'(a) ~O, h(a) =0, h'(a) 0, h"(a) 0, h"'(a) ~ 0

g(z) h(z)

Second order pole

=

=

m is the smallest integer such that lim F(z) exists,

fez)

z-+a

where F(z) = (z-a)-j{z)

h(z)

Atz = a g has a zero of order nand h a zero of ordern+m

g(z) h(z)

g(a) ~O, h(a) 0, ... h(.... J) (a) 0, h(m) (a) ;/:0

g(z)

=

=

h'(a) g(") (a) (m+I) h(-J) (a)

Simple pole Second order pole Second order pole

g(a) ~O, h(a) =0 h'(a) ~O, h"(a) ~ 0 g(a)

g(a)

Simple pole

Atz = a g has a zero of order m, h has a zero oforderm+I

h(z) g(z) (z-a)2

Z-+
2 gl(a)_ 2 g(a) h"'(a) h"(a) 3 h"(a)2 g'(a) g"(a)

3 g'(a) h(4) (a)

h"'(a)

2

3--.

Pole of orderm Pole of orderm Pole of orderm

11m Hit

h"'(a)

F(..-J)(z) (m-I)!

.

F(m-I)(z)

z-+a

(m-I)!

F(z)

=(z _a)m g(z)

11m

, where h(z)

Residue is given by (2) and (3)

Proof Since (2)

g(z/;.)=f(z"t;)=g(z)

and limlg(z)1 =+00, z--+z"

we conclude that lim Ig(z)1 =+00, which means that zk is a pole of g. z--+zt

Suppose that Fo is a contour around Zo such that there no other singularities in int Fo, and let ~ be the contour obtained by rotating Fo around the origin through angle 2k1r. n The contour ~ encloses the pole zk. Hence (3)

Resg(z) =_1_. tg(z)dz, z=z"

21t1r.

8 (4)

Chapter 2

f

Res g(Z) =~ g(z)dz. 2=Zo 2m It

Putting z =t ~ into (4) we get Resg(z) .....

=~.£g(t~)~dt 2m'1 Ii

since t describes To in positive direction, as z describes lie in that direction. Applying (2) we get (5)

f

Res g(z) =~ ~ g(t)dt -Zo 2m Ii

and (1) follows from (3) and (5). REMARK. This result is due to D.

Z. Dokovi~.

l.l.lfthe function/has a pole of order k at z = a, the standard formula for the evaluation of residue of/at z = a is, of course, (1)

1

dk -

I

Res/(z)=--lim --«z-a)k fez»). z=a

(k--l)!

Z-+lI

dZ k -

1

In [1] it is suggested that the formula (2)

I

dp-I

Resf(z) = ··---lim----- «z -a)p f(z», z=a

(p- I)! Z-+lI

dZp-l

where p is a positive integer> k is often more useful, particularly when / has the form f(z)=

g(z) , (z-a)P

whereg(a) = 0 andp > k. For example, the function eZ-1

Z ~f(z)=----- Z3

(z+ 2)

has a pole of order 2 at z = O. Applying (2) for p (3)

(e

z -I ) . d2 Resf(z) = -1 hm- ,

z=o

2z-+odz2

z+2

= 3, we get

9


whereas the application of the standard fonnula (1) gives

(4)

. d(eZ-I)

Res/(z)= hm- - - - . z--+odz

z=O

z(z+2)

and (4) is more complicated to calculate than (3). REFERENCE

1. R. C. Gupta: A New Fonnulafor Finding the Residue. Math. Educ. Sect. A 15(1981),32. 2.4. The question of practical computation of residues of some functions which have essential singularities at z = 0 is considered in [1]. Such functions are, for instance, Z f-+

zn exp g ( : ),

Z f-+

(a + bz)-n exp g (~),

Z f-+

zn exp (az + :) ,

where n is a positive integer and g is an entire function. The residues are obtained by expanding the relevant functions into series in a neighbourhood of z = 0, and they are expressed in tenns of numerical series which can often be summed. We give a few examples. +«>

EXAMPLE 1. If f(z)

=:~:>.z·, then • .0

ResJ(z) eelz = z=O

+OOacn + 1

2:

_n_ _ •

n=O (n + I)!

EXAMPLE 2. If a ~ 0, b ~ 0, then

(I)

1 Res - - = -eelz

z=O

a+bz

(

b

)n cn+'

1

1

2: ----= -b - -h e- bela; a n=O (n+ I)! +00

--;

EXAMPLE 3. We have

(2)

2:

b) + 00 an bn ( b ReszP-'exp ( az+-- =bp ------=- z=O z n=O n! (n + p)! a

where Jp is the Bessel function of order p.

)P12 J (2V-_ab ) p

10

Chapter 2

REMARK. Using computers the desired results can be obtained in a relatively short time. For

example. in [1] it is recorded that applying (1) we get

(e- 16/' -1) e- 1 i 16

in 1316 milliseconds. while applying (2) we get exp

Res

(2

+

Z

Sz -

1) = -

(3i+1)z·

z=o

2 V2i I, (2 iv'!O)

---:----'-----:--

15 y'Se i i+Sy'Sei

in 5733 milliseconds. REFERENCE

1. T. S. Norfolk: Symbolic Computation of Residues at Poles and Essential Singularities. ACM SIGSAMBull. 16(1982). 17-23.

1.5. We have Res+ (2Z2 -1) &2 (arctgz)2 = 11: (1

z=oo

z

-~) . e

Proof. The function F defined by F(z)

= -Z21

(2 Z2 - I) ez2 (arctg Z)2

has critical singularities at z = i and z = -i. We take the branch of arctg for which

limarctgz =1r12. Putting z = lIw in a neighbourhood of w =O. we get

z-+'"

1

nnw' 2 2 3

arctg z = arctg - = - - arctg w = - - w + w

S

- -W + ... 5

If Fl is the branch of F which corresponds to the chosen branch of arctg. then any other branch ofF has the form

where k is an integer. Let

11


g (z)

=

((2 -~) ez2 +~) (arctg Z)2; h (z) =~ (arctgz)2, Z2

Z2

Z2

so that

FI (z) =g (z) - h (z). First of all, when

z ~ 00, the function h behaves as the function

ZH

Z~ ( :

where & ~ 0, and hence

+ &),

y

Resh (z) = O.

z= co

I'

Furthermore, Figure 2.5.

Resg (z) = -~~.i g (z) dz, z=oo 2:11:/ j r-

x

where r is any closed contour which encloses the points -; and ; indicates that the integration is in the negative direction).

(r-

For the contour rwe choose the contour shown on Fig. 2.5. As the radii of the circles around -; and ; (and also the distance between the lines In and I'n~ tend to 0, we have

f-

g (z) dz -+

r

/FI(z) dz - (FI (z) + (( 2 -;~) e I

z2

The function

is odd, and so its integral from -1 to 1 is O. Therefore,

n

+ z12 ) (2 7(; arctg z + 7(;2 ) ) dz

=n; J((2 + y~)e-Y2-~-)(2arctg;y+7(;)dy. -I

n'

12

Chapter 2

and so Res F) (z) = Resg (z) - Res h (z)

z=oo

z=oo

z=oo

=_I_.J:g(Z)dz=-I-.2n2;(1-~)=n(1_~) 2n,j

2nl

r-

e

e

.

REMARK. This result is given, without a proof, in Volwne 1 (2.1.2, p. 13). REFERENCE 1. F. Tisserand: Recueil complementaire d'exercices sur Ie calcul infinitesimal, deuxieme edition, augementee de nOUlleuax exercices sur les variables imaginaires par P. Painleve. Paris 1933.

2.6. If a, bee, then (1)

z-o

Res ez log -

z-b

z=oo

= e" - eb.

Proof. The function F defined by F(z) =

ez log ~--:.~ z-b

bas critical singularities at z = a and z = b. All the branches of F are uniform outside a circle which encloses the points a and b. We consider the branch of the logarithm which is 0 for z = 1. If we putz = l/w when w is in a neighbourhood of 0, then

2

z-o I-ow onw log-·-=log---= -aw (OW 1 + - +a2-w+ ... +-+ ... ) z-b

(2)

I-bw

2

3

n

ntl

n

) bnw bw b w + bw ( 1 +---+-+ ... +-+ .... 2

2

2

3

n+-1

If Fl denotes the branch of F which corresponds to the chosen branch of the

logarithm, then any other branch of F has the form

zH F.(z) + 2ik!re z ,

k eZ. Hence

all the branches of F have the same residue at z = 00, since the integral 2iknJ eZdz taken along a circle of sufficiently large radius vanishes. The residue of Fl at z = 00 is equal to the coefficient of w, with changed sign, in the product ofthe series (2) and the series I

.-

I

1

w

2! w2

I

e W = 1 +_.. +._ .._+

... + J... - I .. + ... n! wn

13


This coefficient is

(b2!

b

n an a -a ( 1 +-+ ... +--+ ... ) +b 1 +-+ ... +._-+ ... ) (n + I)!

2!

(n+ I)!

=(-e"+l)+(eb-I),

and SO

ResFl (z)=et'-eb. Z= 00

Generalization. If n is a positive integer, then z-a Res znez Iog --=e" p (a) -eb p (b). z-b

Z=ao

where the polynomial p is defined by

REMARK. The equality (1) is given in Volume 1 (2.1.2, p. 13) without a proof. REFERENCE 1. F. Tisserand: Recueil complementaire d'exercices sur Ie calcul infinitesimal. deuxieme edition, augementee de nouveuax exercices sur les variables imaginaires par P. Painleve. Paris 1933.

2.7. Suppose that the function / is regular inside a closed contour which encloses the points at. ... , am and suppose that nt. ... , nm are positive integers. If the function F is defined by F(z)

=

f(z) (w-a,)n, (W-a 2)n2 ... (w-a m )nm, w-z z-a, z-a 2 z-am

then (1)

m

2 ResF(z)=P(w),

k= I Z=Qk

where P is a polynomial of degree not greater that nl+ ... + nm - 1, and P(a1)=f(a 1), .,. ,P(am)=/(am), P' (a 1)

=r (al ),

••• ,

p(n,-l) (a 1) =f(nl-l) (a 1 ),

r (am),

P' (am) = ....

p(nm- I ) (am) = jCnm-l) (am).

14

Chapter 2

REMARK. It was implicitly supposed that w ;t at (k = I, ... , m). If w =at for some k = I, ... , m, then the swn on the left hand side of (I) is o. REFERENCE

1. G. Julia: Exercices d'analyse, t. 2, deuxieme edition. Paris 1947.

2.8. Suppose that (1)

f(z+a)=

"

+ ..

k=1

k=O

2 A_kz-k + L Akzk

in the region {zl'i
certain cases the expansion (1) can be useful. So, for example, for Jacobi's elliptic functions sn, cn, dn, we have .

1

1 +k2

kz

6k

sn (z + IK') = - + - - z + 0 (Z3), cn ( Z+l·K') =

i

2 k2 - 1

-k-;+~-Z+

0 (z, 3)

and we conclude that those function have a simple pole at z = iK', and 1

Res sn Z = -

z=iK'

i

Res cn Z= - -, Res dn Z = - i.

k ' z=IK'

k

z=IK'

REMARK 1. Jacobi's elliptic functions also have a simple pole at z = 2K + iK' and Res

z=2K+iK'

1

sn z = - - , k

Res

z=2K+IK'

;

cn z = - , k

Res

z=2K+IK'

dn z = i.

REMARK 2. The functions sn, cn, dn and the constants k, K, K' are dermed by the so-called theta functions St, lh., ~, 84. The constants k, K, K', are connected by the equalities Ilk

K'

=}' V(u I

2

1

-1) (1 _k 2 u2 )

duo

(see Volwne I, p. 304). REFERENCE 1. E. T. Whittaker and G. N. Watson: A Course of Modem Analysis. Fourth edition (reprinted).

Cambridge 1952, pp. 504-505.

15


2.9. Let P",(z) =ao +a1z+ ... +a",z"', Qn(z) =bo+b1z+ ... +bnzn where a".bn the rational function/defined by

-:f;

0 and consider

= p.. (z).

/(z)

Qn(z)

Let rl, ... , rp (p ~ n) be the residues at its poles and let CR enough to enclose all the poles.

= {zllzl= R}

where R is large

I°Ifn>m+ 1 then

2° If n

m + 1 then

=

3° If n < m + 1, let T(z) be the remainder when Pm(z) is divided by Qn(z). Then

i:r.l: .1:=1

= r
Proof We have

i:r.l: =~t/(Z)dZ.

(1)

21t1 C.

.1:=1

1° If n > m + 1, then for sufficiently large R we have

R

laol+ladR+.·.+la",IR'"

~ -lin bn_1 R - ... - bo ~ 0, 2 tr bn R - lin-III as R

~

+00, and so in view of (1),

i>.I: =0. .1:=1

2° Let n

fez)

=

m + l. Then

=

a", + fez) _ am

b..+1

b",+l

16

Chapter 2

and so, by the previous result

§/(z)dz =2m a,.

b_ 1

c.

+0(1)

(R~+oo),

which, in view (1), implies

(n =m+l). 3° For n < m + 1, there exist polynomials Sand T such that P,.(z) =S(z)Q,,(z) + T(z),

where T(z) =Co +CIZ+ ... +c~·, C. t< 0 and s < n. Hence, by the previously established results we have

~:rk =0

if n > s+ 1,

k=1

The last two equalities can be written as

i:rk = T-,,-I)

(z) . (n -1)!b"

k=1

2.10. Up to now we have been concerned with the evaluation of a residue of a given function at a given point. We now consider, in a wayan inverse problem. Namely, instead of determining Res/(z) for a given function/and a given point z = a, we shall now, for r-tl

a given A

E

z=a isA.

C and a given point z = a, determine all those functions whose residue at

Let (/J be the set of all complex analytic functions for which z =a is a regular point or an isolated singularity, and let A be a given complex number. The general solution of the equation in/ (1)

Res/(z)=A z=tJ

is given by

17


A

1

Z-Q

Z-Q

f(z)=- + T(z) - - - Res T(z),

(2)

where T

E (/J is

z=a

arbitraty.

Proof For the function/defined by (2) we have

~:!/(z)=~:! Z~Q +~! T(z)- ~:! C~~~! T(Z») =A ~=e: Z~Q + ~:! T(z)

-(~:s T(z» (~:! Z~Q)

=A + Res T(z) - Res T (z) Z=Q

Z=Q

=A, which proves that (2) is a solution of (1). Conversely, let

10 E (/J

be a solution of (1), i.e. suppose that

10

is such that

Resfo(z) = A. We shall prove that a suitable choice of the arbitraty function Treduces (2) Z=II

to fez)

=fo(z). Indeed, let T =10. The equality (2) becomes fez)

=-

A

Z-Q

A

=Z-Q

+10 (z) -

-

1

Z-Q

Resfo (z) z=a

A +10 (z) --=/o(z). Z-Q

and this shows that (2) contains all the solutions of (1). REMARK. This is a special case of a more general result, proved in [1]. REFERENCE 1. J. D. K~kic: Reproductivity of Some Equations of Analysis II. Publ. Inst. Math. (Beograd) 33(47X1983),109-118.

Chapter 3

Applications of Calculus of Residues in the Theory of Functions 3.1. A Generalization of the Principle of the Argument Consider the integral (1)

1=_I_I./'(z)dz 2m ~ /(z) ,

where c= {zllz-al= r}. Ifz = a is a zero of order n of the regular function/and if there are no other zeros inside C, then 1= n. Similarly, if z = a is a pole of order p of the function/which has no other poles or zeros inside C, then 1= -po By analogy with this, B. Gavrilovic [1) defined the order of an essential singularity as follows: If z = a is an essential singularity of the function / which has no other singularities or zeros inside C, then the number I, defined by (1), is called the order of that essential singularity. In connection with this definition we mention that Weierstrass proved that the value of I is an integer, even if z = a is an essential singularity. Gavrilovic proved the following result: If the following conditions are fulfilled: 10 / is 2°/is analytic in the region int rwhere it regular and does not vanish on the contour can have a finite number of poles and essential singularities, then

r;

(2)

t

I =_I_ l'(Z)dZ = N -P+E 2m r /(z)

,

where N, P, E are the sums of orders of all zeros, poles and essential singularities off, respectively. Formula (2) generalizes the well-known Cauchy formula for the so-called principle of the argument. REMARK. There exist other generalizations of Cauchy's fonnula (and also of Jensen's fonnula) of similar kind. See, for example, Volwne I; 4.2.3, pp. 58-69.

Gavrilovic also proved in [1) the following theorem. Suppose that the function / is regular in the region int where r is a closed contour, and let the function g have in int r isolated singularities (poles or essential singularities) ah ... an. Suppose that in a

r.

18

19

Applications ofCaiculus of Residues in the Theory ofFunctions

neighbourhood of z = ale we have: I(z)

+'"

g(z) =G(z)+

= LA..t(z-akY; 1'=0

f

1'=1

A..t.,

(z-ak )

where G is regular in a neighbourhood of z = ale. Then (3)

1

+'"

n

-.t/(z)g(z)dz= LLA..I,kB"".

2m r

1'=1 k=1

Formula (3) contains, as a special case, formula (2) and also the well known Cauchy residue theorem. REFERENCE

1. B. Gavrilovic: On Residues o/Uniform Functions (Serbian). Rad JAZU 139 (1899), 29-39.

3.2. Runge'S Phenomenon Let I be a continuous complex function defined on the real segment I ai' ... , an E I SO that

and let

Any polynomial Pn of degree ~ n-l such that p,,(ak)=/(ak )

for

l~k~n

is an interpolation polynomial for the function/, and in certain cases it can be taken as an approximation for f If the polynomial Pn of degree n is defined by

then we may define Pn by the Lagrange interpolation formula

It is often taken for granted (particularly when this formula is used in physics) that n -+ +00 implies that Pn(x) -+ I(x) uniformly on I. However, this need not be true, not even for functions which are analytic in a region Dee which contains the real interval I. This fact is known as Runge's phenomenon. Indeed, if a> 0, the function/defined by

20

Cbapter3 1

f(z)=-, Z2+

a2

is meromorphic in the complex plane and has two simple poles at z = ;a and z = -;a. Suppose that R is a positive number such that for

l~k~n,

and let r={z\z=Re", 0~t~2n}. Consider the integral ~_",; /(z) Pn (x) 211:;:r (z-x)Pn(z)

dz

l'

where x:l: ak (1 ~ k ~ n) and x:l: ±ia. It is clear that the absolute value of this integral is not greater thanAIRn+2 whereA is a constant. If the functiong is defined by

g(z)=

/(z)Pn(x) , (z -x) p" (z)

then

Resg(z)=

Resg(z)=f(x),

Z=X

z=ak

Resg(z)= . . Pn(x)

.

2 /Q (Ia - x) Pn (Ia)

z=;a

,

(l ~k~n),

/(ak)Pn(X) (ak-X)Pn'(ak)

Resg(z)=

z=-ia

..

Pn(x)

.

2 la (Ia + x) Pn ( -Ia)

Hence, by the residue theorem, as R --. +00 we get

Res g (z) + z-x

n

L

Res g (z) + Resg (z) + Res g (z) = 0,

k=t Z=Dk

z=;a

z=-Ia

i.e.

f(x)-P (x)+

(1)

n

Pn(x) 2;a(;a-x)Pn (;a)

+

Pn(x) 2;a(ia+X)Pn(-;a)

O.

Suppose now that 1=[-1,1], that n = 2m is an even number and take

2k+l 2m

±--

and SO

(O~ k~n-l)

. for the pomts 01' ...

an'

Then

21

Applications of Calculus of Residues in the Theory of Functions Pn (x)

Pn (x)

2ia(ia-x)Pn (ia)

2ia(ia+x)Pn(-ia)

~~----+----~~----

=

Pn (x)

2 iaPn (ia)

(I

I)

ia - ; + ia + x

=-

1

Pn (x)

x 2 + a 2 Pn (ia) .

The equality (1) becomes (3)

and we shall show that for sufficiently small a the difference 1 (1) - Pn( 1) does not tend to

o as n tends to +00. Indeed, by Stirling's formula we have (4)

1 34m - 1 _ r ( 2 p (1)=----.. ' - - " ' V 2 -

n

2m2m

2m

e

)n ,

On the other hand, from (2) follows

and using Euler-Maclaurin's formula it can be shown that

where c*-O is a constant, while b is defined by

J I

1

o

a

10gb = log(a 2 +t 2 )dt =log(l +a 2 ) - 2 - 2aarctg-. As a

~

0, 10gb ~ -2 and we conclude that there exists an a> 0 such that

2 10gb < log- = log 2 -1. From (3), (4) and (5) follows that

e

1/(1)- p"(l)I-c{ e2b

J

(c' *- 0 constant)

and this absolute value tends to +00 as n

~

+00.

REMARK 1. This text was written according to [1). REMARK 2. Newman and Schoenberg [2] showed that a similar situation may occur when the function J, analytic on an infinite open interval I c R is approximated by polynomial spline functions which coincide with/at an infinite sequence of points from l. A proof by residues of this phenomenon is given in [3].

22

Chapter 3

REFERENCES

1. 1. Dieudonne: Calcul infinitesimal. Paris 1968, pp. 319-320. 2. D. J. Newman and I. J. Schoenberg: Splines and the Logarithmic Function. Pacific J. Math. 61(1975),241-258. 3. W. Schempp: A Note on the Newman-Schoenberg Phenomenon. Math. Z. 167(1979), 1-6.

3.3. Expansion into Hurmann's Series Suppose thatf and g are regular functions in a neighbourhood of z = a, and suppose that a is a simple zero off There exists a neighbourhood of z = a in which the function g can be expanded as an infinite series with respect to the powers of f

Proof Let C be a circle with centre at a and radius r, such that there are no other zeros of fin the disc {zllzl < r}, except the simple zero at a, and let

m = minlf(z)l. zEC Now, let K be a circle concentric with C such that If (z)1 < m for z E K, and hence for z E int K. If u E int K is arbitrary, then If (z)1 > If (u)1 for all z E C, and so by Rouche's theorem the functions z ~ f(z) and z ~ f(z) - f(u) have equal number of zeros in the disc int C. This means that the equation

f(z) - f(u)

=0

has only one simple root, namely z = u in the disc int C. Therefore, by the residue theorem fg(z)

c

f'(z) dz = 2niResg(z) f'(z) , f(z)- f(u) %=11 f(z)- f(u)

implying that ( 1)

g(u)

1 i.g(z)

=2ni~

f'(z) dz f(z)-f(u)·

However,

g(z)

!'(z) f(z) - f(u)

= g(z)!'(z) f(z)

1 1- f(u) f(z)

and integrating (1) term by term we obtain +«>

(2)

g(u) =IAk(f(u)Y, .1:=0

where

= g(z)!'(z) f(f(U»)k f(z)

k=O f(z)

23

AppIicatiOllS of Calculus of Residues in the Theory ofFuncti.OIIS

A =_I_I.g(z)!'(z) dz k 2m ~ (f(Z»),,+l

(3)

(k =0, I, 2, ... )

Formulas (2) and (3) define the expansion of g into a series with respect to the powers of f. It is usually called Bfirmann's series. For k > 0 the expression for the coefficientAk can be transformed as follows: 2·A ( I g(Z») I I. g'(z) d m k = k (f(z»)" c + k ~ (f(z»)" z

_!I. g'(z) dz - k ~ (f(z»)" ,

and putting !(z) = z-a F(z)

where F is regular in int C, we finally get d- ( ) = k!I da k - 1 g'(a)(F(a»)" . k 1

Ak

REMARK. It is not difficult to see that BOrmann's expansion is equivalent to Langrange's expansion (see Volume 1; 4.4.2, pp. n-79). BOrmann's result was published in 1799. REFERENCE

1. L. Cakatov: An Introduction to the Theory ofAnalytic Functions (Bulgarian), Sofia 1972, pp.

244-245.

3.4. Carleman's Theorem Let

!

be an analytic function in the region

{zllzl~r,

- ; ~argz~ ;} and suppose

that it has simple zeros 'ie/It (k = I, ... ,n) in the region D bounded by the semicircles

c={zllzl=r, -; ~argz~ ;},

C={zllzl=R, .-; ~argz~ ;}, where

R>r,

and

the parts of the imaginary axis joining them, and that it has no zeros on the contour. Then

(I)

L"(I) -\ cost =trR1- JIO~!(Rell)lcostdt r R /2

k-1

k

k

-_12

+~ j(~-~)logl!(;Y)!(-;Y)ldY+O(I). 2tr Y R r

24

Chapter 3

Proof Start with the integral (2)

I

=~ .((~+...!....)IOgf(Z)dZ. 2m '! Z2 R2 aD

This integral breaks into four integrals (along semicircles c and C, and two parts of the imaginary axis). The integral along the small semicircle is bounded. On the negative imaginary axis we have Z = -iy and its contribution is

1 R( --1 1) logf(-iy)dy. -J 2n y2 R2 r

On large semicircle C we have Z = Reit and its contribution is

1 1112 (-2it 1) 1 1112 -. J ~+-2 iReltlogf(Re,t)dt=- J(cost) 10gf(Relt)dt. 2m -1112 R R nR -tr/2

The integral along the positive imaginary axis gives

1 R( --1 1 ) logf(iy)dy. -J 2n y2 R2 r

Adding up those integrals, and taking the real part of I, we obtain the right hand side of (1).

On the other hand, integrating by parts, we have (3)

I

=~((-;'-.!.)IOgf(Z») +~ f(.!._-;')f'(Z) dz. 2m

R

Z

aD

2m aD Z

R

f(z)

As we describe the contour, logf(z) increases by 2mn. The integrated term is therefore purely imaginary. Since

we conclude that the integral in (3) is equal to

~(rk!'~ -r~~ ) and taking real parts, formula (1) follows. This is Carleman's theorem [1]. See also [2]. REMARK l. Ifwe replace the region D, i.e. the left half-annulus by the upper half-annulus G=

{z I r ~ Izl ~ R,

0 ~ argz :s;

and if we consider the integral (4)

~"(_l -~)IOgf(Z)dZ, 2m L R2 Z2

tr}

25


we obtain the following version of Carleman's fonnula 1 '\ ) sint L -'i R ft

(

i

i_I

1 " 1 R( 2"--2 1 1 ) 10gl[(x)[(-x)ldx+O(I). =-Jlogl[(Re")lsintdt+-J nR 0 2n, x R

REMARK 2. If[is a meromorphic function in the disc

{~lzIS R}, then by considering the integral

1log [(z) dz IoFR

z

we obtain the Jensen fonnula

I

I

I'

121< ft R .. R -Jlog[(Re u ) dt =logl[(O)1 + 1)og-11- LIOg-1 2n 0 t=1 at i=1 bt

where ak' bk are zeros and poles, respectively, of the function[such that 0 < Ia~
~ 1F'(z) log [(z)dz 2m lIH

where F is a function regular in H such that F(z)

E

R for z

E

R, he obtained the following result:

If ak are zeros of order a k (k = 1, ... , m) of[and if bk are poles of order Pk (k = 1, ... , n) of [ where ak' bk E H, and if [ has no zeros or poles on oH, then

(RIl)

lR -J(P(x)logl[(x~ +F'(-x)logl[( -x~)dx- Irn - J e"P(Relt)log[(Re")dt

2no

2no

= Irn(tatF(at )- tPtF(bt)}

and

REFERENCES 1. T. Carleman: Ober die Approximation analytischer Funktionen durch lineare Aggregate von vorgegebenen Potenzen. Arkiv tOr Mat. Astr. Fys. 17(1922), No.9. 2. E. C. Titchmarsh: The Theory o[Functions. London 1964. 3. D. Mitrovic: On an Equality [or Integrals (Serbian). Glasnik Mat.-Fiz. Astr. (2) 6 (1951), 193-

200.

26

Chapter 3

3.5. Analytic Continuation of Cauchy Type Integrals Suppose that the function 1 is regular in a region D which contains a part Co of the contour C, and suppose that Co divides D into two regions D+ and IF lying on the left and right of Co' respectively. Consider the Cauchy type integral

J

_1 I(t)dt 2m c t-z

(1)

and let F+ and F- be the functions represented by the integral (1) in D+ and D- , respectively. The function F- + 1 is the analytic continuation of F+ in the region D- through Co' Indeed, let C* be the contour formed from C when Co is replaced by a part of the boundary of D (in such a way that the orientation is preserved). See Fig. 3.5. Clearly, this part of the boundary of D is also a part of the boundary of D-. Let L - = fJD- . Then, for any zED such that z II: Co we have

~

J

l(t) dt 2m c.t-z

(2)

=~

J

I(t) dt 2m c t-z

J

+ ~ l(t) dt. 2m r t-z

Denote the integral of the left hand side of (2) by F(z). Since C* is not in the region D, the function F is regular in D. If z e D+, we have F(z) = F+(z), because the first integral on the right hand side of (2) is, by definition, equal to F+(z), and the second integral vanishes by Cauchy's theorem. If z e D-, we have F(z) =F-(z) +I(z), because the first integral in (2) is, by definition, equal to F-(z), and the second is equal to I(z), by Cauchy's integral formula. The function F is therefore the continuation of F+ onto the whole region D.

analytic

Figure 3.5.

REFERENCE 1. M. A. Evgrafov: Analytic Functions (Russian). Moscow 1968, pp. 245-246.

3.6. An Asymptotic Formula Suppose that the so-called Mittag-Leffler function E is defined outside the region

{zl Rez > 0,

G=

IImzl < 1l} by the integral

27


e"

1

E(z) = - . J-dt 2m c t-z

(C= aCT).

We shall determine the analytic continuation of E onto the whole complex plane and we shall establish an asymptotic formula for z H E(z) as z ~ 00. Since the function E is defined by a Cauchy type integral, we obtain an analytic continuation of E onto the whole plane by the method of 3.5. If we denote by J(z) the value of the Cauchy integral (which defines the function E for z Ii!: G u C) the analytic continuation of the function E in G is given by E(z)=J(z)-e"

(z eG).

Hence, in order to find an asymptotic formula for E, it is enough to find an asymptotic formula for the function J defined by a Cauchy type integral. Starting with

1 t-z

1 z

t

t2

--=----+--::--Z2

Z2(t-Z)'

we easily get Co c1 1 =--+-+-F(z), z Z2 Z2

J(z)

(1)

where Co

J 'dt, 2m c

1 = --.

e'

C1

=

Jte" dt, 2m c

e"

t2 2m c t-z 1

-~

F(z) = - . J-dt.

We shall first show that F(z) = 0(1) as z -+ 00. Let C 1 be the part of the contour C which lies inside the circle It - zl < 1 and let C2 be the remaining part of C. We then have F(z)

=1';(z) + 1'; (z)

where

1

t2

e"

1';(z) =-. J-dt, 2m c; t-z

1

t2

e"

1';(z)=-. J-dt. 2m c; t-z

First of all,

In order to find an upper bound for F I , denote by w the point of C 1 which is nearest to z and then

28 eO-

But

(1

1J

F.(z)=-. J-dt+J 2m C, t-z C,

e;-o- dt . t-z

l -11 ~ M

It-zl
e;-ot-z

and

-1

(if

Chapter 3

J-1d t = -1l o gz-a --

2td c, t-z

z-b'

2td

where a and b are the end points of C l (Le. they are the points of intersection of C l and the circle It - zl = 1). Since

Iz -al =Iz -bl =1, we have

dtl ~ 1. I~ J-l t-z 2m

C

Also, since WE C l , we have

lee-I ~ 1.

This means that both functions Fl and F2 are bounded, and so F(z)

=0(1)

(z

We now prove that Co

- Co

~

00).

= 1. Indeed, we have

1

1

1f;

2m c

2m

1
=-. Jeo'dt =- .

1

1

Je'dt + -2m. JeO'dt + -2m. Jeo' dt, -1
11;

-1
_1<;

since the function t H eo' is periodic with period 2td. The first and the third integral on the right hand side annul each other. In the second integral we introduce the substitution et = W and we obtain

1 -Co = - 2td

e e J -dw=-Res-=-l.

Iwl=l

W

W

W

w=0 W

We have therefore proved that Co

= 1,

F(z) = 0(1)

(z

and therefore, in view of (1), we get

I(Z)=;+O(~)

(z~oo).

~

00)

29


This gives us the following asymptotic formula for E:

E(Z)=;+0(:2)

1 . (1)

E(z) =--e +0 -Z2 Z REMARK.

(z ..... oo,

z~G)

(z ..... 00,

z e G).

Using the same method we can obtain a fonnula of the fonn

Co c. c. + ~ -12) I (z ) =-+2"+ ... +-.

zz

where

z'"

z··

f.

c = -1- t eold t •

2m c

which gives a more precise asymptotic fonnula for E. REFERENCE

1. M. A. Evgrafov: Analytic Functions (Russian). Moscow 1968, pp. 254-256.

3.7. Miscellaneous Applications 3.7.1. If (u, v) ~ R(u, v) is a rational function in u and v, we shall determine the conditions which ensure that the function J defined by J(z)=J R(sinz, cosz) dz is uniform and periodic. The function Z ~ I(z) =R(sin z, cos z) is uniform and periodic with period 2n, but its fundamental period can be smaller. Suppose that 0) = 2tr1k is the fundamental period off. If we put t = tg (kzI2), 1 becomes a rational function in t, and if we put u = eikz, 1 becomes a rational function in u, which will be denoted by u ~ F(u). If the function J is uniform and periodic, its fundamental period must be its fundamental period is 0)' < 0), then the equality

0).

Indeed, if

J(z+ru')=J(z) implies

~J(z+ru')=~J(z), i. e. dz

dz

l(z+ru')=/(z).

Hence, the period 0)' of J must be of the form mO) (m e Z). However, since the derivative of z ~ J(z + 0) - J(z) is zero, we conclude that J(z+O)-J(z) = 0

(0

constant)

30

Chapter 3

and SO

J(z+2w)-J(z)=2a, ... , J(z+mw)-J(z)=ma, which means that a = 0, and

0)

is indeed the fundamental period of J.

Since u = eikz, all those values of z which correspond to one value of u differ from each other by an integral multiple of 0). If J is a uniform and a periodic function in z, then the function J 1 is defined by

is also uniform in u. On the other hand, d dJdz dz 1 - J 1 (u)=--=f(z)-=F(u)-. du dz du du iku

~ F(u)

will be a uniform function only if ,k u all the residues of that function are zero; in that case this integral will also be a rational function in u. Hence, we have arrived at the result: The integral of the rational function u H

When J is uniform and periodic, it is a rational function in eikz, and this will take place if and only if all the residues of u ~ F(u)/u are zero. We now apply this result to the case when a+bv u

c u3

R(u, v)=--+-+d, i.e. f() z

c d = a+bcosz +--+ 3 sinz

sin z

(a, b, c, dER).

Since the fundamental period of f is 21l', put U = eiz to obtain

The residue at u = 0 is zero if

The residue at u = 1 is zero if

. 'b +-1C= 1. 0 la+, 2


31

and the residue at u = -1 is zero if

. ·b - - 1. -la+l IC= 0• 2

Hence, the integral

J (Z)=J (a+~cosz +_._C_+ d ) dz smz

sm 3 z

defines a uniform periodic function if and only if b = d = 0, c =-2a. In that case

J(Z)=J(_l _ _ 2 )dz=~. sin z

sin 3 z

sin2 z

REFERENCE

1. F. Tisserand: Recueil complementaire d'exercices sur Ie calcul infinitesimal. deuxieme edition, augementee de nouveuax exercices sur les variables imaginaires par P. Painleve. Paris 1933.

3.7.2. Suppose that P and Q are rational function in x, y. The equality (1)

oP

oQ

oy

ox

-=-

holds if and only if ResP(x,y) does not depend on y for any pole a of the function ~=tI

xI4P(x,y).

This result was proved in [l). The following interesting application is also given there. If the function P is defined by +F P ( x,Y) -_ Ax2 + 2 Bxy + Cy2 + 2 Dx + 2 Ey --, x2-2xy+ 1

determine the coefficients A, B, C, D, E, F so that there exists a rational function Q such that P(x, y)dx + Q(x, y)dy is a total differential. If the function P is considered as a function in x, while y is a parameter, then the poles ofP are at a =y+~y2 -1 and b =y-~l-l. Those poles are simple if y2 :#1. We have (2)

ResP(x, y)= Aa2+2.Bay+Cy2+2Da+2Ey+F x=a

2 (a-y)

and, on putting a =y + ~y2 -1, this residue must not involve y. Equivalently, this residue must not depend on a when we set

Chapter 3

32 02

(3)

+1

Y=--· 20

Substituting (3) into (2) we get ResP x=a

(x, 02+ 1) = (4A +4B+C) a4 +(8 D +4 E) 0 +(4 B+2 C+4 F)02+4E2 +C 3

20

40 3 -40

and the obtained value will not depend on a if 4A +4B+ C=O,

8D+4E 4

4E -4'

4B+2C+4F=O,

C=O,

implying A= -B=F

C=o,

D= -E.

Hence, the function P has the form Ax2-2 Axy+2Dx-2Dy+A A 2D (x-y) P ( x, y ) _ -= + x 2-2xy+l

x2-2xy+l

.

Using (1) we easily get

Q (x, y) = 2 -2Dx 2 1' x - xy+ and P (x, y) dx+ Q (x, y) dy=d(Ax+D log (X2 - 2 xy + 1». REFERENCE

1. G. Julia: Exercices d'analyse. t. 2, deuxieme edition. Paris 1947, pp. 150-157.

3.7.3. Let a l' ... , an be relatively prime positive integers and let M> 0. Denote by N(M) the number ofintegral positive solutions of the equation in xl' ... , Xn:

For large M the expression N(M) is of order (2)

i.e.

33

Applications of Calculus of Residues in the Theory ofFunc:tions

Proof This problem is easily reduced to the problem of showing that the number C(M) of integral nonnegative solutions of (I) is of order (2). Indeed. if M> 0 is large enough and if we put

then the number of integral positive solutions of the equation (I) is equal to the number of integral nonnegative solutions of the equation inYI' ... ,Yn :

(3)

atYt+··· +a"Y,,=M-at - · · · -o,,=Mt •

The equation (3) is obtained from the equation (I) by means of the substitution xk =Yk+1 (k= I, ... , n). The number C(M) of integral nonnegative solutions of (I) is equal to the coefficient of zM in the power series expansion off, where "

/(z)=n

1

(/ z

-0

k=11-zk

/< 1).

This expansion has the form /(z)

=

"

IT

(I

+zOk

+Z2ok + ... )=

k=l

+00

+ ..

kl=l

k n =l

2 ... 2

zO,k,+ ..• +a"kn •

Applying Cauchy'S formula we get C(M)=_l_.1 f(z) dz=_I_.1 g(z)dz 2:n; i j

r

2:n; i j

zM+1

r

,

where r= {z Ilzl= r; r < I}, and

g (z)=

f(z) = _ _ _ __ ZM+l

n

ZM+l

IT (l_z0k) k=l

The function g is analytic and has a finite number of poles on the circle {z Ilzl= I}. The poles are

34

Chapter 3

Besides, the function g has a pole at z = O. Since

Resg (z)= 0

(forM~

z=oa

1)

by the residue theorem we obtain C (M) = _1_. •1: g (z) dz = Res g (z) = 211: I Z_O r

(4)

z

:r

-

2: Res g (z). k,J z=zkj

Since the positive integers aI' ... , an are, by supposition, relatively prime, the poles at '* 1 are simple, and so

= Zkj

(5)

Res g(z)= lim (z-Zkj)g(z)=Kkj, z=zkj z-+zkj

where Kkj are constants. At z = I the function g has a pole of order n and so 1

dn -

1

Resg(z)= ----lim -(z-l)ng(z» (n-l)! z-+I dzn -

z=1

(6)

1

i

[

= ___ 1 ___1_ Mn-I + k=1 n

TIOk

Ok

2(n-2)!

(n-l)!

Mn-2 +

3(

i

k=1

Ok)2 -

i

k=1

24(n-3)!

Ok 2

1

Mn-3 + ....

k~1

From (4), (5) and (6) follows

(7)

lim

C(M)= n

M->+oo Mn-I

(n-l)!

IT Ok

k=1

REMARK 1. It can be shown that the equation (7) is true even if a p

... ,

an are not relatively prime.

REMARK 2. This text was written according to [I). The same problem was considered in [2], Problem 27, and the complete works of Laguerre (tom I, Paris 1898, pp. 218-220) were quoted as references. The problem is solved in [2] without the use of residues. REFERENCES

1. A. o. Gel'fond, L. V. ll'kov: Problem 2. (Russian). Mat. Prosv~cenie 5(1960), 267-270. 2. G. P61ya, G. Szeg(): Aufgoben und Lehrsatze aus der Analysis l. Berlin - G()ttingen - Heidelberg - New York 1964.

3.7.4. If f is a function of positive real part, regular in the unit disc D, with Taylor expansion +'"

f(z) = 1+ Lavz v v=J

then Reak ~ 2 for k

~

1.

(z ED),

35


Proof. We have

I"

=~t(2-Z" -z-")z-lf(z)dz 2maD

J

1 2• =- {l-coskt)f(et')dt fro

(k

~

1).

(k

since tz"dz =0 for n e Z and z *-1. aD

Hence, (k

~

1)

implying 2-Rea"

J{l-coskt)Ref(et' )dt ~ 0

Ib

=-

fro

i. e. (k

~

1).

REMARK. Proved by D. D. Adamovic.

(k

~

1)

~

1),

Chapter 4

Evaluation of Real Definite Integrals by Means of Residues 4.1. Integrals with Infinite Limits 4.1.1. If n is a positive integer, then

Jci:xr +00

L (n) =

dx

o

~-I

( _ 1)n/2 n in 2 -'----'--2: (_I)k ( kn ) (n-2k)n-l, 2n(n-I)! k=O

(1)

n even

n-I

2: (- I)k (n) (n - 2 k)n - II k

(_I)(n-I)/2nin-12

-'----=----2n(n-I)!

I,

k=O

odd.

Proof The following identities are true ~-I

(-l)"/22,,-1 sinn t=

22: (-l)n(n)COS(n-2k)t+~(-1)"/2( k=O

2

k

n-I

( -1)(,,-1)/2 2,,-1 sin" t=

2

2: ( - 1)" (:) sin (n -

2 k), n odd.

k=O

Define the function/" by the equality/" (z)

=Fn(z) + F n_3(z), n

II'

F" (z) =

2: (- l)k ( n) ei (n-2 k) z,

k=O

n' = {

k

2-1, n-I

where

n even

- 2 • n odd

36

n), n even

nl2

37

Evaluation of Real Defmite Integrals by Means of Residues

and where Pn-3 is a polynomial of degree n-3, such that the function regular in the entire plane, except at z = 0, and

Z

H

z-"/,,(z) is

Res (z-n I" (z)) = _1_/~"-I) (0). z=o

(n-I)!

If this function is integrated along the contour consisting of the semicircles {zllzl= R, Imz ~ OJ, {zllzl= r, 1m z ~ O} and the segments [-R, -r), [r, R) of the real axis and if R -+ +00, r -+ 0, we obtain (1). In particular, n 3n n I1S n L(1)=L(2)=---, L(3)=--, L(4)=-, L(5)=-. 2

833M

REFERENCE

1. E. T. H. Wang, R.L.Young: Problem 1064. Math.Magazine 53(1980),181-183.

4.1.2. Boas and Friedman noted in [1) (see also Volume 1; 5.2.3.7, pp.120-l2l) that the semicircular contour used for the evaluation of integrals of the form

Je I(x)dx can be

+'"

ix

replaced by a simpler triangular contour. In note [2) an other simplification is suggested, by considering the Fresnel integrals and for which we usually integrate

Z H

e around a quarter-circle, and the integral

Z-1I2 iz

Jsmx dx

+'"

o

•

x

for which we usually integrate z H show that

Z-l

eiz around a semicircle. We then end up having to

(1)

where k

= 112 for

the Fresnel integrals and k

= 0 for

integrals of the form

Je I(x)dx,

+'"

ix

38

Chapter 4

where f is a rational function with the degree of the denominator greater than the degree of the numerator. The integral involved in the limit is not elementary and has to be estimated by some device. The usual one is to apply the inequality sin t ~ 2thc. The authors of [2] believe that it is simpler to break the integral in (1) into the sum

Jo

J

.,3

./2

R" e-Roinl dt + R" e-Roinl dt. ./3

The second integral causes no difficulty, and for the first, since 2cost ~ 1 for OStSlr/3 we have

J

.,3

J

./3

R" e- Roinl dt S o

2R" e-Roinl cost dt. 0

The last integral can be evaluated explicitly and it is easily seen that it tends to 0 as R -+ +00. Another approach suggested in [2] is to give up the traditional use of circular-arc contours. For the Fresnel integrals take the contour to be the square with vertices at 0, R, R+iR, iR (initially indented at 0). The integrals along the segments [R, R + iR] and [iR, R + iR] are

J(R +iy)"-'e'(R+IY)dy R

R

and

o

f (x+iR)"-'e'(,,+IR)dx o

and the sum of their absolute values does not exceed R "-I J(R2+y2f2e-Ydy +

R

o

0

"-I

f(x 2+R 2)2 e- Rdx.

\

\

\

But R2 + y2 ~ R2 and x 2 + R2 ~ R2, k -1 < 0, and the above sum does not exceed R

R

o

0

R"-' f e-Ydy + e-RRJ:-I f dx

= R"-I(l-e-R)

+ R"e-R -+ 0 0

as

R -+ +00.

Je f(x)dx the rectangular contour is used in [3] and [4].

+«>

For integrals

bt

-«>

REFERENCES

1. R. P. Boas and E. Friedman: A Simplification in Certain Contour Integrals. Amer. Math. Monthly 84(1977), 467-468. 2. Mary 1. Boas and R.P.Boas: Simplification o/Some Contour Integrating. Amer. Math. Monthly 92(1985),212-213. 3. L. V. Ahlfors: Complex Analysis. New York 1969, p. 157. 4. I. Stewart and D. Tall: Complex Analysis. Cambridge 1983, p. 223.

39

Evaluation of Real Definite Integrals by Means of Residues

4.1.3. If a, b are positive numbers and a, p real numbers such that 0 < a - p < 21r and if z ~ /(z) is a :function of the complex variable z = re i9 which has finite limits /(0) =lim/(z),·

/(00)

.... 0

=lim/(z), 6"'1lO

within the sector from () = a to () = p, and is regular within and on the bounding lines of the sector except for simple poles at cI' ... , cn within the sector, then (I)

J o

/(ala x )- /(blJlx) dx =(f(oo)- /(O»)(IOg!!..+i(a-p» x b

-2nif(Z-CII)/~Z»)1 ' 2=e.,

1'1==1

where x is real.

Proof. Consider the integral

j /~Z) dz P and the circles r = aro

along the contour bounded by the lines () = a, () = when ro --.0, R --. +00. We get a

bR"~

fJ

"'';,

IIIIP

(f(oo)-/(O»)jid()+ j /(z)dz_ j /(z)dz_/(O) Z

lIIfI"

J!dz

,..P

bril' z

Z

21lif(z-clI) /(Z»)I

+ /(00) IIII{ !dz+o(l) = bR"~ z

and r = aR,

>=1

Z

2=',

'

which reduces to (I). REMARK I. This problem was proposed byE. B. Elliott and solved by G. H. Hardy. See [I). REMARK 2. It is not necessary to asswne that cI ' ... , cn are simple poles. The same result holds, with WlChanged proof, if c I ' ... , cn are poles of arbitrary order. We only have to replace the expression

in (1) and (2) by the expression

Resf(z) . .....

Z

REMARK 3. The obtained fonnula (1) is an extension of the Frullani fonnula

j f(ax)- f(bx)dx= (J(O)- f(+oo»)Iog~, o

x

b

40

Chapter 4

where a > 0, b > 0 and

f

is a continuous function for x ~ 0 which has finite limits

f(O) = limf(x) and f(+«J) = limf(x). .......0+

.. ~

REFERENCE

1. E. B. Elliot-G. H. Hardy: Problem 14164. Ed. Times 72(1900),89-90.

{I

4.1.4. If c is a positive and t a real number, then

J

c+i..

-1 2ni

c-I ..

tz e--dz= z

for t>O

-1 2

for t=O

o

for t
This is the so-called theorem of the discontinuous factor. Proof The contour of integration will be the line segment joining the points c - iR and c + iR (R > c > 0) and a semicircle constructed on the left of that segment if t > 0 and on the right if t < o. See Fig. 4.1.4.

y

y

c+iR

c+iR

c

x

x

t<0

c-iR

c-iR Figure 4.1.4.

Along the semicircle we have z = c + Rei~ tr ~ 2 t < O. In the first case

where rp =0- tr. Since Izl~ R - c, we get 2

O~ 3tr if t> 0 2

and tr ~ 2

O~ _ tr if 2

41


J n

eCI

J

nl2

e-IR sin rp

I z'

o

R dqJ;;;;; 2 eCI

.

e-2 iR 'PIn .--;--

Iz

0

ed

~ - - - . - 'J'l - t(R-c)

as R -+ +00. Hence,

Je-dz= 1,

R dqJ neW

(I - e- 'R ) <----.- -+0 f(R-c)

c+iR

. lIm

R-.+oo

-1-

lz

2n;

c-iR

1>0,

z

because the integrand has a pole of residue 1 inside the contour. If t < 0, since Izi > R we have

J

-n/2

! i

~xp (c + ~ei 8) t R dO Z

nl2

i; ; ; eCI l

J nl2

e lR cos 8

R dO

Z

-n/2 nl2

n/2

o

0

< 2 eCI Je 'R cos 8 dO = 2 ec' Je 'R sin 8 dO ;;;;; 2 ec'

J

n/2

e 2 IR 8/n dO

o

= ~ eCI (1 rR

e'R )

as R -+ +00. Since the integrand is a regular function inside the contour of integration we conclude that

Je-dz=O,

c+iR

. lIm

-1 R-++oo 2ni

lz

c-iR

z

1<0.

Finally, for t = 0, by a direct integration along the vertical line we get

J' -

c+iR

-1. 2nl

c-iR

1 z

1 . . 1 R 1 -=--, n 1 dz=-.(log(c+IR)-log(c-,R»=-arctg-_2nl

n

c

n

2

2

as R -+ +00. This completes the proof. REFERENCE

1. G. Sansone and J. Gerretsen: Lectures on the Theory of Functions of a Complex Variable, Vol. 1. Groningen 1960, pp. 128-130. 4.1.5. If 01' .•• , an' b I , ... , bm, b are real numbers such that

42

C~4

"

.

... 1

t=1

b > Lla .. l+ Llbtl,

(1)

then

b)

+J"'(n" sina.. x)(n" cos tX sinbx 1 n" a... - d X=-H o ...1 x t=1 X 2 ""I

(2)

Imz > O}, r= {zllzl=R, Imz>O}, where R > r, and let C be the contour comprising of the semicircles r and r and the real segments [-R, -rJ, Proof. Let

[r,

r= {zllzl=r,

RJ.

We now integrate the function F defined by F(z) = (Ii: sina.. z) (Ii COSbkZ) e"2/ ...1

Z

k=1

Z

along C. The function F has no singularities in the region bounded by C and so fF(Z)dz=O.

c

Since z = 0 is a simple pole of F, we have limJ F(z)dz = -Hi ResF(z) = -Hi Ii: a ... r~r

~

~

We now prove that lim IF(z)dz =0.

R......",

r

Indeed, if z E

r. then z = Reti (0 S t S 1i) and hence

II

~ell))(Ii cos(btRell))tRe6R: Rie'/dt

F(Z)dZI = j(Ii:sin(a.. r 0 ""I Re

k=1

S ;,,[( Ijlsin(a .. Re'/)I) (filcos(bkRe'/)I)lt6R"ldt

S

~(j Ii: ~.~.m,) (Ii elbtlR.m, )e-6R .m'dt, R

0 ""I

k=1

because

ISin(a .. Re'/)I S el.~.m"

ICOS(bk Re'/)I S elbtlR.m, , I~··I = e-6R ... , .

The condition (1) can be written as

43

Evaluation ofRea1 Definite Integrals by Means of Residues

Lla"l+ L~j:I-b = a (a <0). "

Iff

1'=1

k=1

Since sinl

~

!

I

for

I E [ 0, ;

J.

we get

/2 (2aR) dt=-, 1r (fiR ) s 2- trJ exp e -1,

R" o

aR"+1

1r

implying lim JF(z)dz =o.

R-->+ ..

r

Finally, if r -+ 0, R -+ +00, from the equality

J

I. RJ( " sina ~F(z)dz= r F(Z)dZ+,:a x"

J

J'(IT" sina"x) ( IT,. cosbkx) -e-I»dd x = 0

+ F(z)dz+ r R

"",I

we obtain (2). 4.1.6. We have

J Vx +00

o

cosx-sinxdx= -~'. (1 +X2) ev''2

Proof Define the function

x)( D . cosbj:x)~ -;-dx

f

by

X

k-1

X

44 This function has simple poles at i and -i with residues

'I = Resf(z)

= ____ 1 -

2eVi

z=;

(l + i);

'2 = Res fez) z=-;

= -

,e;-;;

2v 2

(1 - i).

The functionfalso has a critical singularity at z = 0. We chose the branch of f in which

JZ takes the value 1 for z = 1.

Let r= {zllzl= R}, r= {zllzl= r}, d = {zlImz = 0, r S Re z:S R}, where R is large enough so that the circle r encloses all the singularities off Let C = ru d u (-d) u r. Then

f fez) dz= 2

c

7e

i ('I + '2).

When R -+ +00, r -+ 0, from the last equality follows +J"COSX +i sinx

y

.

-2.,r;( ) dx=2m(Ij+12), o x l+x2

B

21

or, in virtue of (1), +ao

• •

-~( hI . hI) Jo coSX+ISIDXd .,r;(I+~) xS +IC .

Ii

Therefore, +CIO

F

• •

JCOSX+ISIDX dx=~(shl-chl) = -~. o .,r;(1+ X2 )

Ii

Figure 4.1.6.

eli

REMARK. If

J I

D=

o

cos x

I

d

VX(I+X2)

X,

b=J V xsin(I x+X2) dx• o

Jr

then the integral 1= f(z)dz can be expressed in tenns of a and b, where r is the contour

ABCDEFDA shown of Fig. 4.1.6. Indeed, since

J !(z)dz-P !(z)dz=2nir" ABCDA

we conclude that

C

J !(z)dz=-2nir

2•

DEFA

45


I=.r fez) dz+2n ;(r, -r 2) C

JY t

=

2

_

o

=

cos x + ; sin x dx + 2 n!. (sh 1 - ; ch 1) x (1 +X2) x

Y

(nyi ch 1- 2 a) +; (nyi sh 1 -- 2b).

REFERENCE

1. Nouv. Ann. Math. (4) 20(1920), 222-225.

4.1.7. Letfbe a rational function which may have only simple poles aI' ... , am on the positive real axis and let zI' ... , zn be the other poles off, none of which is at the origin. If p is a real number such that lim zP+ 1 f(z)

z-+-o

=

lim zP+ I f(z) = O. Z~CICI

we have:

f

n

+~

v.p.

o

xPf(x)dx= -

n.e-np;

2:

Res(zp f(z»-ncotgnp

smnp k=1 Z=Zk

I akPResf(z), z=ak

k=1

where xl' > 0, for x> O. 2° Ifp

E

Z. then

f

+CO

v.p.

x Pft x ) dx= -

n

In

k~1 ~~~ (zp f(z) logz) - k~l akP (log ak + n i) z~~:f(z),

o

where the logarithm is defined by logz = log 1zJ + i argz,

0;;;;argz<2n.

REFERENCE

1. L. Volkovysky, G. Lunts, I. Aramanovich: Problems in the Theory of Functions of a Complex Variable. Moscow 1972, p. 93.

4.1.8. Suppose that the following conditions are satisfied: 1° The function z H f(z) is real and continuous on the real axis; 2° For real x we havef(-x)

=f(x);

46

chapter 4

30 fis analytic in the upper half-plane where it can have a finite number of isolated singularities 01' .•. , an;

4 0 lim f(Rel')RlogR = 0 uniformly for t R-++ ..

E

[0, n);

Jf(x)dx is convergent and its value isA.

+..

SO The integral

o

Then (1)

J o

f(x)log(x 2 +1)dx =

7ri(2f~esf(z)IOg(Z+i) -AJ. "*

.1:=1

Proof The branch of the function F, given by F(z) = f(z) log(z +i),

which takes the value rol2 at z = 0, is uniform in the plane cut along the imaginary axis from -i to -ioo. Its singularities in the upper half-plane are the singularities 01' ... , an off If CR is contour consisting of the real segment [-R, R] and the semicircle {zllzl= R, Imz > O}, where R is large enough so that all the singularities lie inside CR , then n

(2)

ff(z)log(z+i)dz = 27ri~~esf(z)log(z+i).

c.

.1:=1

.-a.

On the other hand,

J

J

R

(3)

tr

f f(z)log(z +i)dz = f(x)log(x+i)dx + i f(Rell)Rei1log(R+it)dt c. 0 0 o

-Jf(-x)log(-x+i)dx. R

Since the functionfis even on the real axis and since for x> 0 we have arg(-x+i)+arg(x+i) = 7r then clearly R

(4)

0

Jf(x)log(X+i)dx- Jf(-x)log(-X+i)dx o

R

R

+""

o

0

= jf(x)log(x 2 +1)dx+ro Jf(x)dx.

Furthermore,


47

I<

~ Jlf(Rell)IR(logR + n)dt -+ 0, o

asR -++00. Hence, if R -+ +00, from (2), (3), (4) and (5) follows (1). ExAMPLE. If fez) = 11 (Z2 + 1), all the conditions 10 - 50 are satisfied. This function has only one singularity in the upper half-plane at z = i and Reslog(z+i) .., Z2 +1 Also,

!.r

+..

A=

log(2e'ofl ) = -!..log2 +.!:. 2i 2 4

1 +1 dx=f,

and using (I) we obtain

n n) =nlog2.

-1.......;..::-~dx=ni log(x + 1) (-ilog2+--2

x2 +1

o

2

2

REMARK. Several theorems of this type were proved is Section 5.3.2. ofVolwne 1. REFERENCE 1. D. D. Adamovic: A Collection of Solved Problems in Mathematical Analysis. Ordinary and Partial Differential Equations and Complex Functions (Serbian), Beograd 1959.

4.2. Integrals with Finite Limits 4.2.1. If z = 0 is a simple pole of the function/. then ,,+I>

(1)

lim Jf(rel')rel'dt =b Res fez) r-+O

0=0

"

(0 ~ a ~ argz ~ a +b < 2n).

Proof Since z = 0 is a simple pole of/. in a neighbourhood of that point we have the expansion +co

(2)

fez)

=A_I Z-I + LA.z· . • =0

Suppose that r > 0 is sufficiently small, so that z = 0 is the only singularity off in the region {zllzlS: r}. Integrating (2) along the arc of the circle {zllzl= r}, defined by O~

a s:argz s: a+b <2n, we get

48

C~4

Jj(rel')rel'i dt =A_I Ji dt + LAiri Jieii'dt.

(3)

8+6

8+6

+..

..+6

8

8

i=O

..

When r -. 0, from (3) follows (1).

* °is real number, then 26 cosnx (n+cotga), J-...:...:.~-~dx = {2ne-~ sh a

4.2.2. If a

o (cha-cosx)2

2n

eft..

a>O

(n-cotga),

a
sh2a

Jo (cha-cosx)2 dx=O.

26

•

smnx

The above equalities were proved in [1]. In the same paper the following integral

2J6cha-COS(X-t) log chc-cos(x-t) dt o (chb-cosX)2 cha-cos(x-t)

(a

* 0, b * 0,

c *0)

is also evaluated. REFERENCE

1. H.R.Weber: Berechnung zweier in der P/attentheorie auftretenden bestimmten Integra/e. Jng. Arch. 13(1943),377-380.

4.2.3. Suppose that the function z = x+iy H F(z) = u(x,y) +iv(x,y) is regular in the disc {zllzl~ I} and that F(z) is real for real z. If -1 < t < 1, then

Jo 1-2tcosf'+t tsmf' 2 v(cos f',sinf')df'= n(F(t)-F(O».

2..

•

Proof The integral

tsinf' 1=J26o 1-2tcosf'+t 2F(cOSf'+sinf')df' after the substitution eitp =z becomes

1=-!.r. t(z2-1)F(z)

2 ~ z(z - t)(1- tz)

where

dz,

r= {zllzl= I}. The function ZH

t(Z2 -1)F(z) z(z - t)(1- tz)

49


has two simple poles inside r, at z = Hence,

°and z = t. with residues F(O) and F(t) respectively.

1= -.!.(2ni(F(O) - F(t») = ni(F(t) - F(O», 2

implying

J•

tsinqJ

J•

tsinqJ

2

-----!.--=-2

o l-UcosqJ+t

2

-----!.--=-2

o l-UcosqJ+t

. u(cos qJ,slnqJ)dqJ =

°

. v(cosqJ,slnqJ)dqJ= n(F(t)-F(O».

REFERENCE

1. R.Albrecht, K.Zuser: Ubungsaufgaben zur Funktiontheorie. IV. MUnchen 1962, p.57.

4.2.4. If n, keN and lal < 1, then (1)

J(a,n,k) =

j

cos(n - k)t ft dt _.(1-2acost+a 2 )

= 21faln-A:1 ~(n-l+ln-kl)(n-l+ v) a2v (1- a2 o n -1- v v (1- a 2

r..

r'

and a similar equality holds for lal > l. Proof We have J(O, n, k) =

°(n '* k) and J(O, n, n) = 21f. Suppose now that a '* °and

p=ln-kl'*O. Putting tI' =z andC={zllzl= I}, we obtain

--f ( n r dz = -ifc (z-anl-azr dz , 2 c z-a l-az ;

J(a,n,k) =

zR+ p - 1 +Zft-p-I

Zft+p-I

since the integrals

f

zR+ p - 1

and

c (z-ar(l-azr dz

J. zR+ p - 1 dz , 1( c z-a )ft( l-az

r

are equal, which can be seen by changing the variable z = lIw in the second integral. The function/defined by fez)

=

zR+ p - 1 ft

(z-a) (l-az)

ft

has two poles: zl = a and z2 = (11 of order n, and inside the unit circle C is either zl or z2' depending on whether lal < 1 or lal > 1. By the residue theorem we have

so

Chapter 4

J(a,n,k) =2trResf(z)

(lal< 1),

J(a,n,k) = 2trRe~f(z)

(Ial> 1).

6=41

,=tI

Putting z = a + t intof(z) we get f(a+l)=

(a +t),,+p-I ( )". t" I-a 2 -at

The residue at the pole t function

= 0 is the coefficient of ,,-1

in the Taylor expansion of-the

t 1-+ (I-a 2 r"(a +/)n+P-I(I _ _ a_t)-rI

I-a 2

about 1= O. We get Resf(z)=Resf(z)= 1=0

'=G

r

aln- kl I(n-I+ln-kl)(n-I+V)

( 1- a2

v=0

n - 1- v

v

a 2v (1- a2

r

,

and (1) follows immediately. The corresponding formula is obtained for

lal > 1 on noticing that

J(a- I,n,k) =a-2" J(a,n,k). REFERENCE 1. D. S. Mitrinovic - D. C. B. Marsh: Problem 4967. Am. Math. Monthly 68(1961), 510 and 69(1962),675-676.

4.2.5. If n is a positive integer, then

n=2k (1)

n=2k+l where m-I

(21' -I)!!

.=1

(2 v)!!

Sm=l+2:(-l)"

Proof The function F defined by Z2n

F (z) = ------===

(1 +Z2)y I - z 2

51


has two critical singularities, at 1 and -l. Let Fl be the branch of F which corresponds to the value of the square root which is positive for z = ia (a > 0), i.e. to the value of the

square root for which

.Jl +a

2

> 0 (a> 0). The function Fl is uniform outside a closed

contour C which encloses the segment [-1, 1] of the real axis. Let F= {zllzl= R}, where R > 0 is large enough so that the contour C is contained in the region int r. By Cauchy's residue theorem we have 1 F I (z)dz-1 F I (z)dz=2ni(ResFI (z)+ ResFI(z»). r c z=; Z=-;

This equality can be written in the form 1 FI (z) dz= - 2 n i (Res FI (z) + Res FI (z) c z=i Z= - i

Define C as the contour comprising of two line segments PN and P'N' parallel to the real axis and at a distance b from it, and of two circular arcs c and c' with centres at -1 and 1 and radii r (see Fig. 4.2.5). Then

~

+ Res FI (z»). y

c

P'

M'

N'

P

M

N

Figure 4.2.5.

J

FI (z) dz=J,

..-.0 MN

where J denotes the integral on the left hand side of (1). We also have

On the other hand ,

2"

If FI(2)dzl~J 0

c

yrdt=2nyr

!If FI(Z)dz!~J yrdt=2ny,.

2"

I

c'

and so

z=oo

0

c'

x

52

Chapter 4

lim

r.....o

JFI (z)dz=lim JFI (z)dz=O. r.....o c.

c

We therefore conclude that 4J =

-

2 3fi (Re.s FI (z) + Res.FI (z) + Res Fl (z», Z-='

Z=-I

z=OO

and it remains to evaluate those residues. Since we have chosen the branch of the functon f for which

we have ResFl (z)= z=;

;2n

y' , -2 21

(_ i)2n

ResFl (z) = ---_--.,

z=-;

-2y2 ;

i.e.

In order to evaluate the residue at 00, put z = l/w. We get

Hence, if the function

is expanded into series around w = 0, the coefficient of wln-2, with changed sign, is the required residue. We first determine the sign in the expression ~ w2 -1 =±; ~1- w2 which corresponds to the branch Fl' Namely, for z=-;a we took the value +~I+a2 of the function

z~~I-z2. Since

putting w = -Ilia we should get +~1 +a 2 ; we therefore take the sign +. Therefore

53


1

- - = 1 - w2 + ... + ( - l)n w2n + . . ., 1 + w2

1

(2 n-l)!!

2

(2 n)!!

(1-w2)-1/2=1+-w2+ ... +

w2n + ...

The coefficient of w2n- 2 is

p

n~ 3)!! _ (2 n - 5)!! (2n-2)!! (2n-4)!!

+ ... + ( _ l)nL _ ( _l)n. 2

and SO

4 J = 2 n ( _I)n_ + (2 n - 3)!! __ £ n - 5)!! + ... + ( _ I)n ~___ ( _ I)n) ,

V2

(2n-2)!!

(2n-4)!!

2

implying (1). REMARK. Fonnula (1) was recorded in Volwne 1 (5.4.3.20, p. 205) without a proof. REFERENCE 1. F. Tisserand: Recueil complementaire d'exercices sur Ie calcul infinitesimal, deuxieme edition, augementee de nouveuax exercices sur les variables imaginaires par P. Painleve. Paris 1933.

4.2.6. If n is a positive integer, then

J I

(1)

d

x2n

fX(1-x 2 )

o

11:

X= y3

1·4 .. ·(3n-2)

3.6 ... (3 n)

.

Proof The function F defined by F (z)

z 2n

=

-,;;====

fz(1-z2)

has three critical singularities 0, 1, -1 and anyone of its branches, denote it by F I , is uniform outside any closed contour which encloses the segment [-1, 1] of the real axis. Hence,

1 FI (z) dz = - 2 n i Res F) (z). c z=oo

Let C be the contour shown on Fig. 4.2.6, where the segments PL, MN, N'M', L'P' are

S4

chapter 4 y

at a distance b from the real axis, while the radii of the circles c, c', c" are r. Let J denote the integral of the left hand side of (l).

M'

N'

c' M

N

P' c

As b -+ 0, r -+ 0, we have

-I

P

c

.. "

Figure 4.2.6.

IJ

o

J

rn

FI (z)dz-+--;

1 fx(l-r)

N'M'

J

I

J -1

I

Fl(Z)dz-+-62 6

L'P'

J

I dx= --J,

0

JI

x 2n

I II

6

1

Fl (Z) dz

-+ -

~6

PL

,

rn

xl/l (l-r)l/!

- I

Idx=-(-J)=J, 3

xII! (I - X2)1/! •

6

Idx = - ~6 J,

0

where .'

2n 3

2n 3

8=COS-+J S t n - .

Since the integrals along c, c', c" tend to 0, we conclude that

and it remains to evaluate this residue. Putting z = l/w we get

I -I F(z)=F (-I) =-~-, W

W2n-I'I_w2

t

and expanding the function w ~ (I - w2 ll3 into series, the coefficient of wln will be the required residue (we take that branch of the square root which has the value 1 for w = 0). We have (1 -

and so

W 2 )-1/3 =

I

1 + - w2 + . . . + 3

1·4·· ·(3 n-2) 2n w + 3.6··.(3n)

...

ss

Evaluation of Real Defmite IntegJll\s by Means of Residues

1·4·· ·(3 n-2) . , 3·6·· ·(3 n)

ResFJ (z)=a

z=co

where a = 1, &-1, &-2, depending on the chosen branch. But the integral J is a real number, and so from the equality 2 ( 1 -~) J = - 2 n i a _1._4_._••..:..(3_n_-_2....::.) 3·6···(3n)

E

we conclude that a =&-2 which implies (1). REMARK. The equality (1) is recorded in Volume 1 (5.4.3.20, p. 205) without a proof. REFERENCE 1. F. Tisserand: Recueil compiementaire d'exercices sur Ie calcul infinitesimal, deuxieme edition, augementee de nouveuax exercices sur les variables imaginaires par P. Painleve. Paris 1933.

4.2.7. If n is a positive integer, then 1

(1)

J ~x3(1_X2}i

~-dX=

o

_1I:_2.9 ... (7n-5) 2sin~ 7.14 ... (7n)

7

Proof The multiform function

has critical singularities at z = 0, z = 1, z = -1. Outside of the contour C (see Fig. 4.2.6) the function f is uniform and

(2)

f f(z) dz =

-

c

2 n iRes f(z). Z=OD

Starting at M we integrate that branch of the function C in the positive direction. As b -+ 0, r -+ 0, we get 1

(3)

0

ff(z)dz= J f(x)dx+C

0

f which is real for real z along

~1I:iJ f(x)dx

eXP-7-1 -I

4Rill(exp'Ri)! If(x)ldx

+ (exp7

7

0

S6

Chapter 4 u

+(exp -4ni)( ~ni)( exp -4ni) JIJ(x)ldx, exp- 777-1

because the integrals along small circles tend to 0 as r -+ O.

IfJ denotes the integral on the left hand side of (1), then from (2) and (3) we get

J-~J+J-~J= -2niResJCz), e

E

z= 00

i.e.

4n

i

where e=exp-. 7

In order to evaluate the residue at 00, put z = IIw and we get

= -1- ( 1 - W2)-2/7 exp ( - -2 n i) 7 .

W 2n - 1

exp

(

2n

i)

-7- (1+-w2+ 2 7

W 2n - 1

... +

2·9·· ·(7 n-S) w"+"', 2 ) 7·14 .. ·(7n)

and hence ResJ(z)= Z=OO

2·9 ... (7 n -

5)

7·14 ... (7n)

i)

n exp (2 -. 7

Therefore, in virtue of (4) we find J=ni 2·9 .. ·(7n-S) _ _ _ __ 7.14 ... (7n)

e

2,,; 7

2,,;'

-e --7-

and this is equivalent to (1). REFERENCE

1. D. Kurepa - D. Mitrovic: Problem 99. Glasnik mat.-Hz. 5(1950),45-47.

4.2.8. Suppose that n is a positive integer. Integrating the function ZH

(Iog(l

_z»n

iz

along the unit circle, indented at z = 1, applying the standard substitution z = eit and

57


Cauchy's theorem we get

J

2,.

(1)

~)+ ~

(log(2sin

i(t-n)fdt=O,

o

because for z = eit we have

(0<1<2 n). Putting n = 1,2,3 into (1) we obtain

J

2"

log(2sin

o

J 2"

~ )dt=O,

t log ( 2 sin

~) dt = 0,

o

f

2,.

t log2 ( 2 sin

~) dt =

o

f

2"

(2)

4

log3(2sin

~)dt=3

o

f

2"

t 2 log (2 sin

:4 ,

~)dt.

0

This result was published in [1]. Integrals of the form

(3)

" flog" (2 sin

~) dt

o

were considered by Bowman [2] and Beumer [3], but they did not apply contour integration. It was found that the integrals of the form (3) can be expressed in terms of the Riemann zeta-function of integral argument. Indeed, if we substitute the Fourier expansion log (2 sin

~) =

-

cos t

-

~

cos 2 t

- ...

into the right hand side of (2) we get

"

2 fIog3(2sin

~)dt=

-3n ~ k- 3= -3nC(3).

o REFERENCES

1. M. Lutzky: Evaluation of Some Integrals by Contour Integration. Amer. Math. Monthly 77(1970), 1080-1082.

58

Chapter 4

IIIl

2. F. Bowman: Note on Integral J(log sinO)"dO. J. London Math. Soc. 22(1947),172-173. o

3. M. G. Bewner: Some Special Integrals. Amer. Math. Monthly 68(1961),645-647.

4.2.9. We have (I)

..J2 (5,r Jo t 2Jtii sin2tdt = ~ + 2tr- 2n-Iog2 - 4 + 410g2 - 210g2 2 J. 32 6

.12

Proof Let C be the positively oriented contour consisiting of the upper semicircle {zllzl=l, Imz~O} and the real line segment [-I, 1]={zl-I~Rez~l, Imz=O}. By Cauchy's theorem we have

)1/2 I f(lOgz)2 ( -z dz=O, c I+z which after the standard parametrization z = eit reduces to (2)

M

e·il4 J t 2 tg-ei'dt = 2 t<

o

2

t)2 I+t I+t J(I~ og dt+2tri Jlogt--dt- ~ J--dt. ·JI-t JI-t I

I

0

0

I

2

0

2

The integrals on the right hand side of (2) can be expressed in terms of the beta-type integral (i.e. in terms of the gamma function).

J_t

P -

o~

dt =.!.

J

Si(P-I)

(1- s)

20

-i ds =r( t p + t )r( t), 2r(tp+l)

and its first and second derivatives in p, evaluated at p = 0 and p = 1. Using the known values

V/O-) =-r- 210g 2, V/(I) =-r, V/(i) = 2- r- 2Iog 2,

where ~z) =T'(z)/ F(z) and ris Euler's constant, we find

I ~tgfei'dt t2

= ~ ( - 51~ -

~ + trlog2 2J+ ~ (_tr2+log2 -2tr+2trlog2).

Replacing tby 2t and taking imaginal)' parts, we obtain (1). REFERENCE

1. J. Gilles - J.van Casteren: Problem 646. Nieuw Archiefvoor Wiskinde (4)1(1983),93-94.

59


4.2.10. If b > a > -I, then trl2

Jo cos"tcosbtdt =

(I)

r(a

+

I)r(!..=.!.)

sin

2

2"+lr(I+,,;b)

tr{b-a)

2

.

Proof The function / defined by

=(Z2 +I)a zb-I-a

/(z)

has critical singularities at z = 0, z = i, z = -i. The branch of this function which takes the value 2Q at z = 1 is uniform and regular in the plane cut along the segment of the imagiruuy axis which joins the points i and -;. For this branch we have, at

y

A

z= I, arg(z+i)

= t,

arg(z-i)

= -t,

arg z = 0

o

and so, on the right hand side of the cut: arg(z2 +1)

1 x

= arg(z+i) +arg(z-i) = 0;

-i on its upper part we have arg z = N/2, and on its lower part: argz = -N/2. Let Cbe the contour shown on Fig. 4.2.10. where the small circles with the centres at 0, i, -; have radii r, and the Figure 4.2.10. larger circle with centre at 0 has radius l. Denote by rI' r2' r3 the arcs of the small circles and by K the arc of the larger circle which comprise the contour C. Let e(r) be the angle between OA and the imagiruuy axis.

Since the chosen branch of the function/is regular inside C, we have

(2)

0

.( =r/(z)dz =-i e-!(b-,,-I) 2

Jr (

i

c

I - l )" y. P-,,-I dy

+

~r

~b-"-I)/

+ i e2

~

J 1- y2

I-r(

J/(z) dz

1-,,-1 dy

)"

+

r

J/(z) dz

r,

-trl2u(r)

+ 2°i

Jt!blcos"tdt + J/(z)dz. r,

tr/2-s(r)

The condition b > a > -I ensures that

(3)

lim J/(z) dz =0 r-+O

(k

= I,

2, 3).

r.

Hence, if r -+ 0, from (2) we get

(4)

0

-!(b-,,)/) 1 ( ) " 2 =( e~b-,,)/ -e 2 J 1-1 1-,,-1 dy-2"i tr/2Jcos" tcosbtdt.

o

-tr/2

60

Chapter 4

But - - = r(a+l)r(~) ll=.L JI(l -y 2)" y. .b-,,-ldy=-1JI(1 -u )" utr-d u=-I B ( a+ 1,b-a) o 20 2 2 2r (a + 1+ 2 )

and from (3) and (4) follows (1). REMARK 1. Fonnula (1) was proved in [1]. The same integral, with the same condition b > a > -1,

was considered in [2] and the following result was given 1rr(a + 1) Jcos"tcosbtdt= 2G+ r(.s:;+I) r('1 +1)'

td2

(S)

1

0

The right hand sides of (1) and (S) are fonnally equivalent, but (1) is more correct, since the expression r(.at-+ 1) which figures in (S) may not be dermed, and we must assume that it denotes the corresponding limit. For instance, if a = 0, b =4, from (1) we get the correct result: 0, while (S) becomes

1r

4r(-I)

which reduces to 0 if we assume that __ 1r_ means lim 1r 4r(-1) ....... 2r(l+t)r(l-t)

REMARK 2. Fonnulas (1) and (2) hold provided that b > a > -1. Ifwe let b ~ a, from (1) follows: "'2

Jcos"tcosatdt =~1 2"+

(6)

.

o

The same result is obtained from (S) merely by putting b = a; the limit was already taken when

r

(b -2 a ) sin (b2- a) was replaced by r 1r

(1r

?+1

)'

REMARK 3. Fonnula (6) was proved by Cauchy for a > 0 using residues; see Volume 1; S.4.3.11.,

p.197. REFERENCES

1. D. D. Adarnovic: A Collection of Solved Problems in Mathematical Analysis, Ordinary and Partial Differential Equations and Complex Functions (Serbian). Beograd 1959. 2. L. Volkovysky, G. Lunts, I. Aramanovich: Problems in the Theory of Functions of a Complex Variable. Moscow 1972, p. 88.

4.2.11. Integrals of the form

J(cosnx)(P(cosx)Y'2 dx I<

(1)

1=

(n eN)

-I<

where

are expressible in terms of hyperelliptic or elliptic integrals, depending on the positive integer r.

Proof If there exists a polynomial Q of degree r such that P = Q2, then (1) becomes


61

J K

1 = (cosnx)Q(cosx)dx, -K

and this is a standard form. This integral reduces to the computation of

~s Q(~(Z+Z-I)}"-I. We consider the interesting case when there is no polynomial Q of degree r such that

p=Q2.

Since

= J(sinnx)(P(cosx)Y'2 dx =0, tr

H

-K

we have

J K

1 = I +;H = eW(P(cosx)y/2 dx, -K

and putting z = eix we obtain

1 =~fz"-Ip ( !(Z+Z-I) ) Ie 2

112

dz,

where C = {zllzl= I}. The expression

p(~(Z+Z-I»)

can be written as K(z) I z2r, where K is a polynomial

of degree 4r. Now, the main problem is to show that there exists such a positive integer r that the algebraic equation

r:(K(Z)1I2) = 0 which corresponds to the integral (R rationalfucntion)

on the Riemann surface S generated by (K(z)t 2 has a hyperelliptic field. In order to conclude that we have to find the genus g of the Riemann surface S. The hyperelliptic surface of genus g is obtainable as a two sheeted Riemann surface of the function z 1-4 (K (z) t2 , where K is a polynomial of degree 2g+2:

K(z) = (Z-ZI)(Z-Z2)'" (Z-Z2,+2)' Comparing this representation (which involves the genus g of the Riemann surface S)

62

Chapter 4

with the representation K{z)

=(z-e,){z-e2 ) ... (z-e4,),

we conclude that g =2r-l. The last equality relates the genus g of the Riemann surface S to the degree 4r of the polynomial K. This relation can also be obtained using the Riemann-Roch theorem, which implies that the branching index of the Riemann surface v = 2{n+g-I), where n is the number of sheets of the Riemann surface andg its genus. In the case considered we have v = 4r·l, all branch-points being of the first order. There are 4r such points (we made the assumption that the polynomial K has no multiple zeros): the corresponding Riemann surface has n = 2 sheets: hence, we have 4r = 2(g+ 1), where from we again obtain g = 2r-1. We now show that our assumption concerning the zeros of K does not restrict the problem. Let U be any function on the Riemann surface S taking every complex value twice. Then there exists a function u on S, satisfying the following relation: u2 +R,(z)u+~(z) =0,

where R 1, R2 are rational functions. Putting 1 v=u+-R,{z), 2 we obtain a new meromorphic function v on S satisfying the relation v

2

R{) 1 R {)2 {z-a,){z-a2) .. · (z-a,,) =-,'2 Z +- ,z = . 4 {z-b,){z-b ) ... (z-b,,) 2

If we introduce the function w by the relation w = v{z-b,){z-b2 )

...

(z-b,,)

we obtain the meromorphic function w on S with the representation w2

={z-a,){z-a 2 ) ... (z-a,,){z-b,)(z-b2 ) ... (z-b,,).

Assuming that al

= a2

say, and substituting

w

y=-z-a,

we obtain y2

=(z-a 3 ){z-a4 ) ... (z-a,,){z-q)(z-b2 ) •.. (z-b,,)

and all the zeros are different. Therefore we may start with the function z ~ (K{Z»)Jl2 generating the Riemann surface S and having all zeros different, i.e. we shall take K{z) = (z-e,){z-e2 )

•••

(z-e4,)

all the zeros being branch-points ofthe surface S. In view of the relation connecting the genus g (of the surface S) and the degree r of the

63

Evaluatioo of Real Definite Integrals by Means of Residues

polynomial, we have the following conclusion:

Ifr> 1, to the equation F;(K(z)t '2 = 0 (which is algebraic relative to the field on the surface S) corresponds a hyperelliptic field, i.e. the corresponding integral

I

=~frr-'(K(Z)r2 dz Ie

is a hyperelliptic integral.

If r

= 1, then g = 1 and the field is elliptic.

These conclusions follow from the Riemann-Roch theorem.

In the case when the integrals (1) are hyperelliptic, they have the following form: The differentials of the first order have the form

L(z)dz (K(Z»)J12 ' where L is a polynomial whose degree does not exceed g - 1. Their derivatives 2g - 1 linearly independent quadratic differentials of the form

M(z)dz 2 K(z) where the degree of the polynomial M does not exceed 2g - 2. All other quadratic differentials (which do not have this form) have the form

N(z)dz 2 (K(Z»)J12 where the degree of the polynomial N does not exceed g - 3. It is interesting to find in which cases the residue at the pole z = 0 of the function

z 1--+ z"-r-I(K(z»)J12 is equal to zero. It follows from the above that

{o

(n ~r+l)

1= 2trB

(nSr)

where B denotes the residue at the pole z = 0 (whose order is r - n + 1) of the function

z 1--+ z"-r-I(K(z»)J12. On the other hand, if we consider the integral I obtain

I -iH =I

=~fz-(n+r+I)(K(Z»)J12 dz, I

which implies that

c

=I -

iH, instead of I =I + in, we

64

Chapter 4

I =2trB*,

where B* denotes the residue of the function degreen + r+ 1.

Z H z-
at the pole z = 0 of

If we compare the last result with the result above we see that

B*=O for

n~r+l.

This last conclusion was obtained in an elementaIy way; it is more difficult to obtain the same conclusion using Abel's differentials of the second order, which is necessary in this

case.

REMARK. This result was only briefly mentioned in Volwne I; 5.4.3.24, p. 206. REFERENCE

1. D. S. Mitrinovic and Z. R. Pop-Stojanovic: About Integrals Expressible in Terms of Hyperel/iptic Integrals. Glasnik mat.-flZ. astr. (2)18(1963),235-239.

4.3. Cebysev's Approximation of the Integral of a Positive Function Let f be a positive real function on [a, b) such that the integrals (k=O, 1, ... )

(1) a

exist. In papers [1], [2], [3] CebySev considered the problem of determining the upper and

lower bounds for the integral (2)

J f(x)dx

(a~u; v~b)

u

in terms of the constants (1). The short note [1] contains only the result, without a proof. The problem is treated in much more detail in [2] and the solution is formulated by means of residues. We shall expose Ceby§ev's results from [2] and [3]. 1. It should be mentioned that some twenty years before the publication of [I] Ceby§ev thought that there is no justification for the use of residues (and complex fimctions in general). This view of Ceby§ev delayed the acceptance of Master's thesis of 1. V. Sohocki for two years. This thesis will be reviewed later (see Chapter 7).

REMARK

If the integral

J' b

a

I(x) dx

z-x


65

is expanded in negative powers of z, we get b

b

b

f(X) 1 j f(x)dx+1 j xf(x)dx+ ... , j -dx=-Z

a

-x

Z

Z2

a

a

i.e. b

f(X)dx=Ao+A,+ .. . , z-x Z Z2

j a

or (3) a

with accuracy up to r2m. Let --..... Q, Z

1

+ b, - a2 z+b;- -

(4)

be the m4h continuous fraction which corresponds to the right hand side of (3). Then

again with accuracy up to r2m. Let gm(z)lhm(z) be the value of the expression (4) and suppose that zl' ... , zm are all the roots of the equation hm(z) = O. Ifv E {zl' ... , zm}' for instance v =Zq (1 < q s; m) then we have v

(5)

q~l Resgm(Z);;;;jf(x)dx;;;;

k=l z=zkhm(Z)

a

i

Resgm(z).

k=l z=zk hm(Z)

Assume now that zp and Zq (p < q, Zy < Zq) are two roots of the equation hm(z) = O. If we put v = zp and then v = Zq into (5), after subtracting we obtain

66

chapter 4

Therefore, Cebyiev determined the bounds of (2) under the condition that u, v are solutions of the equation hm(z) = O. In paper [3] Cebyiev obtained a better result. Namely, if the expression on the right hand side of (3) is expanded as the infinite continuous fraction 1 -------

1

a.z+b. - - - - a2 z +b2

which converges to F(z) , where F and G are polynomials, then G(z)

(6)

y

L Res F(z) _~ G (z)

F(v) ~Jj(X) dX~L 2 G' (v)

Res

F(z) +~ F(v) G (z) 2 G' (v) ,

II

where the sums of the residues are taken over all the zeros of G which belong to the interval [a, v). Hence, concludes Ceby§ev, if we know nothing about the functionjexcept that it is positive on [a, b) and also the numbers (1), then LReS F(z) gives the value of G(z)

the integral J" j(x)dx with accuracy up to "

±.!.

F(v) . 2 G'(v)

REMARK 2. In [3] Ceby§ev examined the value of the expression F(v) and some consequences of G'(v) the obtained fonnula (6), but he did not use residues. REMARK 3. In papers [2], [3] in which he ~pplied residues, Ceby§ev did not quote the relevant literature (Cauchy, for instance). He only mentioned in (1] a result of Bienaime [4] which inspired him to work on this problem.

67

Evaluation oCReal Defmite Integrals by Means oCResidues

REFERENCES

1. P. Cebyiev: Sur les valeurs limities des integrales. J. Math. Pures Appl. (2) 19(1874), 157160. E Collected Papers 0/ P. L. Gbylev (Russian), t. 3(1948), Moscow - Leningrad, pp. 6365. 2. P. Cebyiev: Sur la representation des valeurs limites des integrales par les residua integraux. Acta Math. 9(1886), 35-56. == Collected Papers o/P. L. Gbylev (Russian), t. 3(1948), Moscow - Leningrad, pp. 172-190. 3. P. Cebyiev: Sur les residus integraux qui donnent des valeurs approchees des integrales. Acta Math. 12(1889),287-322... Collected Papers o/P. L. Gbylev (Russian), t. 3(1948), MoscowLeningrad, pp. 191-225. 4. J. Bienaime: Considerations Ii l'appui de la decouverte de Laplace sur la loi de probabilite dans la methode des moindres carres. J. Math. Pures Appl. (2) 12(1867).

4.4. A Note on some Papers of Ostrogradski and Bouniakowski As will be seen in Chapter 7, the Master's dissertation of Sohocki from 1868 was the first paper in Russian language which mentions residues. Of course, Russian mathematicians knew about residues long before Sohocki's dissertation was published. We noted in Volume 1 (p. 335 and pp. 351-352) that Cauchy himself mentioned in his paper from 1841 the contributions of Ostrogradski and Bouniakowski. We also noted that we identified all the papers of other mathematicians quoted by Cauchy, except those of Ostrogradski and Bouniakowski.

In connection with this we contacted Professor A. P. about the following facts.

Julkevi~

and he informed us

Ostrogradski lived in Paris from 1822 till 1827 and attended Cauchy's lectures delivered at the College de France. Cauchy talked about the results of his research even before their publication, and naturally he lectured about integrals and residues. Inspired by Cauchy's lectures, Ostrogradski wrote two articles: 1. Memoire sur la difficulte qui se rencontre dans Ie calcul des integrales dejinies, lorsque la /onction a integrer est discontinue entre les Iimites de !'integration (paper is dated: 24 July 1824); 2. Remarques sur les integrales dejinies (the paper is dated: 7 August 1824).

J/(x)dx, b

In these papers Ostrogradski considered improper integrals of the form

•

where the function/is not continuous on [a, h). In the first article he applied a procedure similar to Cauchy's principal value, while in the second he explicitly used residues, although the term residue (i.e. residu) was not mentioned. It should be noted, however, that Cauchy used word residu for the first time in his paper [1) from 1826, that is to say after the two papers of Ostrogradski were written. In fact, Ostrogradski in certain cases used the expression

68

(1)

Chapter 4

1

d"- 1

- - l i m -(z-a)II/(z), (n-l)!

Z-+Q

dz"- 1

i.e. the residue off at the pole z = a of order n. Ostrogradski applied his results to the integral (2)

J'"

o

xsinx

cos t-cosx

d x,

and he then used the value of (2) to evaluate the integrals ",/2

~dx J tgx

o

J ",/2

and

log sin x dx.

o

Ostrogradski submitted his papers to the Academie des Sciences de Paris. In the opinion of Professor Julkevic, Ostrogradski was not satisfied by the obtained results and so his manuscripts remained in the archives of the Academy. Julkevic found them there and analysed them in [I}. Formula (1) is really due to Cauchy, but Ostrogradski proved it in the second paper. In the period 1824-1825 Bouniakowski also attended Cauchy's lectures and already in 1825 he successfully applied the calculus of residues in his Doctor's dissertation: Sur Ie mouvement de rotation dans Ie milieu resistant d'un systeme de plans d'une epaisseur constante et d'un contour determine, autour d'un axe incline, par rapport a I'horison, suM de la determination du rayon vecteur dans Ie mouvement elliptique des planetes. Paris 1825. A translation of this dissertation into Russian is published in Istor.-Mat. Issled. 28(1985), 247-260. Bouniakowski skilfully applied the calculus of residues, but he made no original contribution. In fact, he was the first to apply residues to a specific problem of celestial mechanics, which was considered among other questions in his dissertation. See [2} and [3}. REFERENCES

1. A. P. luSkevif: On Unpublished Early Works of M V. Ostrogradski (Russian). Istor.-Mat.

Issled. 16(1965),7-48. 2. V. S. Kirsanov: V. J. Buniakowski's Dissertation and A. Cauchy's Theory of Residues (Russian). Istor.-Mat. Issled. 28(1985),261-266. 3. N. S. lennolaeva: On Doctor's Dissertation of V. J. Buniakowski (Russian). Istor.-Mat. Issled. 29(1985),241-255.

Chapter 5

Evaluation of Finite and Infinite Sums by Residues 5.1. Gauss' Sums Suppose that k and n are positive integers and let 2nik2)

Tk=exp ( -n- . The sums

are called Gauss' sums. In Volume 1, at the end of 6.4.2 (p.232) the equality

_ 1_;n _ /S --yn

(1)

n

1 _;

,

was proved under the condition that n is an odd number. We now determine the formula Sn' where n is any positive integer. y

At----_"

Since Tn_k = Tk • we have

S"=2U,,. where x

(n even) (n odd)

(2)

Integrate the function D~------'E

2niz2 )

exp ( --;;-

(3)

Figure 5.1.

zHf(z)=-~-

exp (2niz) - 1

69

chapter,

70

along the contour r shown on Fig. 5.1 where AH = nl2, OA = OD = R and BC, FG are semicircles of radius r. Since the function (3) has poles at z = 1, 2, ... , n' and Tk 2:n;/

(k= I, 2, ... , n'),

Res!(z)=-.

z=k

we conclude that

i

..1 !(z)dz=271;; Res!(z)=Tt +T2 +··· + Tn· J k=1 z=k

l'

=Un -~T. 2 0 --~TI 2 n 2·

Along the semicircles BC and FG we have

r

•

1 2:n; iTo 2 2:n;/

1 2

!(z)dz= -----.--= --To;

BC

J!(z)

dz = -

~ Tn/ 2 •

FG

Hence, the sum Un is equal to the sum of the integrals along line segments HA, AB, CD, DE,EF,GH.

Along the segment HA we have z = x + iR (x varies from nl2 to 0). The function f becomes 2:n; i ) exp ( -n-
!(z) =---

-.

exp(2:n;i(x+iR»)-I -

and so n/2

4:n;RX) exp ( - ----

I J!(Z) dz I~J ---------- ____ n I

HA

:

0

1 - exp ( - 2:n;R)

dx 'F _n__ . 4:n; R

On the line DE we have z = x - iR (x varies from 0 to n12) and similarly we get

J !(Z)dzl~-.!!-· I,DE ; 4:n;R Along the y-axis we have z = iy, and so

J !(Z)dz+ J!(Z)dz=(J' + J--R) AB

CD

R-,

( 2:n;; ) exp idy exp ( - 2 :n; y) - 1

_--;;-y2

71

Evaluation of Finite and Infmite Sums by Residues

.(1 J , R

= -

I

e-2 7r y - 1

2:;

J, R

=

i

exp ( -

I) (2

+ - - - exp - - n; y 2) d y e2

7r

Y- 1

n

y2) dy.

Along the segments EF and GH we have z = n/2 + iy, and after some calculations we find

J

J

J

EF

GH

,

J(z)dz+

Hence, as r ~ 0, R

R

J(z)dz=i 3 n+1

~ +00

ex p ( -

2:;

y2)d y .

we get

Sn = 2 Un = 2 (i + j3 n+ I)

J

+ ..

exp ( _

o

2:;

y2) dy,

v-;;y =x,

.f2;

or putting

It remains to calculate the value of the constant

Je-/z'dx. Ifn = 4, then

+'"'

o

However, from (4) for n = 4 we get S4 = 2 (i + i

13)

v~ n

J

+ ..

e-ix2 dx,

o

and so

Je+00

2 (1 + i) =J2in

o

ix2

dx,

chapter s

72

implying

J +00

e-1x2 dx =

o

v;n

(I - i).

Substituting the obtained value into (4) we obtain the Gauss sum _ (1 + iln) (1 + i) _

(5)

Sn -

2

r-

V

n.

REMARK. If n is an odd number, it is easily verified that fonnulas (4) and (5) coincide. REFERENCE 1. C. Jordan: Cours d'Analyse de I'Ecole Polytechnique, t.2; troisieme ed. Paris 1913, pp.334-339.

5.2. The Riemann Zeta Function If the Riemann zeta function is defined by (1)

(z)

1

=L--; +'"

,,=1

(Re z > 1)

n

and if Bernoulli's numbers Bn (n

1

z

+'"

=

0, 1,2, ... ) are defined by the expansion

R

- - = l--z+ L_"'2_lr_Z2k ~-1 2 lr=d2k)!

(lzl<21r),

then, (2m)

= (_1)",-1 22",-1 1r 2", B.z (2m)!

(m=l, 2, ... ).

m

Proof. Let

z

qJ(z)=--, eZ -1

In view of (1), for (2)

f(z)

V/(z) =

z e

2,,/.

-

I'

/(z) = V/(z) z2m

(m

= 1,

2, ... ).

Izi < 1 we have

1 = 2m Z-2m-IqJ(2mz)

=~(z-2m-I(I_ mz)+ I(-I) lr 22lr trlr~Zllr-2"'-I). 2m

lr=1

(2k)!

The function/has a pole of order 2m+ 1 at z =0, and using (2) we conclude that the residue is (3)

Res/(z) =_I_(_I)"'2 2"'1r2m B.z", . 2m (2m)!

.-0

73


The function/also has simple poles at ±n (n = 1,2, ... ) with residues Res/(z) =_1___1_1

(4)

2m e2ldz2111

• -0

,.=:1::11

1

1

(n

=2m n 2..

=I,

2, ... ) .

(n+t)(±l±i).The singularities at 0, ±I, ±2, ... ,

Let en be the square with vertices at ±n of / lie inside en and so, by (3) and (4),

§/(z)dz = 2t ~+(-I)-22-,r~", t" V (2m).

(5)

~

.-=1

On the other hand, on the horizontal sides of

z=x±{n+~).

en we have

-(n+~J~x~(n+~).

and so, z

1V'()I

~

=

I

1 11 1 11 e2m • + 11 1 e2• 1• -1 +-+-=2 2 2 e21d • -1 +2

~(lcotg1lZI+I) = i(lcotg n( X±i(n+i)JI+I)

~i( cot{n+i}r+I)si(coth 3; +1). Similarly, on the vertical sides of en we have

z=±(n+fJ+iY, -(n+iJsys(n+~). and so

Therefore, IV'(z)1 and hence (6)

~ M = max( i(coth 321r + 1). 1)

for z e en. (n = I, 2, ... ),

74

ChapterS

sincem ~ 1. From (5) and (6) follows

o=2f-I-+(-l)-2:z",,r- B:z", ...1

(2m)!'

V-

and finally (m= I, 2, ... ).

(7)

REMARK 1. Bernoulli's nwnbers can be calculated from the recurrent Connula

"I(n+l,.-B"I _ L

Bo =1,

V

(n = 1, 2, ... )

~

or

(8)

(B+lyo+l =B..1

Bo=l,

(n = 1, 2, ... ).

Bernoulli's nwnbers with odd indices, exceptB1, are zero, and from (8) we get I I I B. =- 30' B2 = 42 , ...

B2 ='6'

Using (7) we obtain «2)=

... 1

,r

L"2=-' n 6 ....1

REMARK 2. Bernoulli's nwnbers are sometimes defined as the coefficients in the expansion z = 1 --z+£... 1 ~(I)"'1 B. 211 -~. e' -1 2"1 (2m)! In this case we have the so-called Hennite's Connula

1

(2n)2M

f; k 2M = 2(2m)!B•. +..

5.3. Miscellaneous Summations S.l.l.lfweC we have (1)

L;-n.n)2 =-Je +'" (

n=O

1

2Ir

2n

0

2vCOS1dt.

Proof If C = {zllzl= I}, then

75


However, since

1 _ew(6+1I2)

= _1 '" ~z"'" ~Z-"

Z

+«>

"

+«>

L." I z,,=on.

..

L." I ,,=on.

'

we conclude that 1 w(6+1I2) Res -e 2-0 Z

w" w" ~(W =~ L . , , - - =L." ,,=0

n! n!

. )2

..=0 n!

'

and (1) follows directly. REFERENCE

1. D. D. Adamovic: A Collection of Solved Problems in Mathematical Analysis, Ordinary and Partial Differential Equations and Complex Functions (Serbian). Beograd 1959.

5.3.2. If m, n are odd numbers, then

1t gktr- t gktr- =tr- (mn L-2 k m n 2 q

1)

2

+«>

--2 - - ,

mn

k=1

where q is the lowest common factor of m and n. This result is obtained by integrating the function/defined by tr z

trz m

trz n

/(z) = 2'"tg-tg-cotgtrz

along the square en with vertices

and letting n -+ +00. REFERENCE

1. M. B. Gregory, 1. M. Metzger, P. Zwier: Problem 1039. Math. Magazine 52(1979), 260-261.

5.3.3. Using residues Cauchy [l) proved the formula +00_1 _ =_1_+_1_+ m

L (n+k)O k=O

nO-1

2no

~ k-I

(_l)k-I~ r(a+2k-l) (2k)!

nO+ 2k - 1

where a > 1, 0 < () < 1, and c 1, c2' ... are Bernoulli's numbers:

76 C2

I =-,

I

c3 =-,

30

42

REMARK. If 1 < a < 2 and if n is a very large nwnber, Cauchy claimed that the approximation obtained by suppressing the term which involves () is very good and easy to apply. REFERENCE 1. A. Cauchy: Sur un emploi legitime des series divergentes. C. R. Acad. Sci. Paris 17(1843), 370-376.

5.3.4. The equality (1)

f (2m1+ 1)3 (.!.th (2m+l)m +zth (2m+I)R') =!{ z 2 2z 16

_0

holds for all complex numbers z, except those for which at least one term in the infinite series is singular. Proof. Start from the function w H few) defined by

R'

few) =-3 tgR'W thR'ZW w

(weC, z eC).

The value tgnw th1Z%W is bounded on any circle e" = {wllwl= n, n ENol if z satisfies the condition zw ".# (2k+ 1)i12, k e Z when ween' Hence, jf(w)dw < ~, c;, n

where C is a positive constant, implying (2)

lim 1. f(w)dw

n....,.+GO

1

c;,

=O.

Therefore, by Cauchy's residue theorem it follows that the sum of residues offenclosed by en decreases towards 0 as n -4 +00.

The only finite singularities of f are simple poles at w = 0, w = (2m + I) 12, w = (2m +1); 12z, wherem e Zandz".# 0, and we have Resf(w) = tfz, .... 0

Res few) = w=2111+1 2

..J~~!I)I 2z

f

8 th (2m+ I)m (2m+I)3 2 '

8z2

(w) = (2m + 1)3

From (2) we get, for z ".# 0,

th (2m+ I)n 2z'

77

Evaluation of Finite and Infinite Sums by Residues

(3)

f

8

.=_. . (2m + 1)

3(th(2m+l)nz+z2th(2m+l)1r)_~z=0. 2z

2

Dividing (3) by z, and letting z -+ 0, we see that the resulting formula is also true for z = O. It is then easily reduced to (1).

*

The equality (1) was proved by Grosjean [1], under the condition that z iy (y irrational real number). The above proof by residues (which is much shorter and also holds for z = iy, Y irrational) is due to de Doe1der [2]. A special case of (1) corresponding to z = 1, is recorded in [3]. A summation similar to (1) is given in Volume 1, p. 246. Grosjean [1] rediscovered this formula by his own methods. REFERENCES 1. C. C. Grosjean: Proofofa Remarkable Identity. Simon Stevin 58(1984), 219-241. 2. P. 1. de Doelder: A Note on "Proof ofa Remarkable Identity". Simon Stevin 61(1987), 193-195. 3. E. R. Hausen: A Table ofSeries and Products. Englewood Cliffs 1975, p.282.

5.l.5. IfRe(a+b-c-d) < -1 and if a, b are not integers, then + .. r(a+n) r ( b + n ) : r c 2

n=~ .. r(c+n) r(d+n) =

sin:rca sin:rc b

r(c+d-a-b-l)

r(~-~) r(d-a) r(c-b) r(d-c)

This formula was proved by 1. Dougall. It can be obtained by integrating the function

z 1-+ n cot g n z r(a+z)r(b+z)r(c+z)r(d+z)

along a suitably chosen circle. REFERENCE 1. H. Bateman and A. Erdelyi: Higher Transcendental Functions, YoU. New York - Torolito London 1953.

5.l.6. If 0 < 0< n; then we have the Fejer-Jackson inequality (1)

" 1

L-sinkO>O. k=1

k

Proof. Let F(U)

1 =L" -sinkO, k=1

k

l-z" /(z)=l-z

(z

=x+iy).

We integrate the function/along the arcAB of the circle

c= {zllzl= I}, where A

are the points e-i1J, eilJ, respectively. On the circle C we have z = eit, and so

and B

78

Chapter 5

1- e J/(z)dz =; J--I-' elldt l-e l '"

fJ

AS

-fJ

fJ

=; J(EI' + ell' + ... +el"')dt -fJ

.(eifJ_e-1fJ

_

- 21

2;

+

1 e2IfJ_e-2IfJ 1 ElllfJ- e-In6J + ... + 2 2; n 2;

= 2; F(O). We now integrate the function/along the circular arc re {zllz -11 =r} where 0 < r < 1, with end points A and B. On the arc r we have z = 1 - reiu where -p 5 U S P and p is determined so that 1 - reip is the point A. Hence

-f 1-(I-re'''

J/(z)dz=-;r J r

re'''

r.

P

.

II

e'''du=; J(I-(I-re''') )du.

P

~

By Cauchy's theorem we have

J/(z)dz + J/(z)dz =0 r

AS

implying

Jr f(z)dz =2; F(O),

i.e.

2; F(O) =;

1(1-(I-re'''r)dU

-p

=;

1(

1

1- Re(l-re''')'' )dU + Im(l-re''')'' du

-p

-p

and therefore 2F(0)

= 1(1- Re(I-rEl")" )dU. -p

Now, the integrand in the last integral is an even function, and so

(2)

1(o 1- Re(I-re''')'' )dU.

F(O) =

However, since II-re'''1 < 1 we have Re(I- re''')'' < 1, and the integrand in (2) is

79


positive. Hence, we have F(D) > 0, i.e. (1). REMARK 1. This proof of the inequality (1) is due to P. Turan. REMARK 2. Inequality (1) and the corresponding cosine inequality

Lk

1 1+ • -cosk6> 0,

(O
t.1

were generalized by L. Vietoris [1] who proved that (O
provided that a o ~al

~ ... ~a. ~O

and (1 S k

s!!.). 2

REFERENCE

1. L. Vietoris: Ober das Vorzeichen gewisser trigonometrisher Summen. Sitz. Ber. Ost. Ale. Wiss. 167(1958),125-135.

Chapter 6

Applications of Calculus of Residues to Special Functions 6.1. Polygamma Functions of Arbitrary Order The polygamma functions 'P (n) are defined for nonnegative integers n by 'P(II) (x)

d )11+1 =( dx 10gI'{x). +..

They have various applications~ for example, the sum LR(k) can be expressed in terms k=O

of values of polygamma functions where R is a rational function defined at nonnegative integers and of degree ~ -2. See [1]. It is possible that sums with functions more general than rational might be summed if a suitable generalization of the polygamma function to nonintegral index could be found. A natural generalization, suggested by Ross [2], is to define (1)

lJ'(.)

(x) = 1-.- 1 log

r (x),

where] P is Liouville's fractional integration operator: x

(2)

]P log r(x) =_1_ !(X-t)P-l logr(t) dt. r(p)

o

For fixed x > 0 this integral converges and defines an analytic function of p only for Re p > O. Hence, the definition (1) is correct only for Re v<-1. Grossman [3] obtained an extension of ]P 10gI'{x) to allp. His formula reads: ]Plogr(x)= (3)

xP

r (p + 1)

(IOg~+y+ x

r'(P+l) r (p + 1)

r;xl

i'X)

p

+1

AH ..

- xP 2n;.

A-I ..

80

F(s)C(s) _n_ r(p+l+s) sinns

ds

81

ApplicatioOl of Calculus of Residues to Special Functions

where 1 < A. < 2 and integration is along a vertical line. The function pH]' 10gI'(x) defined by (3) is an entire function in the p-plane for each x in the plane cut along the negative real axis (so that log 1 = 0 and xl = e,I0&r). For Re p > 0, the expressions (2) and (3) coincide. Using residues Grossman [3J, [4J derived from (3) two interesting expansions when the contour is deformed in one direction or the other to lie along the real axis. In the first case, suppose that 0 < x < 1 and deform the line contour into one which comes from +00 on the real axis, loops around +2 in the positive sense and returns to +00. There is a sign change because of the change in orientation in the contour integration. The function

sl4 /(s)

=r

I'(s)(s) n I'(p+l+s) sinns

has simple poles at s = 2,3, ... inside the described contour, and Res/(s)=limr I'(s)(s) (~-k)=(_I)k I1k)(k) ~ . I-+k I'(p+l+s) SlDns r(p+l+k) .-k Hence, __I_AT r I1s)(s) n ds=(-It,(k) 11k) 2m A-/oO I'(p + 1+ s) sin ns I1p + 1+ k)

~

=(_I)k '(k) B(p+ I, k) ~, I1p+l)

implying ]Plogl1x)=

x' r(p+l)

(I

rxlog-+r+ F'(p+l) - x I1p+l) p+1

+~(-I)k'(k)B(P+I,k)~ ). and the summation converges for Ixl < 1 and x = 1. For x > I, deform the contour in the other direction so that it comes from -00 on the real axis, loops around + I in the positive sense and returns to -00. This contour encloses simple poles at s = -2k (k = 1,2, ... ) and double poles at s = 1,0, -I, -3, -5, ... If /has the same meaning as before, we get TY2k-p)rt2k+I). Res/(s)= (-ly+l.I:, ':r' slDnp (2n)2k+1 XU L

r--2k

Res/(s) = ....1

x (logx- V'(p+2»); I1p+2)

(k

=I, 2, ... );

82 Resj(s) =

1 (log2nx- y- ",(p+l»); 21"{p+l)

.-0

Res j(s) = (_I)k (2nx)l-u 1r

I=I-U

(2k) (IOgnx- ",(p+2-2k) ('(2k») 1"{p+2-2k) (2k) (k= 1,2, ... ).

The asymptotic expansion for ]P logr(x) as formula is rather complicated.

Ixl~

+00 follows by summing residues. The

REFERENCES

1. M. Abramovitz and I. A. Stegun: Handbook o/Mathematical Functions. New York 1965.. 2. B. Ross: Problem 6002. Amer. Math. Monthly 8(1974),1121. 3. N. Grossman: Polygamma Functions 0/ Arbitrary Order. Siam 1. Math. Anal. 7(1976), 366372. 4. N. Grossman: Errata: Polygamma Functions 0/Arbitrary Order. Siam 1. Math. Anal. 8(1977), 922.

6.2. A Connection Between the Exponential and the Gamma Function Suppose that a, b and 0} are real positive numbers and let R be the rectangle with vertices at a ± iO}, b ± iO}, so that the gamma function r is regular inside R. If s E intR, by Cauchy'S theorem we have

1"{s) =~t 1"{z) dz. 2m R z-s If a and b remain fixed, and if 0} ~ +00, the integrals along the two horizontal sides of R tend to 0, and for a < b we get (1)

J

J

1 ,1+/01> F(z) 1 b+ltD F(z) F(s)=-. -dz+-. -dz 2m ,,_/01> s-z 2m b-ItD s-z

where a
so

In the first of these integrals we have Re(s-z) >

J

°

and in the second Re(s-z) < 0, and

-- = JxS-Z-'dx.

I

1 = xs-z-'d x, -s-z 0

1

+ao

z-s

I

Hence, (1) becomes (2)

1

J I

J

,1+/00

1

F(s) = - . r'dx F(z)x-zdz+-. 2m 0 ,,_/01> 2m

Jxs-1dx Jr(z)x-zdz.

+ao

b+/OI>

I

b-/OI>

Since the function z ~ F(z)x- Z is regular inside R we have

83

Applications of Calculus of Residues to Special Functions

J1\z)x- dz = J1\z)x-zdz,

a+1ot>

)'

b+1«>

p

Z

b-Iot>

a-lot>

and (2) reduces to

1 +'" a+1«> 1\s) =-. xl-1dx 1\z)x-Zdz,

J 2m

J

0

()

-n+l

--n

a

x

a-I«>

or

1\s) =

JH(x,a)r1dx,

+'"

Figure 6.2.

o

where the function

X 1-4 H(x,a)

Jr(z)x-Zdz

1

a+1«>

2m

a-lot>

=-.

converges for positive x. Now, inside the rectangle P shown on Fig. 6.2, the function x 1-4 1\z)x-Z has poles at the points 0, -1, ... -n+1, and

. Res1\z)x- Z = hm (z + v)1\z)x- Z z=-v z-+-v Since lim (z+ v)1\z+ v) ;~-v

(z+ v)r(z+ v)x- Z . z-+-vz(z+1) ... (z+v-1)

. = hm

=1, we get

XV Res1\z)x- Z = z=-v -v(-v+1) ... (-1)

(_1)VxV

="":""--'--v!

Hence, for all a > 0,

H(a,x)

=n-++oo lim L v=O n

Res1\z)x- Z z=- v

=n-++CO lim L v=0 n

(_1)VxV vi

=e- x •

In other words, we have proved the formula (3)

e- x

Jr(z)x-Zdz

1

a+1«>

2m

a-lot>

=-.

(a> 0),

which connects the exponential and the gamma function. REMARK 1. Of course, the ganuna function can be defined by Euler's fonnula

J

+«>

(4)

r(z) = e-'xz-1dx

(Re z>O),

o

and so (3) is the inversion fonnula for (4).

84 REMARK 2. If z e C, and if the fimctionfis such that the integral

-J

f(x)r-1dx is convergent, it is

o

said that the fimction F dermed by (5)

Cbaptcr6

j

F(z) = f(x)r-1dx o

is the Mellin transfonn off. The inversion fonnula is then (6)

1 ,..,.

f(x)

=-. JF(z)x"'dz. 2m,....,.

For f(x) =e-', F(z) =r(z) the fonnulas (5) and (6) reduce to (4) and (3). REFERENCE 1. R. Campbell: Les integrales Euleriennes et leurs applications. Paris 1966, pp. 244-248.

6.3. Residues of Some Functions Related to the Gamma Function We shall now evaluate the residues of some special functions: the gamma function and certain functions related to it. Some of them will be needed later (see 6.4). 6.3.1. The gamma function r is regular in the entire complex plane except for simple poles at z = -n~ n = 0, 1, 2, ... We have Resn:z) z=-n

r(l) (-I)" = lim (z + n)n:z) =z-+-nz(z+l) lim r(z+n+l) --....:.....:;.--=-... (z+n-I) -n(-n+l) ···1 n! Z-+-IJ

6.3.2. For a positive integer n Kurepa [1] defined the left factorial !n by (1)

!n =O! + I! + ... + (n -I)!

The left factorial function L is defined for complex z, such that Re z > 0, by (2)

L(z)

= Je-'-dt +..

(z_1

o

(-I

(Rez > 0).

It is easily shown that (3)

L(z + 1) - L(z) =n:z + 1),

and that (1) follows from (2) on putting z = n

11 =L(I) =1.

E

N. In particular, 10 = L(O) = 0,

We now define the left factorial function L by (2) and (3) for all z Resn:z)

z=-"

from (3) we get

=(-1)" n!

(n

=0,

1, 2, ... )

E

C. Since

8S


ResL(z) =-ResF(z) z=-I

;=0

=-1,

ResL(z) = ResL(z) -ResF(z) = -1- (-1)1 = 0, ;=-1 ;=-1 I!

;=-2

(1)2

1

ResL(z) = ResL(z)-ResF(z) =0---=-, ;=-3 0=-2 0=-2 2! 2!

1 1 ResL(z) = ResL(z) - Resr(z) =--+-, 0=-4 0=-3 0=-3 2! 3! and by induction we find that

ResL(z) =

o=-n

l)k-1 I---k! n-I (

(n

k=2

= 3,

4, ... )

The left factorial function is meromorphic with simple poles at z = -1, -3, ~, ... The points z = and z = -2 are regular (L(O) = 0, L(-2) = 1).

°

REFERENCES

1. D. Kurepa: On the Left Factorial Function In. Math. Balkanica 1(1971), 147-153. 2. D. Kurepa: Left Factorial Function in Complex Domain. Math. Balkanica 3(1973),297-307. 6.3.3. The function

z H F(z)I'(p-z)

(Rep>O, x>O)

Xo

is regular except for the simple poles at z = -n and z = p + n (n at those poles are (-lr r(p+n)xn

n! respectively.

and

(-lr+ 1 I'(p+n)

n!

xp+n

= 0,

1, 2, ... ). The residues

(n

=0,

1, 2, ... )

6.3.4. The functionfdefined by

f(z)

= r(z)I'(z-l) x'

(x> 0)

has a simple pole at z = 1, and second order poles at z = 0, -1, -2, ... We easily get

1 Resf(z)=- . • =1 x In order to find the residue at z = 0, note that

z2f(z) =z 2r(z)r(z-1)x-;

=zr(z + l)r(z -l)x-' =(z -1)-1 r(z + 1)2 x-·.

86

Cbapter6

But

(z-Irl=-I-z- ... l1z+I)=l1l)+F'(I)z+ ...

and multiplying the above series we obtain z2/(z)=-I+z(logx-2F'(I)-I)+ ... , implying that

Res/(z) = logx-2F'(I)-1. r-O

Applying a similar procedure, we obtain the residue at any second order pole:

(2

1)

xk Res/(z) = 10gx--F'(k+I)-r--J: k!(k+I)! k! k+1

(k

=0,

1, 2, ... ).

6.3.5. The function/defined by

fez) = l1z)l1z+n) Xz

(x> 0; n

=0,

I, 2, ... )

has simple poles at z = -k (k = 0, 1, ... n-I) and second order poles at z =-n-k (k = 0, 1, 2, ... ). For the simple poles we easily get

Res/(z) = (_I)J: l1n-k)xJ: z=-J: k!

(k =0, 1, ... , n-I).

As in 6.3.4. we expand (z+n+ki fez) into series, and we conclude that the residue

at the second order pole z = -n-k is equal to the coefficient of (z + n + k expression

( 1),,+J: ( It (I + &V/(l + k +n) )---(1 + &V/(I + k»)(1 + elog x) x"+J: , &(k+n)! k! i.e. that

Res /(z) r--,,-J:

( 1)"~J: = (k - +n)!k! (V/(I+k)+V/(I+k+n)-logx),

where V/(t) = F' (t) I ret).

r

l

=

&-1

in the

87


6.4. Some Integrals Involving the Gamma Function 6.4.1. Consider the integral

_I_Cj F(s)r(p-s) ds

(x>O;

x'

2m c-Iot>

Rep>O)

where 0 < c
1=-"

=(-I)" F(p+n) x";

Res /(s)

n!

=(-1),,+1 F(p+n) x- p -". n!

I=p+n

Let L_N be an arbitrary closed contour which encloses a part of the vertical line Re s = c, where 0 < c < Re p, and which encloses only the "left" poles s =-n; n = 0, I, ... , N (Fig. 6.4.1.1.). Then, by the residue theorem

~

f r(s)r(p-s) ds= f(-It F( P7n) x".

2m L..N

x'

n.

,,=0

y y

-2 -I

Rep

0

x

x

Figure 6.4.1.1.

Figure 6.4.1.2.

We now let N ~ +00, so that the contour L_N deforms into the contour L-aJ shown on Fig. 6.4.1.2. which is always at a distance &> 0 from the poles, and we get (1) =r(p)I(-It p(p+l) ... (p+n-I) x" ..=0

= F(p) (1+x)P

n!

88

Chapter 6

The above series converges for Ixl < l. In a similar way, taking the contour LN which encloses the "right" poles s =p + n;

n = 0, I, ... N, and letting N -+ +00 we get

J

_1_ I'(s)I'(p-s) ds= 2m L.. x'

_..!.. I(-I)n I'(p+n) =_ I'(p) x P n=O

n!xn

(l+xY'

and this series converges for Ixl > I. The contours L-«J and L+«;, are arbitrary (subjected to the prescribed conditions) and they can both be, in a special case, straightened up to become the line Re s =p. Hence, if o< c < Re p, we have I'(p) r(s)r(p-s) ds= { (l+x)P' 2m c-/oO x· _ I'(p) (1+xY'

_1_Cj

O1

= 0, the loop of the contour L_ should be between the points s = -I and s = 0, and -1 < c < o. Hence, we let p -+ 0 and we delete the tenn corresponding to n = 0 in the sum (l). This gives

REMARK. If p

I <+J/oo TT s) TT s) +.. TTn) +.. x" - . ~\ ~\- dS=L(-lt-~\-1 x"=L(-lt-=-log(l+x), 2m <-I.. x' ...1 n. ...1 n where

Ixl < 1.

6.4.2. Consider the integral (1)

I (z) n

=_1_<+J/oO r(s)r( -n 2. m c-/oO

Z•

s) ds

(n = 0, 1, 2, ... )

where c is a real constant such that c E Z. For n = 0, -1 < c < 0 we have (see Remark in 6.4.l.) IO(z) = -Iog(l+z)

(Izl< I).

Similarly, if 0 < c < 1 and n = 0, the integral (1) can be expressed for the sum of "right" residues at s = k: Io(z)

Izi > 1, by means of

1 (1)

+<0 I'(k) =L(-I)k_,-k = -log

k=1

1+- .

k. z

z

If for n = 0 we take -m-I < c < -m (m = 0, I, 2, ... ) then from the expression -Iog(l+z) we have to take the m-th section of the Taylor expansion, i.e. we have m

Io(z)

k

= -log(l+z)+ L(-I)k-I ~ k=1

k

(Izl< I; -m-I
89

Applications of Calculus of Residues 10 Special Functions

Now let n = 0, 1, 2, ... and -m-l < c < -m (m = -1, 0, 1, ... ,n). The integral (1) can be considered, on one hand, as the sum of the residues at the left second order poles s = -m-l, -m-2, ... , -n and the residues at the left simple poles s = -n-l, -n-2, ... , and on the other hand as the sum of the residues at the right second order poles s = -m, -m+ 1, ... , and the residues at right simple poles s = 1, 2, .... If

°

= F(s)F(:n-s) ,

/(s)

z

then Res/(s) =(-lr l V'(I+k)- V'(I+n-k)-logz Zk r--k

k!(n-k)!

(k

=0,

1, ... , n)

where V'(t)=F'(t)/F(t); Res/(s)=(_l)kF(k-n)zk r-k k!

(k=n+l, n+2, ... );

~

Res/(s) = (_I),,+k+1 F(k)

(n+k)! z

r-k

and hence I,,(z) = (-1)"

t

(k

=1,

2, ... ),

logz+ V'(l~n-k)~ V'(I+k)~ k.(n-k).

k=_1

+(-Z),,+I

kl

(_Z)k, L k=O (n +k +1)! +GO

Izl
and

= (-I)"f V'(l+k)- V'(l+n-k)-logz ~

I (z) "

k=O

k!(n-k)!

kl .

+'"

+ (-I)"L

k=O (n +k+ I)!

(-zr k - I ,

Izl> 1.

The above expressions for I,,(z) are analytic continuations of each other. 6.4.3. Consider the integral K (z)

"

=_1_ c+J/oO F(s)F(s)F( -n - s) ds 2· fa C-/oO

Z

,

(n =0, 1, 2, ... )

The function/defined by /(s)

=F(s)F(s)F(-n-s) z'

has simple poles at s = 1, 2, ... , second order poles at s =-n-l, -n-2, ... , and third order

90

chapter 6

poles at s = 0, -1, ... ,-n. We have Qk(Z)

=Resf(s) =(_I)n+k+1 r(k)r(k) 6=k

1

(k

(n+k)! Zk

=1,

2, ... )

bk(z) = Resf(s) = (2'1'(k+I)-'I'(k-n)-logz)r(k-n)zk .=-k k!k!

(k=n+l, n+2, ... )

(1

( 1)lt-k-1 ck(z) = Resf(s) = -log2 z+ ('I'(l+n-k) -2 'I'(l+k»)logz+ 2 'I'(l+k)2 6=-k . k!k!(n-k)! 2

+ 3 If" (1) +.!.( ",(1 +n _k)2 -If" (1 +n - k) )-If"(1 + k) 2

(k

- 2 '1'(1 +n - k) '1'(1 +k») Zk

=0,

1, ... ,n).

Hence, for m < C < m+l, m = 0, 1, ... we have "

(1)

Kn(z)

=~>k(Z)+ k=O

+00

WI

k=n+1

k=1

~)k(Z)+ LQk(Z),

3n2

where Izl<+oo and largzl<-. If m = -1, -2, ... then it is necessary to omit the last sum in (1) and also those terms in the first two sums which correspond to k = -m-l, 0,1,2, ...

6.4.4. Start with integral L (z)

=_1_<+J/oO r(s)F( -s) (-z)-Ods

n

2m

<-/00

(-S),....I

(n

=0,

1, 2, ... ).

The integrand SH f() S

( )-0 = r(s)F( -s) -z I

(-sr-

has simple poles at s

= ± k (k E N) and a pole of order n+ 1 at s = 0.

The case n = 1 is treated in 6.4.2. Using the same procedure we conclude that +'"

Ln(z) = LResf(s), k=lo=-k

Izl< 1

and +'"

Ln(z) =- LR_esf(s), k=O o-k

But it is easily seen that

Izl> l.

91


zle

(k eN)

Res/(s)=k"

.=-Ie

z-Ie

Res/(s)=--

(k eN).

(-kr

r-Ie

The residue at s = 0 is more complicated. In order to evaluate it, we write/in the form /(s)

=(-I)" s-,,-II"(I +s)r(l- s)( -zr',

and we look for the coefficient of S' in the Taylor expansion of the function s 14 (-1)" F(I +s)F(I- s)( -zr·. Multiplying the series for F(l+s), F(I-s) and (-zr', we get Res/(s) = r-O

~l

log"-2V(-z)

",,0

(n-2v)!

~cv--=:-"":'-"':"

where

Therefore, for -1 < c < 0 we have

Izl
(2)

and +'"

-Ie

Izl> 1

L,,(z) = - L-z--Res/(s), 1e=1 (-k)" r-O

implying (3)

L,,(z) +

(-I)"L,,(.!.) = Res/(s) z ...

(n = 0, 1, 2, ... )

0

Formula (2) defines the function Ln which is called polylogarithm of order n in the unit disc. Formula (3) defines the analytic continuation of Ln outside the unit disc. REMARK. Polylogarithms were briefly mentioned in Volwne 1; 8.2.5, p. 301.

It can be shown that polylogarithms satisfy the recurrent relation (n

=0,

1, 2, ... )

and that certain definite integrals can be expressed by means of polylogarithms, e.g.

1 Jou-log"-2 1

U

log(l- uz)du = (-I),,-I(n - 2)! L,,(z) (n = 2, 3,4, ... ; larg(l-z)1 < n)

92

Chapter 6

+JoO.!.e-'I2(~ 1

o

(-I)" £... klk n - I

)dl = L (-z)

(n

n

"=1'

=1,

2, ... ;

Rez > 0)

logl 1 ( -1) --log 1 2X+-, ,r --=--dt = --L2

x>1

log2 t 1 ( 1) 1 ,r dt=-~ - --log x+-Iogx, Jt(l-x) x x 6x 3x

x>l.

+JoO 1

t(l-x)

+00 1

x

x

2x

3x

3

REFERENCE

1. 0.1. Mari~v: A Method a/Evaluating Integrals a/Special Functions (Russian). Minsk 1978.

Chapter 7

Master's dissertation of J. V. Sohocki 7.1. Introduction This section is devoted to Master's dissertation of the Russian mathematician 1. v. Sohocki which was published under the title The theory of integral residues with some applications (Russian), St. Petersbourg 1868, VIII+ 135 pp. This dissertation is very important since it is the first research paper written in Russian and devoted to complex functions. Namely, the leading Russian mathematician of that time, P. L. Ceby~ev thought that it is completely unnecessary to use the methods of complex functions. For that reason Sohocki had to wait for two years before Ceby~ev accepted his proposed dissertation. Later on Ceby§ev changed his views and published two papers in which, using residues, he proved some formulas on the bounds of certain integrals. Those results of Ceby§ev were reviewed in 4.3.

REMARK.

However, Sohocki's dissertation is -not only important from the historical point of view, but also because it contains some serious results which were unjustly overlooked in mathematical literature. For instance, Sohocki was the first to apply residues to Legendre's polynomials, but the procedure was wrongly ascribed to Laurent (see, for example, [I)). It is very difficult to obtain Sohocki's dissertation. After many years of unsuccessful attempts and waiting, two copies finally arrived in Belgrade from Moscow. We believe that this is the first time that Sohocki's results are reviewed in a monograph. and so we shall expose them with some detail. The dissertation is divided into two parts: theory and applications. The first, theoretical, part contains two chapters:

l. General properties of uniform functions; pp_ 1-24 2. Theory of integral residues; pp. 24-43. The second part contains the chapters: 3. Expansion into series of inverse functions; pp. 44-50 4. Symmetric functions; pp_ 50-53

93

94

Cbapter7

5. Continued fractions~ pp. 54-57 6. On functionsXn (Legendre's polynomials)~ pp. 77-107 7. Expansion of a function in terms of continued fractions; pp. 108-123. At the end there is an Appendix (with two sections); pp. 124-135. Chapter 1 contains a short exposition of some parts of the theory of complex functions and we shall not comment it, with the exception of the following important question. The theorem on the behaviour of an analytic function in a neighbourhood of an essential singularity is well known and can be found in almost all the text books on complex analysis. It is usually formulated in one of the following variants:

VARIANT 1. Suppose that a is an essential singularity of an analytic junction f. /f ~ 6 are arbitrary positive numbers, and if A is any complex number (or A = 00), then there is a point z in the disc

{zllz-al< 6} such that If(z)-AI e).

Suppose that a is an essential singularity of an analytic function f. For any complex number A (finite or infinite) there exists a sequence (zn) which converges to a,

VARIANT 2.

such that lim fez,,) n-++aa

=A.

In mathematical literature this theorem is known as the Weierstrass theorem (see, for example, (2) or (3) where Variant 1 is given, or (4) where Variant 2 is given), and it is said that it was proved in 1876 by Weierstrass (5). However, this theorem can be found in Sohocki's dissertation from 1868 and also in the book (6) by the Italian mathematician Casorati. Therefore, Sohocki and Casorati (clearly independently from each other) published this theorem eight years before Weierstrass, and the theorem should be called the Casorati-Sohocki theorem. Until the publication of Markuievi~'s paper (7) the theorem was in Russian literature also ascribed to Weierstrass (even Markuievi~ did so in his book (4)) but after the publication (7) it is called Sohocki's theorem or the SohockiWeierstrass theorem. In Western literature the theorem still carries the name of Weierstrass. We quote the original formulation of Sohocki from his Master's dissertation:

/fa given function f(z) at some pOint a becomes 00 of infinite order, then the junction fez) must take all possible values at that point. Certain explanations are necessary. First of all, the assertion "If the function fez) becomes 00 at a, then it takes all possible values at that point" appears to be contradictory. However, it is only a question of inadequate terminology. Namely, when Sohocki considers a function in a neighbourhood of an isolated singularity, he uses the following terms. If the Laurent expansion of f around a is

f(z)=

+ ..

2:

k=-"

Ak(z-a)k

95

Masta's dissertation of1. V. Sohocki

Sohocki says that / becomes 00 of order n at z = a, and if the Laurent expansion around a is +00

/(z~ = LAk(z-a)k k=-oo

he says that it becomes 00 of infinite order at z = a. In other words, the term "function/ at z = a becomes 00", does not mean that/(a) =00 but we have two cases: (i) if "the :function becomes 00 of order n", it means that it has a pole of order n; (ii) if "the function becomes singularity.

00

of infinite order", it means that it has an essential

Therefore, there is no contradiction, but only inappropriate terminology. The other term which should be explained is "the :function / (z) at z = a must take all possible values". This does not mean that/(a) can be any number, but that limf(z) ..... can

take any value. This can be seen from Sohocki's proof of the theorem, and also from the example that "the :function sin

f-h at z = b takes all possible values" .

It is interesting to note that Sohocki himself did not think that this theorem is particularly important In the preface he emphasized some other results from his dissertation, but this theorem is not mentioned. Nevertheless, the theorem should be known as the Casorati-Sohocki theorem. REFERENCES 1. Encykloplkiie der mathematishen Wissenschaften, IIA 10, Leipzig 1904-1916, pp.705-706. 2. E. T. Copson: An Introduction to the Theory of Functions of a Complex Variable. 2nd ed.

London 1935; Reprinted 1946. 3. E. C. Titchmarsh: The Theory ofFunctions. 2nd ed. London 1939; Reprinted 1964. 4. A. I. Marku§evi~ Elements ofthe Theory ofAnalytic Functions (Russian). Moscow 1944. 5. K. Weierstrass: Zur Theorie der eindeutigen analytischen Frmktionen. Abh. Preuss. Akad. Wiss. Berlin (Math. K1asse) 11(1876). Reprinted in the Collected papers of Weierstrass: Math. Werke, Bd. 2, Berlin 1895, pp. 77-124. 6. F. Casorati: Teoria dellafunzioni di variabili complesse. Pavia 1868. 7. A. I. Marku§evi~ Contribution of J. V. Sohocki to the General Theory of Analytic Functions (Russian). Istor.-Mat. Issled. 3(1950), 349-406. 8. J. D. Keane: On Some Forgotten Results from Master's Dissertation ofJ. V. Sohocki (Serbian). Istorija mat. meh. nauka 4(1991), 85-94.

7.2. Properties of Residues At the beginning of Chapter 2 of his dissertation Sohocki defined the residue at z = a (a finite) and also at z = 00 in the usual way, and derived the known relation between the residues and the coefficients of the Laurent expansion. He then formulated and proved 15 theorems in which he established the main properties of residues in a rather modem way,

96

chapter 7

treating rRes as an operator. We quote all those theorems (but with modernized notation) where we suppose that the functions/, g, h, ... are analytic functions with finite number of isolated singularities. We shall give proofs of some lesser known theorems. THEOREM

1. We have

n

L

Res (f(z) + g (z») =

k=1 Z=Zk

THEOREM 2. n

L

n

L

n

Resf(z) +

k=1 Z=Zk

L

Resg (z).

z=1 Z=Zk

If a is a constant, then Res af(z) = a

k=1 Z=Zk

n

L

Resf(z).

k=l Z=Zk

THEOREM 3. If z = Zo is a singularity of the function (z. w) 14f(z. w) where Zo is independent ofw. then

-

d

Resf(z, w) = Res

dWz=q

df(z w)

• dw

z=q

;

f(Resf(z, w»)dw=Res ff(z, w)dw. Z=Zo

THEOREM 4.

Z=Zo

If the function (z. w)l4f(z. w) has isolated singularities at z

=

a andw

=

b.

then Res(Resf(z, w»=Res(Resf(z, w»). z=a

THEOREM 5.

w=b

Ifz

=

w=b

z=a

a isa singularity off, then

Res!'(z) = O. z=a

THEOREM 6. If f and g are analytic functions such that z = a is a singularity of the function zl4 f(z)g(z). then

Resf(z)g'(z) = z=a

THEOREM 7.

-

Res!'(z)g (z). Z=Q

Ifz

=

a is not a singularity off, then

f(a) = Res z=a

f(z) .

z-a

97

Master's dissertation of J. V. Sohocki THEOREM 8.

Jfz

=

a is not a singularity off. then

j
/(z) Z=Q (z - a)n+ I

THEOREM 9.

Jfz

=

a is a pole ofordern off. then

1 (d---(z-a)n/(z»)

Res/(z)=-(n -I)!

Z=Q

n-

dzn -

I

.

I

Z=Q

.

THEOREM 10. Jfz = a is a pole or an essential singularity off. then the principal part of Laurent's expansion around z = a is:

Res /(/) . I=Q

z-I

Proof If z = a is a pole of order n off, then the principal part of Laurent's expansion around that point is

i~·

k=1 (z-a)k

Since A_k= Res/(t) (t-a)k-l

(k=l, ... , n),

t=Q

we have A_k

--= (z-a)k

R

es /(/) (/_a)k- , t

t=Q

(z-a)k

and so ~

L.,

A_k

--=

R

k=1 (z-a)k

es

~ /(/)(/_a)k- 1

L.,

t=Q k=l

(z-a)k

R

= es

/(/) «/-a)"-(z-a)n)

t=Q

(z-a)n(/-z)

,

implying ~

L.,

A-k

--=

k=1 (z-a)k

R es

/(/)(t-a)n

Z=Q (z-a)n(t_z)

/(/) + R es-. t=Q z-I

However, the first summand at the right hand side of the above equality is equal to 0, and the theorem is proved. In a similar manner we conclude that the theorem is true in the case when z = a is an essential singularity off

98

Chapter 7

THEOREM 1l.1f z

= a is a zero or a pole of f·of order n, then

dlogf(z) Re s = ±n; dz

z=a

if the function g isflnite at z = a then Res z=a

dlogf(z) g(z)= ±ng(a). dz

We take the sign + if z = a is a zero, and the sign - if z = a is a pole of f. THEOREM 12. 1fthe function fhas aflnite number ofsingularilies zl' ... ,z". then

THEOREM 13.1fthefunctionfhas a singularity at z = a, and if the function g is analytic in a neighbourhood of w = b, where g(b) = a, then 1

Resf(z) = - ResJ(g (w))g'(w), m w=b

z=a

where m is the order of the zero z = a ofthe function z H g(z) - g(a). Proof. Start with the Laurent expansion f(z) =

+00

2:

Ak (z - a)k

k=-co

and put z = g(w), a = g(b) to obtain J(g (w») =

+00

2:

Ak (g (w) - g (b»)k,

k=-co

i.e. J(g(w))g'(w)=

+co

2:

Ak(g(w)_g(f!))kg'(W),

k=-oo

or (I)

where

J(g (w)) g'(w) = A_I

g'(w) g (w) -g (b)

+ ~ G (w), dw

99

Mutcr'I diaatatioo of J. V. Sohoc:ki

From (1) we get

Resf(g(w»g'(w)=A_1Res

g'(w)

w-b g(w)-.r(b)

w=b

+ Res.-!. G (w). w=b dw

Since the function G is analytic in a neighbrouhood of w = b, using Theorem S we conclude that d

Res-G (w)=O, w=b dw

and so

Resf(g (w»g'(w) w=b

= A_I Res

g'(w)

w=b g(w)-g(b)

•

Also, using Theorem 11 we get R es

g'(w)

w=b g (w) -g (b)

=

R

es

dlog(g(w)-g(b»

dw

w=b

m,

and hence

Resf(g(w»g'(w) =mA_ 1 • w=b

On the other hand,

A_I = Resf(z), Z=Q

and therefore

Resf(g (w»g'(w)

w=b

= m Resf(z), Z=Q

which proves the theorem. THEoREM

14. If

Iimzf(z)=O, ~

then Resf(z) =0. Z=QO

THEOREM

15. If the/unction/has afinite number a/singularities zl' ...• zn then

100

(2)

Chapter 7

J(z)=

~ Res

f(~)

f(w) +Res . z-w w=o w(l-wz)

k=l W=Zk

Proof. Starting with the equality (Theorem 12) 1 (1 ~ ResJ(w)=Res~J -), II

k=l w=zk

w=o

w

w

and replacingJ(w) by J(w) we get

w-z

(3)

iRes f(w) + Res f(w) = Res f( k=l W=Zk

w-z

W=Z

w-z

w=o

~

) . w(l-wz)

On the other hand, we clearly have Res f(w) =J(z), w=z

w-z

and (2) follows from (3). REMARK. Theorems 1 and 2 mean that Res is a linear functional. Theorems 7, 8, 9 and 14 are standard theorems for the evaluation of residues, while Theorems 11 and 12 are also standard theorems related to the principle of the argument. Theorem 6 corresponds to partial integration and Theorem 13 to the change of variables. Theorem 13 is connected to Theorem 7 from 2.1.2 in Volwne 1 (pp. 15-16). Theorems 10 and 15 were proved by Cauchy who often used them.

At the end of this chapter Sohocki introduced a generalization of the residue, i.e. residues with respect to a line. Namely, if an analytic functionJis "discontinuous along a line I", then the residue with respect to I is defined as the integral ._1. .iJ(z)dz,

2:n,J c

where C is a closed contour which encloses the line /, whileJhas no other singularities in int C (except those on I). Sohocki stated that the most of the previous IS theorems hold for this residue, but he did not develop this idea further.

7.3. Two Formulas of Lagrange In short Chapters 3 and 4 of his dissertation Sohocki derived two known formulas of Lagrange by means of residues. The first formula concerns the expansion into series of inverse functions. That formula

101

Master's dissertation of 1. V. Sohocki

,'was proved in Volume 1 (4.4.2, pp. 77-78). In the remark on p. 79 it was stated that Lagrange proved formula (2) and a more general formula (3) without residues. It seems that Sohocki gave the first proof of those formulas by means of residues. The second Lagrange formula has an important role in the theory of symmetric functions. Sohocki claimed that there was no proof by residues of this formula. The result in question is as follows. Let u -:t; 0 be an arbitrary constant and let x H j{x)

be a polynomial of degree n divisible by x, but not by x2 nor by x-u. Furthermore, let the function g be defined by B, x

B2 x2

g (x) =-+-+ ... and let Xl'

...• Xn

Bm +xm

be the roots of the equation in x:

f(x)-x+u=O. Then the sum (the symmetric function) g (XI)

+ ... + g (xn )

is equal to the principal part of Laurent's expansion of the function Ul-+

+ 00 1 dk-'

L - - (f(u)g'(U))

k=ok! dU k -

1

in powers of u.

7.4. Continued Fractions In Chapter 5 of his Master's dissertation, Sohocki proved some formulas for continued fractions by means of residues. He proved, in a different way, some results of Cebysev [1], and obtained more general results than Roucbe [2] and Heine [3] who treated the same problem without residues. We give two main results obtained by Sohocki. 10 Let the function z H f (z) be regular for all z satisfying Izi > r, where r > 0 is a constant, and let z = 00 be a regular point, a removable singularity or a pole off Suppose that the function/can be represented by an infinite continued fraction

/(z)=%+!!.2 Q 2 q,+q2

+

102

Chapter 7

which is shortly written 2 f( z ) = %a, + -a... _an ... q, + q2+ qn+

Ifwe introduce the sequences (Pn) and (Qn) by the equalities: p)=%,

P 2 =%Ql+a),

P n+2 =Qn+)Pn+)+an+)Pn

(n= 1, 2, ... )

Q)=I,

Q2=ql>

Qn+2=Qn+)Qn+)+an +)Qn

(n= I, 2, ... )

then

Sohocki proved the following formula

where zl' ...•

zp

are all the singularities of the fraction Z H _1_ and q,,(z)

r

is the circle

{zllzl= R}, where R > 0 is sufficiently large. 2° If b

f(Z)dW=~~ .• J z-w q, + q2+

"

• ~ •••

q,,+

then

where the notations are the same as before. REFERENCES

1. P. L. Tchebychef: Sur les fractions continues. J. Math. Pures Appl. (2) 3(1858), 289-323.= Collected Papers ofP. L. Gbykv (Russian), t. 2, Moscow - Leningrad 1947, pp. 103-126. 2. E. Rouche: Memoire sur Ie developpement des fonctions en series ordonnes suivant les denominateurs des reduites d'une fraction continue. Journal de lEcole Imperiale Polytechnique, eahier 37, 21(1858), 1-34. 3. E. Heine: Mittheiling aber Kettenbrache. J. Reine Angew. Math. 67(1867), 315-326.

103

Master's dissertation of 1. V. Sohocki

7.5. Legendre's Polynomials In Chapter 6 of his Master's dissertation Sohocki considered Legendre's polynomials and applying the calculus of residues derived many of their properties. This is the first time that residues were applied to this class of special functions. We shall expose all the results of Sohocki obtained by residues.

As usual, Sohocki defined the Legendre polynomial Pn of degree n as the coefficient of t" in the Taylor expansion of the function t~(l_2xt+t2)-J/2

around t

= O. Hence, if

then +00

In + I v'1 - 2 xt + (2

I

P k (x)

Ik-fl-l,

k=O

and so (1)

and this was the main equality used by Sohocki in his proofs. 1° Ifwe apply the formula (see Theorem 6 from 7.2) dv

du

Res u-= -Resv1=0

dt

1=0

dt

and if we notice that

we get

n Res _1_ y'l - 2 XI + 12= Res y'1 - 2 xl + I 2 ~ 1=0

t n+ I

(~)

dt t n

1=0

1 d

= Res -

1-0 (n

--c::-----:::

- v' 1 - 2 xl + I 2. r:.

dt

i.e.

nRes-1-y'1-2xt+/2=Res-1_ 1=0

t n+ 1

1=0

t(l-x) t n+ 1 v'1-2xt+t2 '

104

Chapter 7

which implies Re s1- n (I +12 - 2 xI) 1=0

12 +Ix

YI-2xI+/ 2

1,,+1

= 0,

or (2 n - 1) x Res --::-!7='=I~~

1=0/"yl-2XI+/ 2

1 =0, 1=0/" IYI_2xI+/2

+(n-I)Res i.e.

nP" (x) - (2n -1) XP,,_I (x) + (n-I) P,,-2 (x)=O

(2)

(n= 1, 2, ... ).

Since from (1) follows Po (x) = 1,

(n= 1, 2, ... )

using (2) we can evaluate Pl(x), P 2(x), ... successively. Formula (2) is usually called Bonnet's formula.

2° Putting x = 1 into (1) we find 1

1

P,,(l)=Res--= 1; 1=0/,,+11+1

and putting x = -1 we get 1

1

P,,( -1)=Res--=( -I)". 1=0 I"+! 1 + 1

3° Let us determine the coefficients ao. al' ...• an such that P,,(x)=oo+o\ x+··· +a"x". Clearly 0k=

1

Res-P" (x)

x=O

(k=O, I, ... , n),

Xk+1

i.e.

= Res 1=0

(_1_ Res --;:--;-;-171~===:) x=ox k yl-2Ix+t 2 . 1,,+1

+1

lOS

Muter'J dissertation of J. V. Sohoc:ki

In order to evaluate the residue at x = 0, put

Vl- 2 tx+t 2 =...[l+ti+u. Then

and we get

(3)

1(2kk) Res

0k=-

2k

1=0

1.

t n - k +1 (1 +t 2)

k+2

The expansion of the function t H (1 + t 2

t-t clearly involves only even powers of t,

and from (3) we see that ok = 0 if n - k is odd. This leads to the important conclusion: The

function Pn is even if n is on even number, ond odd if n is on odd number. Return now to the coefficient n - k = 2m. Then Ok

=~ 2k

(2kk) Res 1=0

1 (2k)

= 2k k

1

°n=

1

=~ 2k

(2kk) Res _ _k-..!.. ----,-__ 1=0

(1 +t)

n-k ( m=-

2

'

k=O, 1, ...•

n we have m = 0 and so

(2 n -I)!!

n!

~-.

4° Ifwe integrate (1) with respect to x we get

J

2

tm+1

1 (2k+l)(2k+3)· .. (n+k-l) (-I)m 2m m!

2mk!m! =

Suppose that n - k is an even number and put

k+t 2m +1 (I +t 2) 2

=( _l)m(n+k-I)!!

In particular, for k

Ok.

1 Pn (x) dx=Res-

1=0 t n +1

J

1 dx v'1-2xt+t 2

= Res-1- (F(t) 1=0 t n+ 1

-~ vi 1-2 tx+ t

(2),

n).

106

Chapter 7

where F is an arbitrary differentiable function.

Let 1

F(/)=-, t

Then (4)

and we get

J

-~Res Z.!(~)=~Res~ dz

PII (x)dx=Res-z-= 1=0

n

t,1+ I

1-0

dt til

n

1-0

where we have used Theorem 6 from 7.2. Using (4), by Theorem 13 from 7.2. we obtain

(Z2-1)"dtdz 1 (Z2-1)" ~-=-Res -- ,

1 dz 1 Res--=-Res -1=0

and so

2" z-x z-x

t"dt

J

didt

2" z=x

(Z2-

1 )" . 1 Res - P" (x) dx=2"n z=.\

z-x

However 1 d" - I ZZ - 1 )" Res ( ---= - -..----. (x 2 -1)", Z=X

z-x

(n-l)!dx,,-I

and hence

implying

(S)

P (x)= __I_~_(x2_1)". "

2" n! dx n

Formula (S) is known as the Rodrigues formula. SO Introducing the notation

I - 2 xt + t 2 = T, we have

z-x

t" dt '

107

Master's dissertation of J. V. Sohocki

J

1 TI/2 '

P (x) = Res - - n ,=0 TI/2 tn+ I '

u=du

d J --P (x)=Res--,
d2 dx2

df

3

d2 U dt 2 -

-Pn(x)=Res---····, ,=0 T'/

2

x-f TJ/2

--=-,

fn-

1 TJ12'

(X-t)2 T'/2'

._- ----1-3--

1

Applying Theorem 6 from 7.2. we get dU Idu U Re s1- -d(2 1 - ) = -(n+ 1) Res-=n(n+ I) Res-n ,=0 t +I dt

,=0 t n dt

dt

,=0 tn+ I

=n(n+ 1) Res

1

,=0 t n +1 Vl-2xt+t2

,

i.e. 1 -d ( 12_ dU) =n(n+I)Pn(x). Res-

(6)

,=0 t n + 1 dt

dt

On the other hand, we have

Res_1_ ~ (/2 dU) = _ Res

1

+ 3 Res

= - Res

T

+ X2~p (x) + 3 Res

,=0 t n +I dt

,=0 tn-I TJ/2

df

,=0 tn-I T'/2

(X-t)2

,=0 t n - I T'/2

+ 2 Res~ ,=0 tn TJ/2 t 2 -2tx

,=0 tn-I T'/2

dx 2 n

+2x~P (x)-2Res dx

If (7)

n

T ,=0 tn-I T'/2

,2 - 21 x is replaced by T - 1, we easily get 1 -d (dU) d2 d2 d Res12 - =X2_P (X)--P (x)+2x-P (x). 2 n 1 ,=0 t + dt

dt

dx

n

dx2 n

dx n

Finally, from (6) and (7) follows

(I-x 2 ) Pn" (x) - 2 x Pn' (x) + n (n + I) Pn (x) =0; in other words, Legendre's polynomial Pn is a particular solution of the differential equation

(1-x2)y" -2xy' +n(n+ l)y=O, which is called Legendre's differential equation.

108

Chapter 7

Those were the main properties of Legendre's polynomials which Sohocki proved by the calculus of residues. Sohocki himself emphasized in the preface that the results are not new. Nevertheless they are important since Sohocki was the first to apply the calculus of residues to this particular class of special functions. Later on Laurent [1] (see also [2]) also applied residues to Legendre's polynomials. and noted that in this way it is much simpler to obtain their properties. earlier proved by Heine. without the use of residues. but there is no mention of Sohocki. Also. Jordan in his book [3] developed the theory of Legendre's polynomials using residues. without mentioning Sohocki. REFERENCES

1. H. Laurent: Memoire surlesfonctions de Legendre. J. Math. Pures Appl. (3) 1(1875). 373-398. 2. H. Laurent: Traite d'Analyse. t. 5, Paris 1890, pp. 188-198. 3. C. Jordan: Cours d'Analyse de l'Ecole Polytechnique, t. 2, troisieme ed. 1913 (nouveau tirage 1959), pp. 344-349.

7.6. Expansion of a Function by Means of Continued Fractions Suppose that the function z~g(z) can be, for sufficiently large !zl. expanded into Laurent's series in powers of z. and also that for sufficiently large !zl we have the continued fraction expansion a. g ( z ) =qo+- a2 ... q. + q2+

where qo, ql' q2' ...• al' a2' ... are polynomials. Let the functions PI' Ql' P2, Q2' ... be defined by

(n=2. 3•... ) and define the sequence (Rn) by

(11= 1, 2•... ) The functions R l • R 2• .•. were studied by Cebyiev [1]. In his Master's dissertation Sohocki considers the problem of expanding a function/into series with respect to Ql' Q2' ••• and also with respect to Rl • R2• ..•. In solving this problem the central role is played by the following formulas. claimed to

be new

:L Res HI

qn(z)-qn(w) Cm(W) Rm(w) Qn (w)=O

z- W

(m1=n).

where en is an arbitrary polynomial with degree lower than the degree of qn and residues

109

Master's dissertation of J. V. Sohocki

are taken for all the singularities of Rm and R", respectively. Suppose now that the function/can be expanded as follows

/(x)

=

+00

2: Pk (x) Qk (x),

k=l

where PrJ is a polynomial with degree lower than the degree of q", then the coefficients PrJ are given by

p,,= (_1),,-1 LRes

°1°2 ... °"

z

q,,(x)-q,,(z) x-z

Q,,(z)g(z)/(z)

and residues are taken for all the singularities of g. Also, if

/ (x) =

+00

L Pk (x) Rk (x),

k=l

then the coefficient PrJ is defined by

REFERENCE

1. P. L. Ceb~ev: Expansion of a Function into Series by Means of Continuous Fractions (Russian). Collected Papers ofP. L. GbyS'ev, t. 2 (Russian). Moscow-Leningrad 1947.

Chapter 8

On the Principal and the Generalized Value of Improper Integrals DRAGAN S. DIMI1ROVSKI,

University of Skopje

8.1. Substitution in Complex Integrals The following theorem justifies the change of variable in Cauchy's principal value of an improper integral between arbitrary, but finite, limits. See [1] and [2], and also Volume 1, p.187. THEOREM

l. Suppose that:

10 The function / is analytic in the whole plane where it may have a finite number 0/ singularities~

20 On the real axis the/unction/may have only simple poles cI' ... , cn which belong to the open interval (a, b).

Then the principal value given by the equality

v.p.

0/ the improper integral all between a and b exists and is

J f(x)dx=(b-a)v.p. J+oo/ (aX+b) l+x b

dx

0

a

. =

2 n

k=l

where a k

1 (l+x)l

b-c =__ k

ck -a

(k

l nl ) Resf(z)+ (cg Uk+'

=1,

Z=Ck

2 Res ( (a-b)

]ogz

(1 +Z)l

~R

(az+b))

/ -- • 1 +z

... , n).

We only give an outline of the proof. By the definition of the principal value we have b

(1)

Ck+l- r

cl-r

V'P.Jf(X)dx=!!!(J f(X)dx+kt J a

a

110

f(x)dx+

J b

f(X)dx).

111

On the Principal and Generalized Value oflmproper Integrals

where r > o. Each integral on the right hand side of (n is proper, and we can change the variable as follows at+b l+t

(2)

X=-- --

b-x x-a

1=--.

The transformation (2) maps the simple pole ck off into the simple pole a"

of the function 1 1-+

y

=b - e"

e,,-a

f( 1+1

al +b ). The point a is

mapped into +00, the point b into 0, and real numbers outside the segment [a, b) are mapped onto the negative t-axis.

"

We obtain the result by integrating the function z/-+(a-b) logz f(az+b) (1 +Z)2

1 +z

along the contour shown on Fig. 8.1. where C={zllzl=R}, r={zllzl=r}, k =1, ... , n. ExAMPLE

1. Let 0 < a < c < b < +00 and consider the principal value b

V.P.J~dx. x-c a

Since b-c log ak·Resf(z)=c log-, z=c c-a "" Res {(a-b) logz az+b } =b-a-c n i, £.. (1 +Z)2 (a-c)z+b-c

C',..R

we have

J b

v.p.

a

x b-c -dx=b-a+clog-. x-c c-a

Figure 8.1.

112

ChapterS

EXAMPLE 2. If 0 < 0 < c) < C2 < C3 < b < +00, we easily obtain

J

1

b

V.p.

(X-C 1) (X-C2) (X-C3)

;=1

a ·

b-c;

1

3

dx-' - L.,

IT ( 3

)

l oc._Q g-.

~-~

I

k=1

k#o;

REFERENCES 1. D. S. Dimitrovski, D. D. Adamovic: Sur quelques formules du colcul des residus. Mat. Vesnik 1(16XI964),113-117. 2. D. S. Dimitrovski, M. Rajovic: Sur Ie valeur principale de I'integrale impropre. IIl-ieme note. Ann. Fac. Sci. Univ. Skopje 2S-26(1975n6), 41-52.

8.2. The Principal Value for Higher Order Poles It was shown in [1] that the method given in [2] can be applied to the Cauchy's principal value of an improper integral between finite limits not only when the integrand has simple poles on (a, b), but also when those poles are of higher orders. I~

fact, integrating the function b-z

zl-+f(z) logZ-Q

along the contour shown on Fig. 8.2. we obtain the following THEOREM

1. Suppose that:

1° The function / 0/singularities;

is analytic in the complex plane where it may have a finite number

2° On the real axis the/unctionfhas poles Xl' which belong to the open interval (a, b);

3° The Laurent expansion 0/ / around Xj is +0.

/(z)

= LA,.,(Z-XJ)II

(j

=I,

... , k)

n=-CIO

Then the Cauchy's principal value b

v.p.

Jf(x) dx a

exists if and only if k

~A-2P.J=O J=1

(v=I,2, ... ,p),

••. 'Xk

%rder n l ,

•••

,n", respectively,

113


where

and 2pj = mj if mj is even, and 2pj = mj - 1 if mj is odd. y

x

Figure 8.2.

In that case we have

+

2:

C'-...[a.

hJ

b-z Res J(z) log finite distance

where B-IJ are residues of the function z by

EXAMPLE l. The principal value

J 8

v.p.

o

x2-x-l

(x-2) (x-3)3

z-a

~ f(z)

log (b - z) / (z - a) at Xj and are given

114

Chapter 8

exists, since 1

5

I(x)=--+ x-2 (X-3)3

1

+-(x-3)

and there are no tenns with even indices. Applying the above fonnula we conclude that the principal value is v.p. I -log 5 -2 log 3 + 8/45. ExAMPLE 2. The principal value

J 8

v.p.

1

(x-2) (X-3)2

o

dx

does not exist, since there are even tenns in the Laurent expansion. EXAMPLE 3.

If a < c < b, n e N the principal value

J b

v.p.

1

(x_c)2n

a

dx

does not exist. EXAMPLE 4.

If a < Ck < b, the principal value

exists. REFERENCES

1. K. AdZiev, D. S. Dirnitrovski: On the Principal Value of an Improper Integral VI (Macedonian). Bull. Soc. Math. Phys. Macedonie 31-32(1981-82),3742. 2. H. Behnke, F. Sonuner: Theorie der analytischen Funktionen einer komplexen Veranderlichen. Zweite Auflage. Berlin-GOtingen-Heidelberg 1962.

8.3. The Principal Value in the Case when the Limits of Integration are Singular Points If a and b are poles off, we define the principal value of the improper integral as b

def

b-r

v.P.!f(x)dx=Iim! f(x)dx. a

r-+O a+r

The folowing theorem was proved in [I).

115

On the Principal and Generalized Value ofhnproper Integrals

THEOREM

l. Suppose that:

1° The function f is analytic in the complex plane where it can have a finite number of singularities;

2° The function f has poles at z = a and z = b of orders k and n respectively where a, b e R and a < b; 3° The functionfhas simple poles at z = xv' v = 1, ... , m which lie inside the open interval (a, b). Then the principal value

.~. p.

Jf(x)dx b

"

exists if and only if

2° The principal parts of the Laurent expansions around a and b have the same coefficients, i. e. they have the form: "

B

~(z_~)V

and

In that case

.!

b

- v. P

/(x)dx=

Res (IOgz-a /(z)) + Res (/(z) In b-z

Z=~

z-a) b-z

II

B + 2 {'+ b-a

where 21 = n

if n

B

•

3 (b-a)3

+ ... +

} B 1 2 (21_1)(b_a)21-1

+

m

Res /(z)· .~lZ=X.

1

x.-a og -. b-x.

is even and 21 = n - 1 if n is odd.

We list a number of corollaries and examples. CoROLLARY

l. If one of the points a, b is a regular point of f, and the other is singular,

J b

v.p. f(x)dx cannot exist.

" CoROLLARY 2.

If the only singularities offon the real axis are poles at z = a and z = b of the same order n, and if other conditions of Theorem 1 are satisfied, then the principal value exists and

116

ChapterS b

-v.p.!I(X)dX= a

~

kanat. dal}

Res (/(Z)IOgz-a) + Res (/(z) logz-a)

+ 2 { _B2 + b-a

z -b

B4 3 (b-a)3

+

B' + 5 (b-a)'

Z= 00

... +

z-b

B }• 21 (2/-1)(a-b)21-1

where 21 = n (n even), 21 = n - 1 (n odd). COROLLARY 3.

If z = a and z = b are simple poles of/such that

Res/(z) = Res/(z) z=a

z=b

and if/has no other singularities inside [a, b), then the principal value exists and b

-v.p. !I(X) dx= a

~

finite '

Res

(I (z) log

z-a) b-z

+ Res (/(z) 10gZ -a) z=oo

distance

EXAMPLE 1. The principal value

.! 2

v.P

1

1 dx (x-l)(x-2)

does not exist, since

Res/(z) = -1 "eRes/(z) = + 1.

z=l

z=2

EXAMPLE 2. The integral

J 2

V.p.

1

x-3/2 dx (x-l)(x-2)

exists, since Res/(z) = Res fez) = 1/2, z=l

z=2

and its value is O. EXAMPLE 3. The principal value

!( b

v.p.

-A+ -B) - [ f(x)-f(a)-(x-a) feb) - f(a) +c ] dx

a

exists.

x-a x-b

b-a

b-z

.

117

On the Principal and Generalized Value of Improper Integrals

EXAMPLE 4. IfF and t:p are analytic functions and if

L

Ak- ])(F(z)-F(a)- F(b)-F(a) (tp(z)-tp(a» ) f(z)= ( n [Ak --+k=O (z - a)k (z - b)k tp (b) - tp (a)

the principal value b

j f(x)

v.p.

dx

a

exists.

EXAMPLE 5. The principal value

J 1

v.p.

o

4 X 3 -4.x2+1 [eX-(e-l)x-l+c]dx x 2 (x-l)2

exists. EXAMPLE 6. We have 4

1 1 v.p. {[- + - - + •

1

x-I

1] 1 1 2 2 3 dx =-]og2+-]og---arctg-. x-4 (1 +x2)(x-2) 2 10 17 5 5

REFERENCE

1. D. S. Dimitrovski, M. Rajovic: Sur la valeur principale de /'integrale imprapre, V-ieme note. Bull. Soc. Math. Phys. Macedonie 26(1976),19-24.

8.4. Generalized Value of an Improper Integral with Infinite Limits Suppose that the functionfhas a finite number of singularities Xl' The value of the improper integral is defined as +00

%1-'11

v. jf(x)dx= -00

lim

%3-'31

j

on the real axis.

%1+'12

(jf(x)dx+ j f(x)dx+

'iJc->O Rf<-++oo i=l, 2, ••• ,m; k=I,2 -R1

+

... , Xm

f(x)dx+ ... +

%1+'12

xm-'m,l

J

R2

f(x) dx+

J

f(x)dx)=

Xm+'m.2

if this limit, which depends on 2m + 2 variables, exists. This limit should be interpreted as the standard "e-b" definition, i.e. the value of the above integral isA if

118

ChapterS

Instead of the above definition which involves one multiple limit, we can define various generalized values by considering iterated limits in some prescribed order. DEFINITION

1. Let f be a continuous function on the real segment [a, b] except at the

points c l ,

,c". Then on each ofthe subintervals

...

[a+eUc1-rU]' [cl+r12,c2-r21]' [c2+r22,c3-r3I1,···, ... , [Ck+ rk2, Ck+l-rk+l.l],"" [c,,+r,,2' b-e21. the integrals Ci+l-ri+hl

J

f(x)dx.

CH ri2

exist. As rik, &k-'O (i

= 1, ... , n; k = 1, 2) suppose that there exists a relation oftheform

(L)

iii = q1/(1i2)' i
(Q)

'".I,.

or

or

=Z(Ii,k)'

i, j

< n.

Generalized value of the improper integral. denoted by v. g. (L, P, Q)

oJ.

is defined as the following iterated limit. taking into account the

conditions (L). (P) or (Q): b

v. g.

b

J f(x) dx = (L, P, Q; Jf(x) dx

a

J. oJ or

d~

=

a

r~~ (,j~ .... (r~:!' [ J f(x) dx + t~1

J

,,-1 ci+l-ri+hl

cl-rU

a+BI

ci+rtz

b- 82

+

J

c,,+r"2

f(x)dx

f(x) dx +

l·· .)).

119

On the Principal and Generalized Value ofImproper Integrals

The above generalized value depends on the path (L, P, Q) along which ril and ri2 tend to zero, and can therefore be multiform. In physics we come across such "integrals along trajectories" . If the conditions (L, P, Q) reduce to r il = r,"2 = rj l = rl2 = r and if the iterated limit from the above definition is interpreted as the standard "e-t5" definition, then the generalized value reduces to the principal value ofthe improper integral.

DEFINITION 2.

Definition 1 is easily extended to the case when the limits are infinite. Suppose that the function f is continuous on (--00, +00) with the same exceptions as in Definition 1. Ifwe take a + &1 = -Rl' b - 62 = R2, then we define the generalized value as DEFINITION 3.

+00

J f(x) dx

v.g.

+00

= (L, P, Q) •

J f(x) dx

-00

-00

def

'ik- O

'jk ..... O

+:t J

R.--~-oo

1 ci+l-'i+l.l

R 2-->+00

J

f(x)dx+

f(x)dx]' .. ))).

Cn+'n2

We give a few examples to illustrate the above defnition. EXAMPLE 1. Let -00 < a < c < b < +00. The value

6, = 62 (condition Q) the principal value

b

v.p.

J~=IOg Ib-c ,. x-c

a

c-a

exists. Ifwe introduce the condition

then the generalized value

J b

v.p.

a

exists.

-dx -= • x-c (L)

J b

a

I

f(x)dx+

-RI

R2

Ci+'i2

does not exist. If

J

ai-'ll

lim ( ... lim ... (lim ... (lim ... [

dx b-c --=Iog - I+logK x-c c-a

120

ChapterS

ExAMPLE 2.

The value of the integral

I

+00

v.

R

I-x dx=Iim ;xl

o

......0

R-++ 00

j I-x dx= lim (--.!..-IOgR+.!.-+log 8) R ;xl

B

......0

8

R.....+ 00

does not exist. For R = lis we have the principal value

I

1)

lIB

v.p.

I-x ( 210g8+--dx=lim x2

•

......o

8

which does not exist. Ifwe introduce the condition R (8) =Cee 1/'(8-+O. R-++co)

(Q)

then the generalized value v.g

.

+j"I-X • j+"I-X --dx= --dx x (Q) ;xl

2

o

0

-lim

......o(R(8)-Cee1/.)

f

•

1/. I-x --dx--IoIC• x2

exists for any posivite number C.

ExAMPLE 3. The integral

+..

v.

I

......0

o

R-++.. •

does not exist, but for R

f

+ ..

v.p.

J R

log x

-x- dx = lim

R-++_

=lis we have the principal value

logx -x-dx=O.

o

Ifwe introduce a condition of the fonn (Q)

I2.>

log x l . - - dx = - lIm (lor R_Io x 2 ......0

R=/(s)

for instance, if (log R)2-(Iog 8)2 - 2 C

121


then we get the generalized value

ExAMPLE 4. The integral

J

J

+ ..

v.

dx x (1 + x 2 )

- ..

=

lim {

Bj-+O R~+ ..

J R2

-BI

I(x) dx+

-RI

I(x) dx}

B2

log

82

VI +8"2 2 does not exist.

For 8\

= '2. R\ =R2 we have the principal value

J -.

+ ..

v.p.

and for &\

dx

---=logI=O x(1 +X2)

= k'2. R\ =R2 we have the generalized value

+Joo

v.g.

-

dx x (1 +X2)

.

•

(Q)

-

J+"

.

dx x (1 + X2)

logk.

ExAMPLES. We have

J J 2

v.p.

o

2

v.p.

o

J

(x-I/2) (x-3/2) dx=O. x (x-I) (x-2) 2x3+x-I dx-nI4. x (x-I) (X2 + 1)

n/2

v.

o

dtp = _I_lim a-costp Vl-a 2 .1-+0 az-+O

J

n/2

v.p.

o

{IOgl~I+IOgl VI+a+ vr+a-vr=al}. Vl-a

I

82

I

dtp 1 V1+a-VI-a • ---log a-COs tp V l-a2 VI +a+ VI-a

and if introduce the condition

}

122

ChapterS

we obtain the generalized value

f

*f

n/2

v.g.

n/2

- -dtp -=

a-coslp (Q)

o

0

dip

a-cos Ip

=

I

VI-cz2

I

I

log A VI +a- VI-a •

VI +a+ VI-a

The following general theorem is valid. 1. Suppose that f is an analytic function in the upper half plane {zlImz ~ O} with the properties: THEoREM

1° The function f can have a finite number ofsingularities in the region {zlImz > O} ~ 2° On the real axis the function f can have poles Xl' ... 'Xm of order nl' ... , nm respectively~

3° max \Zj{z)1

~

0 as z ~ +00, where 0 S arg z S frand

\zI = r.

Then the generalized value of the improper integral b

b

v.g. ff(X)dx= a

* ff(x)dx

(L,P,Q)

a

depends only on the limits

lim ( ... lim

rlk-->4J

1=1,2, •..

t

Rk-++ .. m; k=1.2

where BUe is the coefficient of{z - x.)-k in the Laurent expansion off around Xv which is convergent in the annulus

{z Imin 'vk<1 Z-Xv 1<1 xv-a, I}. where ap is the singularity nearest to xv'

In order to prove this theorem we use the contour shown on Fig. 8.4. which has jumps along the lines {zl Rez =xk } (k =1, ... ,m) and {zl Rez =O} in order to describe different approaches of ril and r,7. to zero. Starting with the Laurent expansions around the poles xv' using the Cauchy residue theorem, and leaving out the technical details, we arrive at the formula

123

On the Principal and Generalized Value of Improper Integrals +~

V.g.

+~

J f(x) dx= ,lim (L, P, Q) • J f(x) dx=

-~

__

I~

Rk-+-

'I~ Rk-+_ i-l.2, ... ,m;k-l.2

(J

1m z>O

-

~

J dx +

RI

=2xi2Res f(z)+xi

XI

J +'12

f dx + ... +

J

J dx) ... )

xm+'m2

m

2 Resf(z)-

,=1 z=x,

Res J(z) log!:!! + lim ( .. lim r'l

.=1 z=X,

R2

X2-'21

XI-'ll

= lim (... lim ...

'I~

~ ~

]J"k

,=1 k-2 k-l

[< r_I)k + _r I-J) k- 1

k- 1

'1'2

which proves the theorem. This theorem reduces the question of the existence of the generalized (and the principal) value of an improper integral to the question of evaluating limits of a rational and a logarithmic function depending on 2m variables. y

x

Figure 8.4.

We give some corollaries and examples. THEOREM 2. If the only singularities off on the real axis are simple poles, while the other suppositions of Theorem 1 remain unchanged, we have

124

Chapter 8

V.g.

.J+OO

f(x)dx=2niZResf(z)+ni 1m (z»o

_00

- Iim...

m Z Resf(z)-

._1 z=x.

(lim . .. 1og n (-r'2 )z=x.

rvfc-l'O .-1,2, ...• III,· k-l,2

Res/(z)

m

• -1

) ...• )

r'l

Since the principal values are the most important, we give a few results in connection with them. The first one is well known. THEOREM 3. If the function f is analytic in the region {zl Imz ~ O}, Xl' ... ,

if it has simple poles

xm on the real axis, and ajinite number ofsingularities in the region {zl Imz > O},

and if

maxlzf(z>l-+ 0, as z -+ 00 then the principal value of the improper integral exists, and

J f(x)dx=2ni ZI~~!{(z)+ni '~I ~e!f(z). m

+00

v.p.

-GO

THEOREM 4. The principal value of the improper integral depends only on the even coeffiCients B v. 2k co"esponding to the terms (z - X .)-2k in the Laurent expansion around

xv· THEOREM 5. Iff has poles of arbitrary order on the real axis, the principal value of the improper integral exists if and only if m

PI

~

2 B.2 =O, Z B.4 =O, ... , Z B.. 2k =O .

• =1

.=

1

.=1

where B lie are dejined in Theorem 1. THEOREM 6. Iff has only one pole of even order on the real axis, the principal value

cannot exist. THEOREM 7. If the function f has only poles of order s; 5 and

Theorem 1 are fUlfilled, the principal value exists if and only if

if the other conditions of

125


ExAMPLE 6. We have

J

+GO

v.g. •

eix :If sin a - - dx=--a2 _x2 a

lim

ri--"' i=l. 2. 3.4

-GO

which implies

f

+GO

v.p.

cosx :If sin a --dx=--, a 2_x2 a

-00

J

+00

v.p.

sin x --dx=O. U 2 _X2

-00

ExAMPLE 7. We have

-

.

lIm

ri--"' i-=l, 2. 3,4

[rl e r4] )og-+-)og-, i

r2

2

which implies

J +00

v.p.

cos x :If dx=--(e-1+sinl), x (x-l)(l +X2) 2

-00 +00

v.p.

J

sin x :If - - - - - dx= - (e- 1 + cos 1-2). 2

x (x-I)(I +X2)

-GO

ExAMPLE 8. Ifg is an entire function and maxlzg(z)l-+ 0 as z -+ 00, 1m z > 0 then

J[ +GO

v.p.

1

(X_I)2

-GO

-

1

(x + I)

2

+g

(X)] dx=O.

r,

126

Chapter 8

EXAMPLE 9. We have

*J

+00

v.g.

-00

=

(

Ii~ ... lim log

Bi~O;

,= 1,

... , n

81 )A (8;:;... [(8 2 3)B (828;~I)N] +

where A=

1 (a-b)(a-c) .. . (a-n)

, B=

1 (b-a)(b-c) ... (b-n)

... ,N=

• 1

(n-a)(n-b) (n-c) ... (n-m)

For the principal value we obtain +00

v.p.

J

--------0.

(x-a) (x-b) .• ·(x-n)

-00

EXAMPLE 10. The value of the integral

J

+00

v.

1 I'1m (R 22 - R 12) xdx= lim x21R2 =RJr++oo 2 -Rl 2R2~+OO Rl~+OO

-00

does not exist. However, if we take

we get

*J

+00

v.g.

(Q)

-00

xdx- lim (R2 -R1)(R2+R1)= lim _1_(2R, Rio R2~+ 00 RI~+ 00 2 Rl

EXAMPLE 11. The value of the integral

+~)-1. Rl

.

127


+00

J

dx

n=#:1

(x-c)" '

-00

does not exist, since by defmition,

J

+00

v.

dx (X-c)"

-00

=

lim 81->0 82->0

_1_ [ _ _ 1__ I-n (-e 1),,-1

Rhr+ OO =

lim _1_ [ _ _1__ 8,....0 I-n

(-e 1)"-1

However, if we take (L):

ez = e1

,,-I

y':;-I-+..... k-.e-:1":---:"1

for odd n, then

J+oo ~=lim

v.g.

(x-c)"

8,....0 (L)

-OD

.J+oo ~=~ -oa

(X-c)"

I-n'

ExAMPLE 12. If -00 < a < -c < c < b < +00, then the value

.

I I' I I I

I

J,. b

v

dx- - l o 1 g (a+c)(b-c) +1- 1m og e2 e, -x 2 -c2 2c (b+c)(a-c) 2c 8i->O e1 e4

does not exist. However, the principal value (for &1

=~ =&:l = &4) is

b

vp

••

J~=-I-IOgl (a+c)(b-c) I. x 2 -c2

,.

2c

(b+c)(a-c)

ExAMPLE 13. Let -00 < A < a < b < C < ... < n < B < +00. The principal value

J B

v.p.

A

dx

(x-a)(x-b) (x-c)· •• (x-n)

128

chapterS

exists and is equal to B

V.p. J/(X)dX=IOa[/;=: rl~=: IP·"I~=:n· A

where a-

1

1

..... ,.. - - - - - - - - - - - . (a-b)(a-c)· .. (a-n) (n-a)(n-b)(n-c) .. ·(n-m)

REFERENCE 1. D. S. Dimitrovski: On the Generalized and Principal Value of an Impraper Integral I (Macedonian). Ann. Fac. Sci. Univ. Skopje 21(1971),5-12.

8.5. Generalized Value of an Improper Integral Between Finite Limits Using the definition from the previous section, we can prove the following general theorem. THEOREM 1. Suppose that

10 The function f is analytic in the extended plane, and it can have a finite number of isolated singularities outside the segment [a, b] ofthe real axis; 2 0 On the real open segment (a, b) the function f can have poles c l , ... 'Cn of order

mI' ... ,mn respectively, thefunctionfbeing regular at z = a and z = b.

Then the generalized value of the improper integral is b

v.g.

b

J f(x)dx=v·g·(L,P,Q;Jf(x)dx

a

a

=

2:

Res (f(z)IOgb-Z) -Res (f(z) IOgb-Z) z-a

C"-.[a,b]

+lim

r·k ..... O

1

( ... lim ... ((-ni+IOg~)(i b_ 1 ••)

.=

'2.

n.

m

'"

'"

+--L. L.

{2 . b 11: ,.

. (1

2.

B-k,. k-l

I

[e(k-l)in/2_1

.

k-I

'2v

1)] + __ «_l)k-l

+Zl-k _ _ _ _ _ ,k-I ,k-I

1.

B

-k,. + -k,.

l-k

211:i.=lk=2

where

z- a

Z=OO

,k-I

1.

e(k-I) in-e(k-I) in/2

+ __-----:~--k-I 'I.

129


1° b_ 1 v =Resf(z) •

..-C"

{

(v= 1, 2, ... , n);

b-Z)

2° B_1.v=Re f(z)log-z=c. z-a

(v= 1, 2, ... , n);

3° rl y ' r2 y are the radii of the semicircles around c y which tend to zero depending on the conditions (L, P, Q); 4° the function b-z z 1-+ log-:--:z-a

is determined as follows: if z =X E R it becomes the real function x ~ log(b -x)/(x-a). and the argument of the logarithm is 0 on the positive x-axis and 2m below the x-axis. The proof of this theorem is obtained by integrating the function

b-z z H f(z)log-z-a along the contour shown on Fig. 8.5.

x

Figure 8.5.

130

Chapter 8

We list a few corollaries. 1. Ifall the poles on (a, b) are simple, then the generalized value is

f f(x)dx= "" b

v.g.

b-z b-z Resf(z)log---Resf(z)log--

L.,

a

z=oo

Z-Q

~[a,bl

Z-Q

+ lim ( ... lim (-ni+lo g !:!!..) ' V2.....o

'VI.....o

r V2

.=i

b_ 1 •• ) . 1

2. If all the poles on (a, b) are simple, the principal value is b

v.p. ff(x)dx= a

2:

C,[a,bl

Resf(z) log b-z - Res fez) log b-z Z-Q

Z=oo

Z-Q

-ni

n

L

Resf(z).

,=1 z-c"

3. Iffis regular on [a, b), we obtain the known formula

f

b

f(x)dx=

a

""

L.,

C,[a bl

b-z b-z ResfCz)log---Resf(z)log-. Z-Q

z=oo

Z-Q

4. If all the poles on (a, b) are simple, and ifwe introduce the condition (L)

rvl = krv2 +o(rv2 )

we obtain b

b

v.g. J f(x) dx=v.g. (L;Jf(x)dx a

a

+

""

L.,

C,[a,bl

b-z

b-z

Resf(z) l o g - - Resf(z) log-- + (-n i + logk) Z-Q

Z=oo

Z-Q

2:n

5. Ifall the poles on (a, b) are simple, and if fez) = 0(1 / Z2) as z ~ 00, then A

B

Z2

Z3

f(z)=-+-+"

.

and

b-z

Resf(z) log--=O %=00

Z-Q

Resf(z).

.=1 z=c.

131


and introducing the condition (L)

'iy

= krZy

we obtain b

v.g.

b

JJ(x) dx=v.g. (L;J J(x) dx a

a

L

b z

ResJ(z) log--- +(logk-~ i) z-a

C'\..[a,b)

L" ResJ(z).

.=1 Z=C.

b

V.g. j J(x)dx= a

L

ResJ(z) log b-z _ ResJ(z)iog b-z z-a

C'\..[a,b)

z-a

z=oo

7. If all the radii are equal, i.e. r 1 v = r2 v = r for all v = 1, ... , n, then b

v.g. j J(x) dx= a

L

ResJ(z) log b-z - ResJ(z) log b-z z-a

C'\..[a, b)

.

z-a

Z=oo

m. (_l)k-l_l

+ !~ k~2 (1 -

k) rk-1

-'lE

i

i ResJ(z)

.=1 Z=Cv

[h_ k • 1 + h_k • 2 + ... + h_k •n ].

8. The limit in 7. exists if b_z P. 1 + h-z P. 2 + ... + b_ 2 P. n = 0 for p

= 1, 2, ... , n.

In that case the pricipal value is b

v.p. jJ(x)dx= a

L

C'\..[a.b)

ResJ(z) log b-z -ResJ(z) log b-z z-a

Z=oo

z-a

-'lEi

i

ResJ(z).

v=1 Z=Cv

If zZJ(z) ~ M as z ~ 00 and if/is regular on [a, hI, we obtain the following equality

132

chapterS b

f l(X) dx =

a

L:

C'\.[a,bJ

Res/(z) log b -z . z-a

REMARK. The results of this section are taken from the M. A. thesis of StojQ: Mitevski, which has not been published.

Chapter 9

Applications of the Calculus of Residues to Numerical Evaluation of Integrals DOBRILO D. TOSIC,

University of Beograd

Cauchy's calculus of residues can be applied to numerical evaluation of certain classes of definite integrals, those which cannot be evaluated by standard methods of numerical integration. Difficulties arise when the poles of the integrand are near the integration interval ([1)- [3)) and also when the integrand is a highly oscillating function ([4)- [10)), particularly if its singularities are near the integration interval ([ 11)). We shall give a few illustrations of the use of the calculus of residues. 9.1. Consider the integral 1

J~dx x2+a2

1=

-1

where a is a very small number. Since a is small, the integrand takes a large value in a neighbourhood of x = 0 and so the standard methods of numerical integration (Newton - Coates' formulas, Gauss Legendre's method, Romberg's method, etc.) cannot be applied. Start with the contour integral

Ic=

f --dz,

y

ez

Z2+

C

a2

where C is the contour consisting of the real segment [-1, 1) and the upper semicircle

r = {zllzl= 1,

Imz ~ O}. The integrand has two simple poles zl = ia and z2 = -ia, and only zl lies inside C.

x Figure 9.1.

133

134

Chapter 9

By Cauchy's residue theorem we have

1C= I + 1

=

l'

2 70. Res (-eZ-)

=

z2+a2

z=la

2 7 t.leia :rr: ia , - . =-e 21a

a

where

Hence,

. a-I. I = -:rr: eia - 1:rr: l' = cos a + 1• - :rr: SID a

a

a

l'

Since IE R, the expression ';cos a is its approximate value, and the correction is contained in Re I y . Notice that the pole ia is sufficiently distant from the semicircle r. which is the essence of this residue method. In order to evaluate the approximate value of the integral, we use the expansion

Since

the integral I y becomes 11' = i

j "e-16 eel6 (I -

a2 r

i26

o

+00 ~

"IF -

(-

k=O

+i

=and so

+00

2:

k=O

I)k a 2k

j

+ a4 e-i46 -

eeos 6

••• )

dO

sin (sin 0 - (2 k + I) 0) dO

0

(_I)ka 2k j

eeos 6

cos (sinO-(2k+ I)O)dO

0

+00

+00

k=O

k=O

2: (- I)k Ik a2k + i 2: (- I)k Jk a2k

Applications of the Calculus of Residues to Numerical Evaluation of Integrals

135

(1)

The integrals Ik can be evaluated by the Gauss - Legendre method and we get

J"

e"0s 8 sin

(sin 0 - 0) dO =

J"

e"0l 8 sin

(sin 0 - 3 0) dO =

10 = o II =

o

-

0.9716595188,

-

1.582397407.

Therefore the expansion (1) has a form suitable for the numerical evaluation of I: 1

I=g(a)= J~dx 2

(2)

-1

x 2 +a

=~- 0.9716595188-~a+ 1.582397404a2.+··. a

2

For instance, for a= 1~, we get 1=31414.9547193153, and all the figures are

correct.

Notice that (2) is the Laurent expansion of the function a H g(a) around a = o. The function g has a simple pole at that point. 9.2. Let us evaluate an approximate value of the integral

Of course, we aim at a highly precise approximate value. Let

and

J--dx +OO eix

o

1 +Xl

•

In order to evaluate this integral we start with the contour integral

136

Chapter 9

1=

f --dz, eiz

1 +z'

c

where the contour C consists of the real segment [0, R] and

y

the quartercircle {zllzl= R, Re z ~ 0, Imz ~ O}, where R > is large enough; see Fig. 9.2. The integrand has only a simple

B

°

pole

Zo

=e l1
e1z

eizo

1 + Z3

3 Zo 2

B1 =Res ( - - = lim (z-zo)--=1 + Z3

z=zo

Z--"Zo

o

' -1) e-V3/2. = - -1 (1 + I. V_ r) 3 ( cos -1 + I.sm 6 2 2

x

Figure 9.2.

By Cauchy's residue theorem we have eix l+x3

R

1= J --dx+ o

J AB

eiz l+z3

JO e- X --idx=2~iBI' l-ix 3

--dz+

R

If R -+ +00, then it is easily shown that the integral alongAB tends to 0, and so

f

+co

II

(1)

=

o

cosx . )--dx=Re (2nIB I l+x'

n

n) 6

2 -V3/2 cos (1 =-e --- 3

2

J --e-xdx +co

X'

0

l+x'

J --e +co

x,

o

1 +x,

-x dx.

Therefore, the evaluation of II is reduced to the evaluation of the integral of the x3 function x 1-4 - - 6 e- x which is positive and is not oscillatory. If we integrate this l+x function over the segment [0, 15], we obtain a value correct up to 10 decimals, since

Applying the Gauss - Legendre method we get

137

Applications of the Calculus of Residues to Numerical Evaluation ofIntegrals

J--eIS

o

x,

1+ x6

x

dx=0.171811227 ...

Hence, from (I) follows II

= 0.70888800 ... where all the figures are correct.

REMARK. The following, more general formula, is also true:

wherep <0.

9.3. A direct evaluation of the integral +00 Ip= _x-cospxdx (p>O) X2+ 1 o is rather complicated because the integrand is highly oscillatory for large p, and the

J

amplitude x / (x 2 + 1) decreases slowly.

y

Since the function Z 1-+ _ z_ Z2+

1

eipz

has a pole at z = i, we cannot use the contour shown on Fig. 9.2, but we take the contour from Fig. 9.3. which does not enclose the poles ± i. Hence, the integral Ip transforms into the integral +00 pxV2 - S. lpxv'2.) dX -x- e-px¥2/2 x cos-D-1 +x· 2 2 o

Figure 9.3.

(2

J

which is not difficult to evaluate by numerical methods, owing to the presence of the exponential term, and the calculation becomes simpler as p increases. For instance, for p = 10 it is enough to replace the upper bound +00 by 4 and the error is less than 10-13 . Applying the Gauss - Legendre method we obtain 110 l'II 0.01079 184327. 9.4. Let

=J

+00

1

n,p

o

cosnx

(x-l)2+ lO-P

dx

(n, p>O).

The integrand is a highly oscillating function for large n, and its pole is near to the integration intelVal for large p. Integrating along the contour of Fig. 9.2. we get

138

~9

Notice that the integrand on the right hand side of (1) is not oscillatory and that the pole is at a sufficient distance. Besides, it decreases more rapidly as n increases. Numerical examples: 110 ,3 IW -60.777375; 110,6 9.5. The integral

..

a =~ JCosnx dx It

n

-.

r+pz

IW

-2609.8102; 110 ,12

cosnz

ZH ---

z2+l

-2635994.6.

(nEN, p>O)

is the Fourier coefficient of x H II (X2 + p2) on [-n; nI. The integrand is highly oscillatory for large n, while for small p the pole ip of the function

IW

JI

JC+;C

-JC+;C

is near the integration ip

interval. e1ltz Integrating the function Z H - 2 - - 2 along z +p the contour shown on Fig. 9.5. as c ~ +00 we

-JC

o

JC

Figure 9.5.

obtain

The presence of the term e-ny ensures that the numerical calculations can be performed with high precision. For instance, using the above formula, for p =0.01 we get alo~90.483336, aloo~36.787794, alooo~4.539952 .10-3 •

REFERENCES 1. E. L. Kaplan: Numerical Integration Near a Singularity. J. Math. Phys. 31(1952),1-28. 2. D. S. Mitrinovi~ and 1. D. Keai~: The Cauchy Method ofResidues. Theory and Applications.

Dordrecht-Boston-Lancaster 1984. 3. D. D. T~i~: Introduction to Numerical Analysis (Serbian). Beograd 1987. 4. R. Piessens: Gaussian Quadrature Formulas for the Integration of Oscillating Functions. ZAMM 50(1970),698-700.

Applications of the Calculus of Residues to Numerical Evaluation of Integrals

139

5. R. Piessens and F. Poleunis: A Numerical Methodfor the Evaluation of Oscillatory Functions. BIT 11(1971). 6. R. Piessens: Gaussian Quadrature Formulas for the Evaluation ofFourier Cosine Coefficients. ZAMM 52(1972),56-58. 7. T. N. 1. Paterson: On High Precision Methods for the Evaluation of Fourier Integrals with Finite and Infinite Limits. Nwner. Math. 27(1976), 41-52. 8. R. K. Littlewood and V. Zakian: Numerical Evaluation of Fourier Integrals. J. Inst. Math. Appl. 18(1976),331-339. 9. P. Hillion and G. Hurdin: Integration of Highly Oscillatory Functions. Journ. Compo Phys. 23(1977),74-81. 10. R. Wong: Quadrature Formulas for Oscillatory Integral Transform. Nwner. Math. 39(1982), 351-360. 11. D. D. Toliic and D. V. Toliic: On the Numerical Evaluation of Certain Classes of Integrals by Residues. VI Conf. Appl. Math. Tara 1988, pp. 236-242. 12. Ph. Rabinowitz and Ph. J. Davis: Methods ofNumerical Integration, New York 1984.

Chapter 10

Inclusive Calculus of Residues MIODRAG S. PETKOVIC,

University ofNis

It is known that a continuous function z H /(z) can only be presented in a computing machine discretely because of the limited number of digits of the arithmetic of the finite precision. The graph of/is only more or less well approximated by a set of points inside a certain interval. For that reason it is practically impossible to determine a value of / without a detailed information about the rounding error. In the case of a complex function, the constants which appear in the expression for that function are not exact complex numbers but discs with small radii (Henrici [3]). Besides, in solving practical problems one often deals with uncertain or approximate numbers as initial data. One of the possibilities to overcome these problems is to use interval arithmetic and interval methods (see monographs [1], [5], [6]). Applying interval arithmetic we can automatically take into account the rounding error and the errors of initial data. The resulting interval contains the exact result or the solution of the considered problem. We apply inclusive calculus of residues to the problem of computing the integral J:J(z)dz= J:K(Z) dz

:r

r

:r h (z)

r

in the presence of a round-off error and uncertain initial data. We assume that zl' ... ,zn are the poles of the function f, that is the zeros of the function h, which lie in the corresponding discs ZI' ... ,Zn which are all in int r. The inclusive calculus of residues, presented in [8], is based on the calculus of residues of a function/taking a disc in the complex plane as its argument. Therefore, a short review of some properties of the disc arithmetic (circular arithmetic) will be described first, including circular complex functions. A complex circular interval (disc) Z with centre c = mid Z E C and radius r = rad Z (~ 0), denoted by Z = {c; r}, is the closed set (1)

Z

={zllz-clsr}. 140

141


The set of all discs of the form (1) will be denoted by K(C). Addition and subtraction of discs ZI

= {c l ; rl} and Z2 = {c2; r2} are defined by

Multiplication of discs can be introduced in various manners. It is customary to use the following definition due to Gargantini and Henrici [2]:

where Zk = {ck; rk} (k = 1,2). Furthermore, the inversion of the disc Z = {c; r} which does not contain the origin (that is, Icl > r) is the disc (4)

Using (3) and (4) we introduce the division by (5)

For the above operation the inclusive isotonicity property is valid ([I»; namely IfA", Bk e K(C) andA k ~ Bk (k = 1, 2), then AI*~ ~ BI*~

where

*e

{+, -, " :}.

Let A (H) be the class of analytic functions in a region H of the complex plane. The closed set f*(Z)={f(z)lzeZ, ZeH}

where f e A (H) need not be a disc, which makes considerable difficulties in numerical calculations. For this reason one often operates with a circular complex function F:D ..... G (D, G ~ K(C» which satisfies (6)

F(Z) 2f*(Z) F(z) =f(z)

(inclusion) (complex restriction)

for each ZeD. In further text we shall use the notation F(Z)

={mp; rp}

142

Chapter 10

where mp = mid(F(Z» and rp = rad(F(Z». The function Fwhich satisfies (6) is called a circular inclusive extension off We also say that F(Z) is an including approximation for r(Z). A suitable form of a circular complex function, the so-called Taylor's centered form, was proposed in [7]. According to this form the following representation of some elementaIy functions, important in practical calculations, are derived:

1. f(z)=zn;

zn={c n; (I c I+r)n_1 c In};

2. f(z)=ez;

eZ'={ee; leel (e'"-I)};

3. f(z) = log z;

log Z

4. f (z) = arctg z;

s.

= {lOg c; -

arctg Z

(±IEEZ)

log

(1 - I: I)}

= {arctg c; - ~ log (1 2

(OEEZ);

.' )(1--.I.-cl '-)} ;

I.-cl

fez) = sin z; sinZ = {sin c; sh r (ch2 y- sin2 X)1/2 + (ch r -1) (Ch2 y- cos2 X)1/2};

6.

f(z)=cosz; cos Z = {cos c; sh r (Ch2 y- cos 2 X)I/2 + (ch r -I) (ch2 y - sin 2 x)1/2};

7. f(z)=p(z) =

n

{n

akzk; P(Z)= p(c);

~ k-O

~

k=O

\p(k) (c) \ ,k} ,

k.

.

The argument of the listed circular functions is the disc Z = {c; r}, where c =x + iy. The property of inclusive isotonicity of a circular complex function is used in canying out the inclusive calculus of residues. A circular function F is said to be inclusive isotone if the implication

is valid. In particular, if Z

E

Z, then

f(z)=F(z)EF(Z) is always true. Let Z H fez) be a regular function on a closed curve r and in int r except at the poles zl' ... ,zn of orders m l , ... ,mn respectively. Consider the class of functions f which can be represented in the form f( z) = g (z) I h (z), where the poles of the function fare,

143


at the same time, the zeros of the function h. In that case for each pole zk (k = I, ... , n) we have

where sk is a complex function such that sk(zk) (7)

"* o. Then

2: Res/(z),

f/(z)dz=2ni

r

k=l Z=Zk

where the residue of/at zk

E

int r(k = I, ... , n) is given by

(8)

In practice it is often impossible to determine the exact value of a pole zk (because of the presence of a rounding error when the computers with limited precision are used, or for some other reasons - for instance, in an experiment), but it is known only that this pole lies inside the disc Zk = {ck; rk}. In such a case, instead of (8) we can use the following formula (9)

The notation RESf(z) is used to emphasize that RESf(z) is an interval. On the basis of the inclusive isotonicity property it is clear that

RES j(z)ERESj(z), Z=Zk

ZkEZk

zEZk

and so from (7) we obtain

(10)

1j(z) dzE 2 n i iRES j(z).

r

k=l zEZk

On the right hand side of (10) we have a circular interval of the form {w; R} which contains the exact value of the considered line integral. The centre w of this disc can be taken as an approximate value of the integral, whereas the radius R determines the upper bound of the error.

144

chapter 10

The inclusive calculus of residues will now be illustrated by a few examples. We use the operations (2) - (5) and the expressions for elementary circular functions. ExAMPLE 1. We shall fmd the circular interval which contains the exact value of the integral

where by

Fa = {~lzl=4}. We take that the (exact) poles zl = td and z2 = -td of the function/defmed /(z)

e'

= (Zl + 11)1

lie in the discs ZI = {3.l4i; 0.01} and Z2 = {-3.14i; 0.01}. respectively. Note that the exact value of the integral II is 1 i-=;0.318309 ... n

Since ZI c int Ij and Z2 c int Ij. we calculate the residues for both poles. Applying the operations of circular aritlunetic and the expression for the circular complex function eZ we obtain eZ I(ZI+ni-2)

e{3.141;0.0I}{ -2+6.281593 i; 0.01}

(ZI +n i)3

{6.281593 i; O.o1p

RES fez) = --'--"----'-

zEZI

= {0.025357 +0.008029 i; 0.OOO663}.

RESf(z) =

eZ2(Z2 -n; -2)

ZEZ2

(Z2 - n

W

=

=

e(-3.14i; 0.01) {-2 - 6.281593 i; O.OI}

{ - 6.281593 i;

0.01 P

{0.025357 -0.008029 i; 0.OOO663}.

According to this and (10). it follows that

J: _e_z-

j (Z2 + n2)2

dz =iO.318309.. . E2n i (RESf(z)+ RESf(z» zEZI

r)

=

ZE Z 2

{0.318645 i; 4.17 x 10-3}.

Ifwe take the centre 0.3l8645i of the obtained disc for an approximate value of II' then the error is not greater than 4.17'10-3. EXAMPLE 2. We shall determine the range of values of the integral

in the case when the variations of the poles of the integrand are not greater then 1% relative to the exact values %1 = I and %2 =-1.

145


Wetakethatthepolesz, andz2 lie in the discZ, ZI

{I; 0.01}

(ZI + 1)3

{2; 0.Ql)2

RES f (z) = --'--:c zEZI

1 RESf(z) = - - - -

ZE Z 2

(z2-1)2

= {I; O.OI} andZ2 = {-I; 0.01} and we fmd

{0.250025; 0.005032},

{-2; O.OW

{ - 0.250025; 0.OO25065},

so that 12 = 0 E 2 n ; (RES f (z) + RES f (z») = {O; 0.0473646}. zEZI

ZEZ2

Hence, the range of values of the integral 12 is the disc {O; 0.0473646}.

ExAMPLE 3. It is known that the pole z, of the fimctionfdefmed by sin 3 z

z~f(z)=- Z-ZI

lies in the disc Z, = {1.57; 0.001}. Applying the inclusive calculus of residues and circular arithmetic we shall detennine the value of the interval integral 13= ff(z)dz,

r

3

={zJJzJ=2}.

r3 Since Z, c int 13, applying the operations of circular arithmetic and the expression for the circular fimction sin Z, we obtain RES f (z) = sin 3Z, = sin { - 4.71; 0.003} = {0.999997; 0.OOOI2}, zEZI

so that

13 -2ni RESf(z) = {6.283166 i; 7.54 x 10-'}. zEZ,

For instance, if z,

=-nil e Z, then

,isin3z

j -n- dz=2 n;= ;6.283185307 .. • EI3 -{6.283166 i; 7.54 x 10- '}. r3 z+"2

REFERENCES 1. G. Alefeld, J. Herzberger: Introduction to Interval Computation. Academic Press, New York 1983. 2. I. Gargantini, P. Henrici: Circular Arithmetic and the Determination of Polynomial Zeros. Nwner. Math. 18(1972),305-320. 3. P. Henrici: Uniformly Convergent Algorithms for the Simultaneous Approximation of all Zeros ofa Polynomial. Proc. ·Constructive Aspects of the Fundamental Theorem of Algebra.· (ed. by B. Dejon and P. Henrici). Wiley-Interscience, London 1969,77-113.

146

chapter 10

4. D. S. Mitrinovic, 1. D. Ketkic: The Cauchy Method of Residues. Theory and Applications. Dordrecht-Boston-Lancaster 1984. S. R. E. Moore: Interval Analysis. Prentice-Hall, Englewood Cliffs, New 1ersey 1966. 6. R. E. Moore: Methods and Applications ofInterval Analysis. SIAM, Philadelphia 1979. 7. Lj. D. Petkovic, M. S. Petkovic: The Representation of Complex Circular Functions Using Taylor Series. ZAMM 61(1981),661-662. 8. Lj. D. Petkovic, M. S. Petkovic: On Some Applications of Circular Complex Functions. Computing (in print).

Chapter 11

Complex Polynomials Orthogonal on the Semicircle GRADIMIR V. MILovANOVIC, University ofNis

11.1. Introduction Suppose that X is a real linear space of functions, with an inner product if, g) :,X2 -+ R such that (a)

if+ g, h) = if, h) + (g, h)

(Linearity)

(b)

(a/, g) = a if, g)

(Homogeneity)

(c)

if, g) = (g,j)

(Symmetry)

(d)

if,j) > 0, if,j) = 0 ¢:> f= 0

(positivity)

where/, g, hEX and a is a real scalar.

IfX is a complex linear space of functions, then the inner product if, g) maps,X2 into C and the requirement (c) is replaced by

(c' if, g) = (g, f)

(Hermitian Symmetry)

where the bar designates the complex conjugate. A system of polynomials {Pk}' where Pk(/) = Ik

+ tenns of lower degree

(k =0, 1, ... )

and (Pk' p",) =0 (k *m),

(Pk' p",»O (k=m)

is called a system of (monic) orthogonal polynomials with respect to the inner product ( , ).

..

The most common type of orthogonality is with respect to the following inner product

(f, g)=

J!(t)g(t)dJ.(t), R

where dA.{t) is a nonnegative measure on the real line R with compact or infinite support, 147

148

Cbapter 11

for which all moments

I'k=

J tkcU(t),

k=O, 1, ...

R

exist and are finite, and Jlo > O. Then the (monic) orthogonal polynomials {Pk} satisfy the fundamental three-term recurrence relation (1)

PHI (t) =(t -

ak) Pk(t) - bkPk-l (t),

k= 0, 1, 2, ... ,

P-l (t)=O, Po (t)= 1,

where the coefficients ak and bk are given by

(k=O, 1, ...),

ak= (tPk,Pk) (PkoPk) bk =

(k = 1, 2, ... ).

(Pk, Pk) (Pk-II Pk-t)

We note that bk > 0 for k

~

1. The coefficient bo in (1) is arbitrary, but the definition

is sometimes convenient. Typical examples of such polynomials are the classical orthogonal polynomials of Legendre, Ceby§ev, Gegenbauer, Jacobi, Laguerre and Hermite. An other type of orthogonality is the orthogonality on the unit circle with respect to the inner product 2n

(/, g)=

J/(ef")g(ef") dp.(O),

o

dl'(O)~O.

These polynomials were introduced and studied by SzegO [1]. Monic orthogonal polynomials { t} on the unit circle satisfy the recurrent relation

cPk + 1 (z) =

Z

cPk (z) + cPk (0) Zk cPk (liz),

k=O, 1,2, ... ,

which is not of the form (1). For more details consult Nevai [2]. Similarly, we may consider orthogonal polynomials on a rectifiable curve or an arc lying in the complex plane (e.g. Geronimus [3], SzegO [4]). Complex orthogonal polynomials may also be constructed by means of double integrals. Namely, introducing the inner product by


(J, g)=

149

JJJ(z)g(z)w(z)dxdy. B

for a suitable positive weight function w where B is a bounded region of the complex plane, a system of orthogonal polynomials can be generated (see Carleman [5] and Bochner [6]).

11.2. Orthogonality on the Semicircle Recently Gautschi and Milovanovi~ [7] (see also [8]) introduced a new type of orthogonality, the so-called orthogonality on the semicircle. The inner product is defined by (1)

(J, g) =

where r= {z

JJ(z)g (Z)(iZ)-1 dz,

r

eel z =£1

8,

0 S (}S tr}. Alternatively, (1) can be expressed in the form

or

(2)

(J, g)=

Jf(e'')g (e'') dO. o

Notice that this inner product does not satisfy the conditions (c' and (d). Namely, the second factor in (1), i.e. (2), is not conjugated, so that this product does not possess Hermitian symmetIy; instead it has property (c). The corresponding (monic) orthogonal polynomials exist uniquely and satisfy a threeterm recurrence relation of the form (1) from 1l.1, due to the property (zJ, g) = if, zg). The general case of orthogonality with the complex weight function w with respect to the inner product (J, g)=

J!(z)g(z)w(Z)(iZ)-l dz,

r

i.e. or

(3)

(J, g)=

JJ(e'll)g(e'6) w(ei6) dO,

o

was considered by Gautschi, Landau and Milovanovi~ [9]. Let w : (-1, 1) -. R+ be a weight function, which can be extended to a function z 14 w(z) which is regular in the half disc

ISO

Chapter 11

Together with (3) consider the inner product 1

(4)

[f. g) =

JI(x)g (x)w(x)dx,

-1

which is positively definite and therefore generates a unique set of real (monic) orthogonal polynomials {Pk}: [Pi' p",]=O for

k*,m and

[Pi' p",]>O for

k=m

(k, m=O, 1, 2, ... )

On the other hand, the inner product (3) is not Hermitian; the second factor g is not conjugated and the integration is not with respect to the measure Iw(e j ')ld6. The existence of corresponding orthogonal polynomials, therefore, is not guaranteed. We call a system of complex polynomials {lrk} orthogonal on the semicircle if [lri

,

[11'."

11'.. ]

=0

11'.. ]*,0

for k

*' m

for k=m

and (k, m=O, 1, 2, ... )

where we assume that lrk is monic of degree k. The existence of the orthogonal polynomials {lrk} can be established assuming only that

,. (5)

Re(I,I)= Re lw(el9)cl6=1=O. o

11.3. Existence and Representation of n;. Assume that the weight function w is positive on (-1, 1), regular in D+ and such that the integrals (3) and (4) from 11.2 exist for smooth/and g (possibly) as improper integrals. We also assume that the condition (S) of 11.2 is satisfied. Let C8 ' & > 0 denote the boundary of D+ with small circular parts of radius & and centres at ±l spared out and let P be the set of all algebraic polynomials. Then. by Cauchy's theorem, for any g e P we have 0= (1)

Jg(z)w(z)dz .

c.

=(r.J+ J

C•• _l

+

J-.

J )g(z)w(z)dz= -1+. J g (x) w (x) dx,

c•. +1

151


where rs and C.!;±l are the circular parts of Cs (with radii 1 and assume that w is such that for all g (2)

J

lim

E

&

respectively). We

P

(VgEP).

g(z)w(z)dz=O

._0 C•• ±I

Then, if &~ 0 in (1), we obtain I

(3)

0=

Jg (z) w (z) dz = Jg (z) w (z) dz + J g (x) w (x) dx, r

C

gEP.

-I

The (monic, real) polynomials {Pk}' orthogonal with respect to the inner product (4) of 11.2, as well as the associated polynomials of the second kind

J 1

qk(z)=

Pk (Z)-Pk(X)

z-x

-I

w(x) dx

(k=O, 1, 2, ... ),

are known to satisfy a three-term recurrence relation of the form (k=O, 1, 2, ... ),

(4)

where (5)

Y_I

=0,

Yo

=1 for

{Pk}

and Y_I

=-1,

Yo

=0

for {qk}.

Denote by mk and Pk the moments associated with the inner products (4) and (3) of 11.2, respectively

where, in view of (5) (6)

bo ="'0.

THEOREM 1. Let w be a weight function. positive on (-1, 1), regular in D+

=

{z E Cllzl <1, Imz> O} and such that (2) is satisfied and the integrals in (3) exist (possibly) as improper integrals. Assume in addition that

J If

(7)

Re(1, 1) = Re w{e(9)dO:t O. o

Then there exists a unique system of (monic. complex) orthogonal polynomials {Irk} relative to the inner product (3) of 11.2. Denoting by {Pk} the (monic. rea/) orthogonal polynomials relative to the inner product (4) ofl1.2, we have

152 (8)

Chapter 11

(n=O, 1,2, ... ),

nn (z) =Pn (z) - i 0n-1Pn -l (Z)

where

°

= l'oPn (0)+ iqn (0) n-l . . i l'oPn-l (O)-qn-l (0)

(9)

(n=O, 1,2, ... ).

Alternatively, (10)

(n=O, 1, 2, ... );

0_1 =1-'0'

where ak' bk are the recursion coefficients in (4) and Po = (1, 1). In particular, all On are real (in fact, positive) if an = 0 for all n ~ O. Finally, (n= 1,2, ... ),

Proof. Assume first that the orthogonal polynomials {Nk} exist. Putting g(z)

=~ Nn(Z)Zk-l, I

1 s: k < n

in (3) we find 1

J

J

0= nn (z) zk (iZ)-l W (z) dz - i nn (x) ,xk-l w (x) dx r - l

(1 :;;.k
and hence, upon expanding Nn in the polynomials {Pk}' (12)

nn (z) = Pn (z) -

i 0n_l P n-l (z)

(n=O, 1,2, ... ).

for some constants On-I. To determine these constants, put g (z)= [nn (z) - nn (0)] (iZ)-l = ~ {Pn(Z)-Pn (0) -i 0n-l Pn-l (Z)-Pn-l(O)} ,z

z

in (3) and use the first expression for g to evaluate the first integral, and the second to evaluate the second integral in (3). This gives (n~

1).

Since (Nn, 1) = 0, (1, 1) = Po, and using (12) with z = 0, we get (9) for n ~ 1. Note that the denominator in (9) (and the numerator, for that matter) does not vanish, since Re Po 0

*

153


by (7) and Pk(O), qk(O) cannot vanish simultaneously, {Pk} and {qk} being linearly independent solutions of (4). For n = 0, (9) yields, by virtue of (5), 0_1 = JIo. To show the first relation in (10), replace n by n + 1 in (9) and use (4) for z obtain

o

= 0, to

_POPII+1 (O)+iqll+1 (0) iPoPII (O)-qll (0)

,11-1 -

Po [-aIlPII (O)-hIlPn_1 (0)] + i [-all qll (O)-hll qll_1 (0)] i POPII (O)-qll(O)

iall [iPoPII(O)-qll (O)]-hll lPOPII-1 (0) + iqll_1 (0)] i Po PII (O)-qll (0)'

=iall+~

(n~ 1).

(111-1

Using (9) with n

8o

= 1, (4) with k = 0 and (5), yields

Po (-ao) + iho

•

mo

•

=Iao+-' Po

'Po

since bo = mo (see (6». Therefore, (10) also holds for n = O.

If all an = 0, then w is symmetric and we can prove that (see Theorem 1 in 1l.6)

Po

=(1,1) = nw(O) > O.

Hence, using (10) we conclude that On is real. Conversely, defining (nn, zk)=

"n by (8) and (9), it follows readily from (3) that for n ~ 2,

J

~

J 1

n,,(z)zk- 1w(z)dz=i

r

n,,(x)xk- 1w(x)dx=O

_I

and from (13), (9) and (8) for z = n that ("n' 1) = 0 for n ~ l. Furthermore, (nil' nil) =

Jnn (z) zn w (Z)(iZ)-1 dz

r

J

1 =[

nn(z)zn-1w(z)dz=i

r

f

1

nil (x)xn- I w(x)dx

_I

1

.. i

J [PII(x)-iO

-I

1

=011_1

Il _

1PII_I(X)]xn- 1 w(x)dx

f P~-l (x)w (x) dx,

-1

'

IS4

Chapter 11

proving (11). We note from (8) that

(14)

(p", 1)=iO"_1 (P"-h 1)= ...

=

" (iO._ )(I, 1), n .-1 1

and, similarly, 1 ~k
(IS)

Here, (11) has been used in the last step. From (14) and (IS) there follows (16) the inversion of (8). When k = n, the empty product in (16) is to be interpreted as 1. EXAMPLE 1. w(z) = 1 + z.

Here, J.lo = (I, 1) = Ir + 2;, Re J.lo Fw1hennore, bo= mo = 2, a "

=

1 (2n + 1X2n + 3)

(n~O),

*' 0,

b = n(n+1) " (2n+1)2

so that by (10)

n-4i 00 = 3 (2 - in) ,

by (4) and (5),

and by (8),

01 =

3n+8i S (4+ in)

so that the orthogonal polynomials {Irk} exist.

, •••

(n~l),

155


in

n

==Z2_.-2

EXAMPLE 2. w(z) =

4+in

z-

4

3 (4+ in)

,

r.

Here P-o =

J" e

2i8

o

dO = 0

so that (7) is violated and thus the polynomials {Irk} do not exist, even though w(x) ~ 0 on [-I, 1] and the polynomials {Pk} do exist. It is easily seen that q~O) = 0 when k is even, so that 0n-I is zero for n even, and undefmed for n odd. For an explanation of this example see Theorem 1 in 11.6.

11.4. Recurrence Relation We assume that

Jwk )d8;t: o. tr

(1)

Re(l, 1) = Re

8

o

so that orthogonal polynomials {7rk} exist. Since (zf, g) must satisfy a three-tenn recurrence relation

= if,

zg), it is known that they

k =0, 1, 2, ... ,

(2) 7'-1

(z) = 0,

Using (8) of 11.3 and (2), for k

~

1 we get

and substituting here for zPk(z) and zPk_l (z) the expressions obtained from the basic recurrence relation (4) of 11.3, yields

[ok + i «()k -

()k-l -

a/] Pk (z) + [b k - {Jk - ()k-l (ak + iOk-1)]Pk-l (z) + i [{Jk 0k-2 - bk- 1 Ok-I] Pk-2 (z)= 0,

By the linear independence of the polynomials {Pk} we conclude that

ak +i(Ok- Ok_l- ak)=O, (3)

k!1;.l, k !1;.I,

k!1;. 1.

156

Chapter 11

From the last equality in (3) and (10) of 11.3, we get (4) for k > 2. The first equality in (3) gives (5) To verify that (4) also holds for k = 1, it suffices to apply the second relation (3) for k = 1, in combination with (10) of 11.3 and (5) for k = 1. With a", Pk thus determined, the second equality in (3) is automatically satisfied, as follows easily from (10) of 11.3. Finally, from ~I

(z) = z- jao =

PI

(z) - jOo =z - ao - jOo'

we find

Alternatively, by (10) of 11.3 we may write (5) and (6) as (7)

k~l,

We have therefore proved: 1. Under the assumption (1). the (monic. complex) polynomials {1l'k} orthogonal with respect to the inner product (3) of 11.2 satisfy the recurrence relation (2), where the coeffiCients ale> Pt are given by (5) (or (7» and (4), respectively, with the On defined in (9) (or (10» ofll.3.

THEOREM

By comparing the coefficients of zk on the left and right of (2), we obtain from (5), (6) that (n ~ 1).

11.5. Jacobi Weight We consider now the case of the Jacobi weight function

157


w(z)=~·II(z)=(I-z)a(1 +Z)II,

(1)

a> -1, P>I,

where fractional powers are understood in terms of their principal branches. We first obtain the existence of the corresponding orthogonal polynomials

{t4 Jl)}. a

THEOREM 1. We have

1'0 = ,100. {J) =

(2)

Jw
dO =

31:

+ i v. p.

o and hence Re Po

J 1

II) (ei 8)

W(O,: (x) dx,

-1

*" O.

Proof. Let CE , &> 0, be the contour formed by {JD+, with a semicircle of radius r about the origin spared out. Then, by Cauchy's theorem

where rand c e are the circular parts of Ce (with radii 1 and (3) we get

J W;X)

&,

respectively). If &~ 0 in

1

O=I'o-i v. p.

dx-nw(O),

-1

which proves (2). The following theorem is also proved in [9]: THEOREM2. We have

31:(fl.O)(z) /I) ( " = ( - 1),,-(0. 31:" . where

frIO

)

- Z ,

denotes the polynomial frn with all coefficients conjugated, i.e. n...(z) = frll(z).

11.6. Symmetric Weights and Gegenbauer Weights THEOREM 1.

satisfies (1)

then

If the

weight function w. in addition to the assumptions stated in 11.3,

w( -z) =w(z) and w(O»O,

158

~ll

,uo = (1, 1) = nw (0),

(2)

and the system %rthogonal polynomials {Kn} exists uniquely. Proof. Proceeding as in the proof of Theorem 1 from 11.5, we find

. J 1

w(x) -;-dx-nw(O),

O=,uo-l v.p.

-1

where the Cauchy principal value integral on the right vanishes because of symmetIy of w. This proves (2). Under the assumption (1) we have ak = 0 for all k ~ 0 in (4) of 11.3. In this case the relations (6), (5) and (4) of 11.4 reduce to k~

1.

Furthermore, by (10) of 11.3

wherem = 1, 2, ... In particular, for the Gegenbauer weight

A> -1/2,

(3)

we have p" (z)

=C" (z)

- the monic Gegenbauer polynomials - for which

known, b =m = o

b -

0

r (i.+l/2) lrn r"' r(i.+l) ,

k(k+2A-l)

k- 4(k+i.)(k+i.-l) '

Therefore, by (10) of 11.3,

k~ 1 - •

as is well

159


()n=

n(n+2A-l) 1 -, 4(n+A)(n+A-l) 8n - 1

n=l, 2, ""

() _ 0-

r(A+l/2)

Vn r(A+l)

i.e. n~

1.

The corresponding orthogonal polynomials can be represented in the form :n;k(z)=C;(z)-i()k-l CLt(z) and their norm is given by

In particular, we have: 1° The Legendre case A. = 112: () = ( k

r

«k

+ 2)/2) r«k+l)/2)

)2

,

k ~ 0;

2° The teby§ev case A. = 0: 00 = 1, Ok = 112, k ~ 1; 3° The teby§ev case of the second kind A. = 1: Ok = 112, k ~ O. The Legendre case is considered in detail in [7], and Gegenbauer case, including various applications to numerical analysis, in [10].

11.7. The zeros of n;,(Z) It follows from (2) of 11.4 that the zeros of 7rn(z) are the eigenvalues of the (complex, tridiagonal) matrix iag

1

PI

ia,

J= n

p,

0 ia z

1

0

Pn-I

ian _ ,

160

chapter 11

The elements of I n are easily computed using (6), (5), (4) of 11.4 and (10) of 11.3, since the recursion coefficients ale> bk for the orthogonal polynomials {Pk} are known.

If the weight w is symmetric, then Pk = oLI is positive, and I n can be transformed into a real matrix. Indeed, a similarity transformation with the diagonal matrix

transforms the complex matrix I n into the real nonsymmetric tridiagonal matrix

o

o with eigenvalues 1J v = -isv . Using the EISPACK subroutine HQR (see [11» we can evaluate all the eigenvalues 1J v (v= 1, ... ,n) of An' and then all the zeros sv= i1J v (v= 1, ... ,n) of 7rn(z). A theorem on the distribution of zeros of 7rn(z) in the case when the weight function w is symmetric, i.e. w(-z)

= w(z)

and w(O) > 0

is proved in [9]. We shall quote here a particular, but important, result regarding the distribution of zeros of the polynomials n;!(z), orthogonal with respect to the complex Gegenbauer weight w(z) = (l-z2

r"

2,

A. > -112.

THEOREM 1.1f A. > -112, all the zeros of ~(z) are simple and for n ~ 2 they belong to

the upper unit halfdisc D+ = {z

E

Cllzl < 1, Imz> O}.

REMARK 1. The problem of distribution of zeros for general weight functions is not solved. The case of Jacobi weights (see (1) of 11.5) is particularly interesting. Numerical experiments indicate that all the zeros belong to the half disc D +. REMARK 2. For the Gegenbauer case a second order linear differential equation is found in [9] whose particular solution is the polynomial 7r:(z). Applications of those polynomials to numerical integration and numerical differentiation of analytic functions are given in [10). REFERENCES 1. G. SzegO: Ober Trigonometrische und Harmonische Polynome. Math. Ann. 79(1918),323-339.


161

2. P. Nevai: Geza Freud, Orthogonal Polynomials and Christoffel Functions. A Case Study. J. Approx. Theory 48(1986),3-167. 3. Va. L. Geronimus: Polynomials Orthogonal on a Circle and Interval. Pergamon, Oxford 1960. 4. G. SzegO: Orthogonal Polynomials. 4th ed. Amer. Math. Soc. Colloq. Publ. Vol. 23, Amer. Math. Soc., Providence, R. 1. 1975. 5. T. Carlemann: Ober die Approximation Analytischer Funktionen durch lineare Aggregate von vorgegebenen Potenzen. Ark. Mat. Astronom. Fys. 17( 1922123), 1-30. 6. S. Bochner: Ober orthogonale Systeme Analytischer Funktionen. Math. Z. 14(1922), 180-207. 7. W. Gautschi, G. V. Milovanovic: Polynomials Orthogonal on the Semicircle. J. Approx. Theory 46(1986),230-250. 8. W. Gautschi, G. V. Milovanovic: Polynomials Orthogonal on the Semicircle. Rend. Sem. Mat. Univ. Politec. Torino (Special Functions: Theory and Computation), 1985, 179-185. 9. W. Gautschi, H. J. Landau, G. V. Milovanovic: Polynomials Orthogonal on the Semicircle. Constr. Approx. 3(1987), 389-404. 10. G. V. Milovanovic: Complex Orthogonality on the Semicircle with respect to Gegenbauer Weight: Theory and Applications. In: Topics in Mathematical Analysis (Th. M. Rassias, ed.), World Scientific Singapore 1989, pp. 695-722. 11. B. T. Smith et al.: Matrix Eigensystem Routines - EISPACK Guide. Lect. Notes Compo Sci. Vol. 6, Springer-Verlag, New York 1974.

Chapter 12

A Representation of Half Plane Meromorphic Functions DRAGISA MnRovIc, University of Zagreb The subject of the present topic is the representation of half plane meromorphic functions in terms of their boundary values on the real axis R and of their residues at the poles. 12.1. For the convenience of the reader we recall some notions and results. The open upper half plane and the open lower half plane will be denoted by E C: Im(z) < O}, respectively; also we put

Lt+ = {z E C: Im(z) > O} and Lt- = {z Lt = Lt+ u Lt- so that ..1 = C \ R.

Let Ip(x) be a continuous function from R to C. The support of Ip(x), denoted by supp(Ip), is the closure of the set {x E R: Ip(x) -:I: O}. The support of Ip(x) is then the smallest closed set outside of which Ip(x) is identically zero. Letf(z) be a function which is analytic in the complex plane C for all z outside a closed set K!;;; R. We shall refer to this assumption by saying that the function f(z) is sectionally (locally) analytic in C with boundary on the real axis R consisting of the set K. In other words the plane C is cut along K which does not belong to the domain of analyticity of f(z). In this situation the function f(z) can be decomposed into two functions:

f

(z) = {/+(Z), I-(z).

zELl +. zELl-.

Observe that K c R implies immediately the equality f+(x) =f- (x) for all x the same time the functionf(z) has a jump discontinuity j+(x) - f-(x) on K.

E

R \ K. At

We shall need the following two simple results. FACT l.Jf the function f+(z) is analytic in ..1+ and continuous in ..1 + with the order f+(z)

o(J..) as Izl

=

1z1 ..... +00 in this hal/plane, then

162

={z E C : Im(z) ~ O}

163

A Representation ofHalfPtane Meromorphic Functions

J +00

_1_

(1)

2n i

f+(I)d/={/+(z),ZEA+, I-z 0 ,zEA-.

-00

To prove (1) observe first that the involved improper integral exists due to the fact that f+ (I)

o(.!.) as I~ ~ III

=

+00.

Now denote by L the contour which consists of the segment

[-1", r] and the semicircle C: IzI = r. Let z be an arbitrary point inside the contour L. According to the Cauchy integral formula for a closed contour, we have _1_J 1+(C) d C=_I_Jf+ dC +_1_ ff+(I) dl =/+(z), C-z 2ni C-z 2ni, I-z

2ni

L

C

~

where Re(~ = I. Evidently, for z e ..1- the integral along L equals zero. Letting r ~ +00 the integral along C vanishes and the representation (1) follows at once. In a similar manner we prove FACT 2.1fthefunctionf-(z) is analytic in ..1- and continuous in ..1-

with the order

J +00

_1_ 2n i

(2)

r

(z)

={z eC: Im(z) s; O}

=rf ..!..) as Izi ~ +00 in this halfplane. then "llzl

1-(1) dt={

t-z

0 ,zEA+, -f-(z), zEA-.

-00

Further, we shall use the following two propositions. PRoPOsmoN 1. Let rp(t) be a continuous function from R into C with the order

o(.!.) It I

rp(t) =

as

I~ ~ +00. Let q,(z) be the function defined by

J

+00

'P(z) =_1_. 2n,

tp(t)

t-z

dt,

zE~.

-00

Then

. lim [qi(x+ie)-qi(x-ie)]=
164

Chapter 12

at all points x e R. PROPOsmON

junctions in

2. (Analytic continuation principle). Letf+(z) and f-(z) be analytic

~+

and ~-, respectively. Suppose that r(x+is) converges to f+(x) and

rex-is) converges to rex) as s-+ 0+ such that f+(x)- f-(x) open set n.

!;;

=0 for all x in some

R. Then there exists a unique function fez) that is equal to f+(z) in

n. u

and to f- (z) in ,d-, and is analytic in ~+ u

~+,

~-.

r

In particular, if n. = R and if (z) or f- (z) vanishes at infinity and has a pole of order m at some point z = a, a e ~+ or a e ~-, then by this Proposition and Liouville's theorem we have

f

(z) = Pm - t (z) , (z_a)m

where Pm-l (z) is a polynomial of degree less then or equal to (m - 1). 12.1. Now we are ready to prove the following theorem.

Let fez) be sectionally analytic in C except for aflnite number ofpoles at 1, 2, ... , n of order a", located in ~+ u ~- and with a boundary on R consisting

THEOREM 1.

a", k

=

of a closed set K!;; R Let fez)

=0C~I)

1z1-+ +00

as

and assume that the boundary

values

lim f(x+i8)=f+(x),

......0+

lim f(x- i 8) = f-(x)

......0+

exist at all points x e R. Then for z E K and z ;It a", k = 1, 2, ... , n the function fez) has the representation (3)

f(z) =_1_. 2:1f I

f

+00

-00

where

/+(1)-/-(1) I-z

/I

Ok

dt+ ~ ~

Ak,p , k=l p=l (z-a)P

165

A Representation ofHalfPlanc Met'OIJlOI)lhic Functions

Proof The hypothesis on the boundaIy values of fez) implies the continuity of the

functions f+(x) and f-(x) on R Put qKx) =f+(x)- rex), x e R First we shall show that supp( f{J) =K. To do this first consider K to be a proper subset of R, K c R The function f (z) being analytic on the open set R \ K we have

for all x e R in R \ K. Thus f{J (x) = 0 for all such x; this proves that supP( f{J) ~ K. If supp(f{J) c K then fez) would be analytic on a set which is larger than R \K contraIy to the hypothesis. Consequently, supp(f{J) =K. Also, it is easy to see that in the case K = R we have supp( f{J) = R Therefore the function F(z)=~

f

+00

211: I

/+(1)-/-(1)

dt

I-z

-DO

is defined and analytic in the domain C \ K A slight meromorphic modification of Facts 1 and 2 shows that F(z) =

o(.!.) Izl

as

Izi -+ +00 and that

F(x ± i&) converges to the

boundaIy values F±(x) for all x e R, respectively, as &-+ 0+. Hence by Proposition 1 we may write (4)

lim [F+(x+ is)-F-(x-i s)] =f+(x) -f-(x)=F+(x) -F-(x). 0--+0+

Now let us put H(z) =fez) - F(z). Obviously the function H(z) is defined and analytic at least in C \ R except at the poles off(z). Further, H(z) =

o(.!.) Izl

as 1z1-+ +00 and

H+(x) - H-(x) = f+(x) - J(x) - [F+(x) - F-(x)l-

According to the relation (4) we have H+(x) =H-(x) for all x e R Hence, by Proposition 2 the function H (z) is analytic everywhere in C except at the poles ak> k = 1, 2, ... , n. By virtue of the generalized Liouville's theorem, H(z) is a rational function in C which vanishes at infinity. Consequently the partial fraction expansion of H (z) is possible and gives the coefficients Ak,p in (3) as can additionally be seen in [1, p. 168]. The representation (3) is proved. CoROLLARY

poles

l. Let fez) be sectionally analytic in C except for a finite number of simple

ak (k = 1,

2, ...

,n)

which are located in .1+ u .1- and with a boundaIy on R

166

Chapter 12

consisting of a closed set K s; R. Let fez)

oC!I) as 1:1-+

=

+00

and assume that the

boundaJy values limf(x + i e) =f+(x)

e-+O+

limf(x-i e)=](x)

e-+O+

exist at all points x e R. For z E K and z E ok (k = I, 2, ... , n) we have

CoROLLARY

J +-

1 ..

fez) =2 _ ,.

--

I+(t)-I-(t)

dt+

t-z

"

Res I(z) -=------

~ z=ak

L.,

k= 1

Z-Qk

2. Let fez) be sectionally analytic in C and with a boundaJy on R consisting

of a closed set K s; R. Let fez)

o( I!I) as 1:1-+

=

+00

and assume that the boundaJy

values lim f(x + i e) = f+(x).

lim I(x - i e) + f-(x)

e-+O+

e-+O+

exist at all points x

E

f(z) =_1_. 2n,

R. For z E K we have

J +-

--

f+(t)-I-(t)

t-z

dt.

REFERENCES

1. J. W. Dettman: Applied Complex Variables. Macmillan, New York 1965. 2. F. D. Gakhov: Boundary Value Problems. Addison-Wesley, Reading, MA 1966. 3. N. I. Muskhelishvili: Singular Integral Equations. P. Noordhoff, Groningen, The Netherlands 1958.

Chapter 13

Calculus of Residues and Distributions DRAGISA MrrROVIc, University of Zagreb

13.1. Test Functions and Distributions For the convenience of the reader, we shall briefly describe the spaces D, E, ~a of test functions and corresponding spaces D', E', ~a' of distributions needed in sections 13.S 13.7.

13.2. The Spaces D and D' The vital concepts of the theory of distributions are the support of a function, locally integrable function and functional. The support ofa functionf: R -+ C is the closure of the set {t E R:f(t):# O}; note that it is the smallest closed set outside of whichf(t) is identically zero. We shall write supp (f) for this. If supp (f) is a bounded set, then f is said to have compact support. In future every set in R (or C) which is closed and bounded, will be called a compact. A function f defined almost everywhere on R is locally integrable if it is measurable and

JIf(t)ldt < +00 for every compact K c R (Lebesgue). For example, continuous K

functions, piecewise continuous functions, integrable functions, measurable and locally bounded functions are locally integrable. In particular, the function t 1-+ log t is locally integrable on R (contrary to the function t 1-+ .!.).

t

We denote by D =D (R) the set of all infinitely differentiable functions rp: R -+ C with compact support (it is not required that all functions rp be zero outside the same bounded interval). The elements of D are called the test (basic) or finite functions. Note that the set D is a vector space.

An example of a test function (in D) is 167

168

Chapter 13 1

tp(t)={e-l-t2.ltl
o

•

Itl~1.

In fact, fP e CW = CW(R) and supp(fP) = [-1. 1). Observe that the zero element (the null element) in D is the function fP (t) == 0 on R; obviously, its support is the empty set. The topology of D is described as follows.

A sequence (fPn) offunctions fPn e D. n = 1,2•... ,is said to converge to zero in D as n -+ +00 if (i) the supports of all functions fPn

are contained in a fixed compact subset K of R;

(ii) for eachp = 0, 1,2, ... the sequence of derivatives

('11.:» converges uniformly to

zero on K as n -+ +00 (qf,,0) == fP). Thus a sequence (fPn) converges to fP in D if the sequence (fPn - fP) converges to zero inD as n -+ +00. A functional T: D -+ C is a rule that assigns to every function fP e D a number T( fP). The value of T at the "point" fP will be denoted by (T, fP). A functional Ton D is linear if

(T, 9'1 +9'2)=(T, 9'1)+(T, 9'2) for all 9'1> 9'2 eD, (T, c9')=c(T. fP) for all fPeD, ceC. A linear functional Ton D is continuous if _lim(T, fP.. ) =0 for every sequence fPn 00 which converges to zero in D as n -+ +00. A distribution (on R) is every continuous linear functional on D =D (R). The set of all distributions on R is denoted by D' =D' (R). The vector structure of the set D' is defined by the formulas

(Tl +T2' tp)=(Tl' tp)+(T2' tp), (cT, tp)=c(T, tp),

tpED, cEC.

A sequence (Tn) of distributions is said to converge to TeD' in D' if lim (T" , fP) =(T, fP) for every fP e D.

11-++00

In this definition the existence of the limit TeD' is postulated. However, if (Tn' fP)

169


converges for every rpeD as n~+OO then rpH(T, rp)= lim(T", rp), rpeD, is a ......+'"

member of D'. For example, the functional 8: D ~ C defined by (8, rp) = rp(O) is the famous Dirac distribution. The second useful distribution (connected with 8) is the principal value distribution

p(-7) defined by the Cauchy principal value as follows:

rp(t) (p(!),t rp) PT _'" t =

dt = ~~ J rp(t) dt III:... t

= lim(-JE rp(t) dt + +J'" rp(t) dt), E-+O+

-'"

t

t

..

rp e D.

The set of distributions that are most useful are those generated by locally integrable functions. For every fixed locally integrable function/: R ~ C the functional T D ~ C defined by

r:

co

(1)

(TI' fP)=

J/(t) fP(t) dt,

fPED,

-co

is a distribution. A distribution T is regular if there exists a locally integrable function/such that (1) holds. All other distributions are termed singular. For example, the distributions 8 and

p(-7) are singular. Let flbe an open subset ofR, fl~ R. We say that a distribution TeD' is zero on flif (T, rp) = 0 for every rp e D with support inside n, that is, for every rp e D (fl). For example, if the function/(t) in (1) equals zero almost everywhere on R, then Tris zero on R; the Dirac 8 distribution is zero on fl= R \ {O}; the regular distribution TH generated by Heaviside's function H(t)= J 1, t~O, lO, t
is zero on the set fl = {t e R : -00 < t < O}. We say that two distributions S, TeD' are equal on an open subset fl ~ R if S - Tis zero on fl and we write S = Ton n, or (S, rp) = (T, rp) for every rp e D (fl). The support of a distribution TeD' is the complement in R of the largest open subset

170

Chapter 13

of R where T is zero. We denote it by supp (7). Equivalently, a point belongs to the support of T if and only if there is no open neighbourhood of it on which T is zero. Also, the support of T is the smallest closed subset of R outside of which T is zero. For example, supp (b) = to}, supp (Tn) = [0,00), sup{

pGJ) =

R.

The following result is often used: If fJ e D and TeD' are such that supp (7) f"'I supp (fJ) = 0, then (T, fJ) = O.

13.3. The Spaces E and E' There are a number of important subspaces of the vector space D' of distributions. The present section deals with the subspace of D' which consists of the distributions with compact support. We denote by E = E(R) the vector space of all infinitely differentiable complex valued functions on R (with arbitrary support). A sequence (fJn) of functions fJn e E converges to a function fJ in E and n -. +00, if for every p

= 0,

1, 2, ... , the sequence (qf:»

compact subset of R (qf0) 00, then fJ e E.

converges to

vip)

uniformly on every

=fJ). The space E is complete, that is, if fJn -. fJ in E as n -. +

A linear functional Ton E is continuous if (T, fJn) -. 0 when fJn -. 0 in E as n -. +00. The set of all continuous linear functionals on E is denoted by E' = E' (R). Every element TeE' is an element of D', that is, T is a distribution. To see this, first note that if TeE'then (T, fJ) is well defined for all fJ e D since DeE. Further, if the sequence (fJn) converges to zero in D as n -. +00 then fJn -. 0 in E also; hence (T, fJn) -. 0 as n -. +00 and TeD'. We conclude that E' cD'. It can be proved that all members of E' are distributions with compact support. One of the most important distributions in E' is the Dirac t5 distribution defined by (0, fJ) = fJ(O), fJ e E. We recall that supp (0) = to}. The following result shows how one can modify a test function in E outside the support of a distribution in E'.

PRoPOsmoN l. Let T be a distribution in E'. Let e(t) be a function in ex'such that e(t) = 1 on a neighbourhood of the support ofT. Then (T, fJ) = (T, efJ)for all fJ e E. Now consider the function in teR 1

k(t; z ) = - - 2ni(t-z)

z

=x + iy being a parameter.

For every

Z E

C \ R the function t 14 k(t; z) is infinitely

171


differentiable on R and hence an element of E. If TEE' then its value at k(t, z) is well defined and depends of z E C \ R. Thus when T operates on E it generates a function of the complex variable z for all z E C \ R The function

T(z)=~ 2:n:1

(Tt, _1_): =_1_. (r" t-z

2:n:1

e(t»

t-z

is called the Cauchy integral of T or the Cauchy representation of T. It can be proved that fez) is analytic in the domain C \ supp (T). For example, the Cauchy integral of the Dirac 0 distribution is the function

1 ( 1) = ---. 1

~(z)=-. A

~t' -

2:n:1

t-z

2:n:IZ

which is analytic in C \ {O}. Also, let f( t) be a fixed continuous function on R with compact support and let Tf be the regular distribution corresponding tof(1). Since in this case supp (Tj) =supp (j), the Cauchy integral of Tfis reduced to the usual Cauchy integral off(t). In fact, putting rpz

1 ( t)- 2:n:;(t-z)

in GO

J f(t)rp(t)dt,

(Tf> rp)=

rpEE,

-GO

we get at once 1 ( T , ,1) 1 JII --dt f(t) T(z)=-. - =-. A

2m

t-z

2m

II

t-z

with supp (j) = [a, b J.

13.4. The Spaces ~a and ~ a' Let a be a real number. We say that a function rp E ()a = ()JR) if rp E C'" = C"'(R) and if for eachp = 0, 1,2, ... there exists a constantMp such that

It is easy to see that the set ()a is a vector space.

172

Chapterl3

A sequence (9'n) is said to converge to zero in t)a if

(ii) for eachp = 0, 1,2, ... the sequence (9'':» converges uniformly to zero on every

compact of R as n

-+ +00;

(iii) for each p there exists a constant M p ' which is independent of n, such that

- oo
t)a

as n

-+ +00.

Note

The vector space t)a' = t)a' (R) is the space of all continuous linear functionals on t)a' where continuity has the usual meaning, i. e. that for T E t)a' (T,rpn)-+O

if

rpn-+O in

t)a

as

n-++oo.

We have D c t)a for every a E R; and if a sequence (rpn) converges to rp in D then the sequence also converges to rp in t) a. Accordingly every continuous linear functional on t)a is also a distribution, that is, t)a' cD'. For example, the distribution

p(7) is an element of

t) a'

for all a < O.

Letf(t) be a fixed continuous function on R with f(t) = oC!I) as Itl-+ +00 and let Tf

be a functional on t)a defined by GO

Jf(t)q;(t)dt,

(T" q;)=

a
-GO

Then Tf is a regular distribution in t) a', a < O. Let z denote any point located in the domain C \ R. Then the function

1

tH----

27d(t -z)

belongs to 0 a for every a ~ -l. Therefore every distribution T E t)a', a ~ -1, acting on this function generates a function of the complex variable z E C \ R. The function

173


1(T,,-, 1) (-z

T(z) = - . A

2m

analytic in the domain C \ supp (T), is also called the Cauchy integral of T For example, the Cauchy integral of

A

(

<

1 1 t1 ) (z)= 2ni t'

-,

J

2z

GO

1)

t-z

a ~ -1.

p(~) is the function -1

P

E () a',

1 = 2ni

dl 1 (/-z) =

-GO

-1

-, 2z

1m (z»O,

1m (z)
Finally let us note the following proper inclusions: If a < pthen()a c ()pand ()a' C

()p'. Also we haveD c

()a

cEandE' c()a' cD'.

13.5. A Distributional Representation of Half Plane Meromorphic Functions We shall give here a representation of half plane meromorphic functions in terms of their distributional values in D'(R) and residues at the poles. In order to prove the main result of this section (Theorem 1), we need the following notations and results. The open upper half plane and the open lower half plane will be denoted by Li+

= {z E c: Im(z) > O}

respectively; also we put Li

and

= Li+ u

Li-

= {z E c: Im(z) < O}

Li- so that Li

= C \ R.

Let K be a closed subset of R. A function f: C \ K ~ C is called a locally analytic function in C with boundary K if it is analytic in the domain C \ K (a synonym for "locally" is "sectionally"). Let f( z) be locally analytic in C (with the possible exception of a finite number of singular points in Li). We say that f( x ± ie) converges to f± in D' (R) as e~ 0+, respectively, if 00

(1)

(f±, f{J)= lim (f(x±ie), f{J)= lim

for every 'P

e ->- 0+

E

D(R).

Jf(x±ie)f{J(x)dx

• ->-0+ -00

174

Chapterl3

Also,j( x ± i &) converges to f± in l?a' (R) as equality (1) holds for every rp E l?a(R).

&~

0+ if the integral in (1) exists and the

The convergence in l?a' (R) implies the convergence in D' (R), but the converse is not true. However the following result holds. PROPOSITION

f±(z) =

1 [7]. Let f±(z) be an analytic function in L1±, respectively, with

0(1.) Izl

as Izl

~ +00. Thenf±(x ± is) converges to f± in D'(R) as &~ 0+ ifand

only if f±(x±i&) converges to j± in l?a'(R), a
as&~O+.

2 [6]. LetJ+(z) andf-(z) be analytic functions in .1+ and .1-, respectively,

0(1.) Izi

with the property f±(z) =

as

Iz I

~ +00. Suppose that J+(x + is) ~ J+ ED' (R)

and f-( x - i&) ~ f- ED' (R) in D' (R) as &~ 0+ such that
3 [4]. Let T(z) be the Cauchy integral of a distribution Tel? a'(R), a ~-l.

Then lim T(x ±i &) =T± exists in D'(R) with &->0+

(2)

r+ -T- = T,

(3)

r

+ T-

PROPOSITION

-1

~ a < 0,

=-

~ ( T * VP';).

4 [4]. Let T(z) be the Cauchy integral of a distribution Tel? a' (R),

and T(z) =

0(1.) Izi

as

Izl

~ +00.

Then lim T(x ±i &) =T± exists in l?a'(R) HO+

with (2) and (3). Observe that the distributional relations (2) - (3) extend the famous Plemelj formulas (here

Vp'; has the same meaning as p(';».

We are now in a position to prove THEOREM 1 [5]. Letf(z) be locally analytic in C except for afinite number ofpoles at a", k = 1,2, ... ,n, of order ak located in .1+ u .1- and with a boundary on R consisting of a

175


closed set

K~R

Izl -+

Let /(Z)=oC!I) as

lim/(x-i &) =/- in D'(R). Then/orz

6--+0+

Ii!: K

+00.

and Z:F- ak' k

Let =

!~/(X+i&)=/+

and

1,2, ... , n, the/unction/(z)

has the representation (4)

1(/,+ -/,-, 1) + ~

n,

/(z)=2---=-

-t-

z

akA

II

k=l

~ (z:~p)P' p=l k

where 1

.

dm dzm

Ak,aTc-m=- bm -[(z-z,Jak/(z)], m=O, 1,2, .. ' , ak-l. m!

z-+ak

r -/-.

Proof Put T = First we shall prove that supp (T) ~ K. To do this, first consider K to be a closed proper subset of R. Since the functionf(z) is analytic on the open set R \ K we have

(/+, fP)= lim (/+ (x+is), fP)= lim (/-(x-is), fP)=(/-, fP) .....0+

.....0+

for all rp E D(R) with support in R \ K. Thus (T, rp) = rp) = 0 for all such rp; this proves supp (T) ~ K. Now if supp (T) c K then there exists an open interval (a, b) c K\ supp (T) on which T is zero in D'(a, b). By the analytic continuation (proposition 2)./(z) would be analytic on (R \K) u (a, b) contrary to the hypothesis; we conclude in the case that K is a closed proper subset of R, that supp = K. By the same reasoning supp (T) = R in the case K = R. By Proposition 1 the boundary values /+ and /- are distributions in "a'(R), for all a < O. Also E "a'(R), for all a < O. Since the Cauchy kernel belongs to "JR) for all a ~ -1, we must take -1 ~ a < 0 in order to form the Cauchy integral of Thus we consider E "a'(R), -1 ~ a < O. Also we can take (J+-/-) E "-1' (R) since "a'(R) c"_I'(R) for a>-l. Therefore the function

(n

F(z)

=_1_ (/,+ _/,-, _1_) 2ni

t-z

is defined and analytic in the domain C \K. It can be shown that F(z)

=o(~) Izl

as

Izl -+ +00. In addition by the Plemelj relations (proposition 3) F(x ± is) converges to a boundary value F* in D'(R), respectively, as &-+ 0+. Now let us put H(z) =f(z) -F(z).

176

chapter 13

Obviously the function H(z) is defined and analytic at least in C \ R except at the poles of

fez). Further, H(z) =

rf ..!..) as Izl-+ +00 and "'Uzi

(H+ -H-, cp)=(f+ -f-, cp)-(F+ -F-, cp) for all f/J E D(R). According to the first relation in Proposition 3 we have (P-+ -F-, f/J) = k = 1, 2, ... ,n. By virtue of the generalized Liouville theorem, H(z) is a rational function in C which vanishes at infinity. Consequently the partial fraction expansion of H(z) is possible and gives the coefficients A k,p. The representation (4) is proved. COROLLARY 1. Letf(z) be locally analytic in C except for a finite number of poles at ak' k = 1,2, ... , n, which are located in .1+ u .1- and with a boundary on R consisting of a closed set

K~R

Let f(Z)=O("!") as Izl Izl

limf(x-i &) = f- inD'(R). Then for z

&--+0+

/(z)

~

2nl

I-z

HO+

K andz:l: ak' k

=_1_. (/,+ _/,-, _1_) + L n

-+ +00. Let limf(x+i&)=f+ and =

1,2, ... ,n, we have

ResJ(z) Z

_=---'-'."k_ k=l z-ak

COROLLARY 2. Letf(z) be locally analytic in C and with a boundary on R consisting of a closed set

K~R

Let f(Z)=oC!I) as Izl

limf(x-i &) = f-. Then for z

..--+0+

f(z)=~ 2nl

~

-+ +00. Let !2J!f(X+i&)=f+ and

Kwe have

(/,+ -f,-, _1_). I-Z

REMARK 1. If we replace in Theorem I the convergence in the D'(R) topology with that of

oa' (R), -1 S a < 0, then by Proposition 1 we get an equivalent result. 13.6. A Generalization of the Residue Theorem Consider the Cauchy integral

177

Calculus of Residues and DistributiOlll

1 (T , ,1) -

T(z)=-. A

2n.

t-z

r

of a distribution T E E'(R). Let be a closed contour enclosing the support of T and lying in the domain of analyticity of a given function 9'(z). Let us evaluate the integral

.J (T"

J

T(z) 9' (z) dz=_1 2n.

r

_1 ) 9' (z) dz, t-z

r

We recall that the function f(z) is analytic in C \ supp (T) (hence it is analytic at evety point of the open set R \ supp (T». Also we remember that

J

-1. -9'(z) dz =-9'(t)

2m r t-z

for t E R

f""\

int (/).

In order to obtain the main result (Theorem 1), we begin by the following lemma.

1. (Stieltjes-Vitali, (3». Let (f,,(z» be a sequence offunctions, each analytic in a domain Gee. and let the set of functions {fn(z)} be uniformly bounded on each compact subset of G. In addition; let S c G be an infinite subset of G with an

LEMMA

accumulation point in G and let lim fn(z) exist for every z n-++ao

E

S. Then (f,,(z» converges

uniformly on each compact subset ofG. Note that the limit function fez) = lim fn (z) is also analytic in G (Weierstrass). n-++'"

LEMMA 2.

Let r be a closed contour in C and let

h (z) =

_1_.

2nl

J

h (C) dC,

r

C-z

be the Cauchy integral ofa continuous function h (~ given on r. Let

be the Riemann sum of h(z). Then for each p = 0, I, 2, ... , the sequence of p-th A

(h~P)(z») converges uniformly to the p-th derivative h(p)(z) subset of C \ r as n -+- +00. derivatives

Proof First observe that

(h~P)(z»)

is the sequence of functions

on every compact

178

Chapter 13

j,(P)(z) =...e!.. ~ h(,.)

2 m. ~ (I' )P+I ._1 ':o.-Z

II

LV

':0.

each of which is analytic in C \ rfor p = 0, 1,2, ... Now let K be an arbitraIy but fixed compact subset of C \ r, and let d = d(r, K) be the distance from K. Denote by I(I) the length of r and put M inequality

rto

=maxrlh(~I.

We get at once for all n and all z e K the

h(Ck) .de I~ Mp! ~ l.de 1~/(r)Mp!~M . 7:. (Ckk 2ndp+l k£:.' k 2ndp+l K.p z)P+1

IhC,)(Z)I=I2!... ~ 2ni k

II

for p

=0,1,2,

... Thus the sequence

(j,!P)(z»)

is uniformly bounded on every compact

r.

subset K of C \ Let {z v} be an infinite set of points in C \ r having an accumulation point in C \ By the theory of integration the limit

r.

lim h (p) (z ) =h(P) (z.) =L _

..

2ni

II·

J

r

h (C) (C-Z.)p+l

de = _1_. 2nl

J r

exists for every v = 1, 2, ... In view of Lemma 1, for every p

(j,!P)(z»)

converges uniformly to

j,(P)(z)

h(p) (C) de C-zp

= 0, 1, 2,

... , the sequence

on every compact subset K of C \ ras n -+ +00.

The proof is complete. Obviously, Lemma 2 holds if the contour function h (z), that is, if 1 h(z)=h(z)=-. A

2nl

f -de, h (C)

r

C-z

1 fh(C) h(z)=O=-. --de, 2nl C-z r

r

lies in the domain of analyticity of a

zEint(l), zEext(l).

A

Thus we have

1. Let 9'(z) be an arbitrary analytic function in the strip S = {z eC: -a < Im(z)
PROPOSITION

9' (z) = _1_. 2nl

J r

tp (C) dC, C-z

zEint(I')

be the Cauchy integral of 9'(z), where Further, let

r

is an arbitrary closed contour lying in S.

179


be the Riemann sum of IP(Z). Then for each p = 0,1, 2, ... , the sequence (qf"P)(z») converges uniformly to q/p)(z) on every compact subset ofint (l) as n -+ +00 (";0)

=11').

CoROLLARY l. Assume that the conditions of Proposition 1 hold. Put 12= R 11 int (I). Then the sequence ( IPn (x» converges to lP(x) in E(12) as n -+ +00.

Proof EvelY function IPn (x) is the restriction of the function IPn (z) on 12 and hence is an element of E(12). In addition, evelY compact in R is a compact in C.

Now we are ready to prove the main result of this section. THEoREM l. Let f(z) be the Cauchy integral of a distribution T e E'(R) and let lP(z) be

an arbitrary function analytic in the strip S ={z eC: -a < Im(z)
fr

t(z)fP(z)dz=_I_. 2n,

f

(T,.

r

_1_) fP(z)dz=(T, -fP(I». t-z

Proof Put 12= R 11 int (l) and let fP(x) be the restriction of lP(z) on R. Then the value of the restriction of T at an element II' e E(12) is equal to (T, 11'). Since the integral in (1) is a Riemann integral it can be approximated by the Riemann sums: J=!t(Z)fP(Z)dz= lim

~

,........ k=l

r

t(zk)fP(Zk)LJzk

We now exchange the integration and the application of T. Since T is linear we can write J= lim (Tt • _1_

~

IP(Zk)

2n i k=l t-Zk

,........

LJZk ).

In view of CorollaIY 1, and from the definition of continuity in E(R), we obtain J=

=

(r ( If ·

~

I m-.L,.--LJZk t.1 n .......

T" - . 2nl

1

2n I

r

IP (Zk)

A

)

k=l t-Zk

tp(z) d z ') =(T, -fP(l)) .

t-z

180

Chapter 13

The proof is complete. COROLLARY 2. Assume that the conditions of Theorem 1 hold. If ~(z);;: 1 for Z

J

fez) dz=_1_. 21/:1

r

J _1_) (T"

t-z

r

E

S, then

dz=(T, -I).

Observe that Theorem 1 and Corollary 2 generalize the residue theorem. EXAMPLE 1. Let c5(z) be the Cauchy integral of the Dirac distribution c5 e E'(R). Since by the defInition (0, fP) = fP(O) for fP e E(R), we have

J8(z) tp (z) dz = <<5, - tp (I»

= - tp (0).

r

On the other hand, we verify directly the same result:

J

1

<5 (z)tp (z) dz= - - . A

21/:1

r

J

tp(z) dz= -tp(O).

r

z

13.7. A Generalization of the Cauchy Integral Theorem for an Infinite Strip In the present section we consider analytic functions which are defined in a half plane by integrals over an infinite straight line (-OO+;a, +oo+;a), a> o. This type of analytic functions frequently appears in problems of mathematical physics. For example, let F(z)

be an analytic function in the open upper half plane

..1: = {z E C : Im(z) > a}

and

continuous in the closed half plane -:1: = {z E C : Im(z) ~ a}. If F(z) ~ 0 uniformly (with respect to arg (z), 0 ~ arg (z) ~ 1r) as Izi ~ +00, then the function F(z) has the representation

J

oo+;a

F(z)=_1_ 21/: i

For z

E

-oo+;a

F(C)

C-z

dC,

..1: ={z E C : Im( z) < a} this integral is zero.

In connection with functions of such type we shall put

. 1. =..1: u ..1: .

We need the following lemma. LEMMA 1 [8]. Let F(z) be an analytic function in the strip S

= {z E C : y, < Im(z) < Y2}

181


and let F(z) in S tend uniformly to zero as \z\--. +00. Then the function F(z) has the representation

J

~+~

(1)

Fz=_l_ ()

211:;

-~+~

J

~+~

F(C)dC __1_ C-z

211:i

-~+~

F(C)dC C-z

for YI < a < b < Y2. The representation (1) can be interpreted as the Cauchy integral formula for the infinite strip a S; 1m (z) S; b. If this strip is reduced to the strip -a S; 1m (z) S; a and if, in particular, z is located on R, then we have the representation

J

~-ia

_F(x)=_l_

(2)

211:;

-~-ia

-1

J

~+~

F(C)dC __1_ x-C

211:i

-~+Ia

F(C)dC. x-C

Our aim is to generalize the formula (2) in the sense of distributions in tJa'(R), a < 0. The proof of the main result (Theorem 1) is essentially based upon

S;

2. Let tp(z) be an analytic function in the strip -b S; 1m (z) S; b satisfying for every p = 0, 1, 2, ... the relation tp(P)(z) = O(\z\Cl), -1 S; a < 0. as \z\--. +00. Let A. be a positive real number and let < a < b. LEMMA

°

(i) If

J

A+ia

1 ((JA(Z)=-

(3)

A

211:i

-A+~

'1'(C) -

C-z

d"'0,

is the Cauchy integral of tp«() and if

is the Riemann sum of q,A(Z). then the sequence (q,A..,(X») converges to q,A(X) in tJa(R),

-1

S;

a < 0, as n --. +00.

(ii) If

(4)

~(z)=~ 211:'

f

.. +Ia

-oo+ia

'1'dC, C-z

is the Cauchy integral of tp( (). than the sequence ( q,A (x») converges to q,( x) in 0 a(R),

-1 S; a < 0, as A. --. +00 (over a sequence (A.n ) which approaches +00 as n --. +00).

182

Chapter 13

Proof (i). According to Lemma 2 in 13.6. we conclude that for every p the sequence

= 0,

(Vft,,!(z») converges uniformly to Vft>(z) on every compact of Li

as n ~ +00. This implies that the sequence

ll

1, 2, ...

=Li: u Li:

((p
(5)

converges uniformly to Vf:)(x) on every compact ofR as n ~ +00. On the other hand, for every p = 0, 1,2, ... , independent of n E N, we derive from (5) an estimate of the form

(x) I~ B". p I ;,(p) TA.n 1+lxl'

- oo
The previous analysis shows that the sequence

((PA,n(X») converges to (p).(x) in O_I(R) as

n ~ +00. Since the convergence in 0_ 1(R) implies the convergence in -1 S a < 0, the assertion follows at once. Observe that

°

a(R) for

(p).(x) is an element of 0_ 1(R), since the space 0a(R) is complete. Also,

all the derivatives of (p). (x) then belong to O_I(R). (ii) The integral (4) exists since the integrand behaves as I,,-l+a when I" ~ +00. Since rp(P)(t;) ~ 0 as I" ~ +00, the differentiation with respect to z and the partial

integration of the integral (4) lead to the equality

On the other hand, the application of these operations to the integral (3) leads to the formula

~
(6)

J

A+la

-A+ia

tp
C-z

fII(k) (J. + ia) ]. (J. + ia - z)P- k

The integral (4) being convergent, we can write

J

~+w

2ni

-oo+ia

fPCP) (C) dC = lim _1_

C- Z

A-..oo

2n i

J

A+w fPCP) (C) dC.

-A+w

C- z

183


Thus it does make sense to form the difference

~ (z) - ~A (z) =

J

-oo+ia

J'

9' (C)

C-z

dC + _1_.

-A

=_1_

211:i

J -J oo+ia

-A+/o

_1_. 211:1

9'(l+ia) (I+ia)-z

211:1

9' (C)

A+ia

dC

00

dt+-1

211:i

-00

C-z

A

9'(t+;a) (t+ia)-z

dt.

Now the relation 19'( f+ia) I s: Aol tl a, a < 0, implies the inequality

I~ (z) -~A (z) I~~ 211:

JI 1-I+adt+~J I

-A

00

t

211:

-GO

A

rIt GO

tl-l+adt=Ao

:rJ •

1- 1 +odt.

1

Since the last integral can be made arbitrarily small for sufficiently large A, independent of zELIa' we conclude that ~,t (z) converges uniformly to ~(z) on every compact of Lla

as..t-. +00. By considering the difference (p = 1, 2, ... )

it is easy to see, by means of the formula (6), that rlt)(z) converges uniformly to 9'(P)(z) on every compact of Lla as ..t -. +00. In particular, from this it follows that the sequence (9'Y) (x») converges uniformly to

= 0,

for every p

frfp) (x) on every compact of R as ..t -. +00. In addition,

1, 2, ... there exists a constant M p ' independent of A, such that M.

\ fPi p ) (x) I:i _ P - • A

1+JxJ

- oo
Consequently, ~,t(x) -. ~(x) in O_l(R) as ..t -. +00. This implies that ~A(X) -. ~(x) in oa(R) for -1 s: a < 0. The proof is complete. Now we can formulate the main result. THEOREM

1. Let T(z) be the Cauchy integral of a distribution TEO a'(R), -1 s: a < 0,

with T(z) =

o(~) Izl

as

Izl-. +00. Let

9'(z) be an analytic function in the strip

-b < Im(z) < b satisfying for every p = 0, 1, 2, ... the relation 9'(P)(z) = O(lzl~, -1 s: a < 0, as Izl -. +00. lfO < a < b, then

184

(7)

Chapter 13

~ 2n,

J

--/11

(TI' _1 ) tp(z)dz __1_ t-z

-_-III

2ni

J

_+/11

(TI' _1 ) tp(z)dz=(T, -tp(t». t-z

-_+/11

Proof. First of all observe that the operation (T, 'P) is well defined for a given a, -1 S a
J T(z)tp (z) dz,

-+/11

J1 =

--+/11

r

-+/11

J2 =

T(z)tp(z)dz,

---/11

First consider the integral J 1. It converges since If(z)'P(Z)1 =O(lzrl+a) on the straight line (~ia,+OO+ia) as Izl A+II/

J

(8)

~ +00.

Therefore we can write

T (z) tp (z) dz.

-A+ia

By approximating the integral in (8) by the Riemann sums we have J 1 = lim lim A......oo

_00

L T (Zk) tp (Zk) L1 Zk k=l n

A

(9)

· l'1m = 11m

.1...... 00 n-+oo

(T

~

qJ (Zk) A ) t' -1. £., --LJZk 2:n I k=l t - Zk

•

The distribution T being a continuous linear functional on ()a(R), -1 S a < 0, by a successive application of Lemma 2 ((i), (ii», from (9) it foUows

J

A+ia

J l = lim (Tt, A-+oo

_1_. 2:n I

-.1+ia

J

oo+ia

qJ(Z)dZ)=(Tt, t-z

_1_. 2:n I

-oo+ia

qJ(Z)dZ). t-z

185


In a similar manner, starting from the integral J 2 we get J2

=(Tt, 2nl ~

J

oo-ia

-oo-ia

rp(z)

t-z

dZ).

Consequently, J=J2 -J1 =

(Tt, 2nl

_1_.

J

""-ia

dz __2nl1_. t-z

rp(z)

-",,-ia

J

",,+ia

-",,+ia

rp(z)

t-z

dz) .

Taking into account Lemma 1 (formula (2» we obtain (with a change of the variables), that is

J= (T,

-,,(t»,

",,-ia

J

oo+ia

T(z)tp(z)dz-

-oo-ia

J

T(z)tp(z)dz=(T, -tp(t».

-",,+ia

The proof is complete. REFERENCES

1. H. 1. Bremermann and L. Durand: On Analytic Continuation. Multiplication and Fourier Transformations ofSchwartz Distributions. 1. Math. Phys. 2(1961), 240-258. 2. H. 1. Bremermann: Distributions. Complex Variables and Fourier Transforms. AddisonWesley, New York 1965.

3. 1. H. Curtiss: Introduction to Functions of a Complex Variable. Marcel Dekker, New York: 1978. 4. D. Mitrovic: The Plemelj Distributional Formulas. Bull. Amer. Math. Soc. 77(1971), 562-563. 5. D. Mitrovic: A Distributional Representation ofAnalytic Functions. Math. Balkanica 4(1974), 437-440. 6. D. Mitrovic: A Singular Convolution Equation in the Space ofDistributions. Publ. Inst. Math. (Beograd)11(1977),151-163. 7. D. Mitrovic: A Distributional Representation of Strip Analytic Functions. Intemat. 1. Math. Sci. 5(1982), 1-9. 8. A. Sveshnikov and A. Tichonov: The Theory of Functions of a Complex Variable. Mir Publishers, Moscow 1978.

Name Index The names are followed by the year of birth or the years of birth and death, if known to the authors. 176,177,178,179,180,181,183 Ceby§ev, P. L. (1821-1894) 64, 66, 67, 93, 101,102,108,109,148,159 Coates, R. (1682-1716) 133 Copson, E. T. (1901-1980) 95 Curtiss, J. H. 185

Abel, N. H. (1802-1829) 64 Abramovitz, M. (1915-1958) 82 Adamovic, D. D. (1928) 35,47, 60, 75, 112 A~iev, K. (1949) 114 Ahlfors, L. V. (1907) 38 Albrecht, R. 49 Alefeld, G. (1941) 145 Aramanovich, I. G. (1918) 45,60

Davis, Ph. J. (1923) 139 Dejon, B. 145 Dettman, 1. W. 166 Dieudonne, J. (19~1992) 22 Dimitrovski, D. S. (1934) 110, 112, 114, 117,128 Dirac, P. A M. (1902-1984) 169,170,180 Doelder, P. 1. de 1, 77 Dokovic, D. Z. (1938) 8 Dougall, J. (1867-1960) 77 Durand, L. 185

Bateman, H. (1882-1946) 77 Behnke, H. (1898-1979) 114 Bernoulli, 1. (1654-1705) 72,74,75 Bessel,F. W. (1784-1846) 9 Beumer, M. G. 57, 58 Bienaime, J. (179~1878) 66,67 Boas,M. 38 Boas, R. P. (1912-1992) 37,38 Bochner, S. (1899-1982) 149, 161 Bonnet, O. P. (1819-1892) 104 Bouniakowski, V. 1. (1804-1889) 67,68 Bo~,F. 57,58 Bremermann, H. 1. 185 B11rmann, H. 22, 23

Elliot, E. B. 39,40 Erdelyi, A (1908-1977) 77 Euler, L. (1707-1783) 21, 58, 83 Evgrafov, M. A (1926) 26,29

~ov,L.(188~1963) 23 Campbell, R. 84 Carleman, T. (1892-1949) 23, 24, 25, 149, 161 Casorati, F. (1835-1890) 94,95 Casteren, 1. van 58 Cauchy, A-L. (1789-1857) 3, 5, 18, 26, 27, 33, 51, 57, 58, 66, 67, 68, 75, 76, 78, 82, 100, 110, 112, 122, 133, 134, 136, 150, 157, 158, 163, 169, 171, 173, 174, 175,

Fejer, L. (1880-1959) 77 Fourier, J. (1768-1830) 57,138,185 Fresnel, A (1788-1827) 37,38 Friedmwn,E. 37,38 Frullani, G. (1795-1834) 39 Gakhov, F. D. (1~1980) 166 Gargantini, I. 141,145 Gauss, K. F. (1777-1855) 69,133, 135,136, 137

187

188

Name Index

149.161 B. (1864-1947) 18. 19 Gegenbauer. L. (1849-1903) 148. 157. 158. 159.160 Gel'fond. A. O. (1906-1968) 34 Geronimus. Ya. L. (1898) 148. 161 Gerretsen. J. (1911) 41 Gilles. J. 58 Gregory. M. B. 75 Grosjean. C. C. 77 GroSSlWlll. N. 80.81.82 Gupta. R. C. 9 Gau~.VV.(1927) Gavrilovi~.

Hardy. G. H. (1877-1947) 39.40 Hausen. E. R. 77 Heaviside. O. (1850-1925) 169 Heine. E. (1821-1881) 101. 102 Hemici. P. (1923-1987) 140.141. 145 Hermite. Ch. (1822-1901) 74.148 Henberger. J. 145 Billion. P. 139 Hurdin. G. 139

n'kov. L. V. 34 Jackson. D. (1888-1946) 77 Jacobi. K. G. J. (1804-1851) 14. 148. 156. 160 Jensen. J. L. VV. V. (1859-1925) 18.25 Jermolaeva. N. S. 68 Jordan. C. (1838-1922) 72. 108 Julia. G. (1893-1978) 14. 32 JuJkevi~. A. P. (1906) 67.68 Kaplan. E. L. 138 Keati~. J. D. (1945) 1.17.95.138.146 Kirsanov. V. S. 68 Kurepa. D. (1907) 56. 84, 85

Lagrange. J. L. (1736-1813) 19. 23. 100.

101

Laguecre,E.(1834-1886) 34,148 Landau, H. J. 149. 161 Laurent, H. (1841-1908) 93. 108 Laurent, P. A. (1813-1854) 3, 4. 94. 95. 97. 98, 101, 108, 112, 114, 115, 122, 124, 135 Lebesgue, H. (1875-1941) 167 Legendre. A. M (1752-1833) 93. 94. 103. 107,108,133,135,136,137,148,159 Liouville, J. (1809-1882) 80,164,165,176

Littlewood, R. K. 139 Lunts, G. (1911) 45,60 Lutzky, M. 57 Maclaurin, C. (1698-1746) 21 Mari~v, O. I. 92 MarkuJevi~.A.1. (1908-1979) 94.95 Marsden. J. E. 6 Marsh, D. C. B. (1926) 50 Mellin. Hj. (1854-1933) 84 Metzger, J. M 75 Milovanovi~, G. V. (1948) 147. 149, 161 Mitevski, S. 132 Mitrinovi~, D. S. (1908) 1, 50.64, 138. 146 Mitrovi~, D. (1922) 25, 56, 162, 167, 185 Mittag-LeIDer, G. (1846-1927) 26 Moore, R. E. 146 Muskhelishvili, N.I. (1891-1976) 166 Nevai, P. 147, 161 Newman, D. J. (1930) 21,22 Newton. I. (1643-1727) 133 Norfolk, T. S. 10

Ostrogradski, M. V. (1801-1861) 67.68 Painleve. P. (1863-1933) 12, 13, 31, 53, 55 Paterson, T. N. L. 139 Petkovi~, Lj. D. (1952) 146 Petkovi~. M. S. (1948) 140, 146 Piessens, R. 138, 139 Plemelj, J. (1873-1967) 174,175,185 Poleunis. F. 139 Poly&. G. (1887-1985) 34 P~tojanovi~, Z. R. (1935) 64 Rabinowitz, Ph. 139 Rajovi~, M. 112, 117 Riemann. B. (1826-1866) 57,61,62,63, 72, 177,178,179,181,184 R.och. G. (1839-1866) 62,63 Rodrigues, B. O. (1794-1851) 106 Romberg 133 Ross, B. 80,82 Rouche, E. (1832-1910) 22, 101, 102 Runge, K. D. T. (1856-1927) 19

Sansone, G. (1888-1979) 41 Schempp, VV. 22 Schoenberg, I. J. (1903Schwartz, L. (1915) 185

) 21,22

189

Name Index

Smith. B. T. 161 Sohocki, J. V. (1842-1927) 64, 67, 93, 94, 95,100,101,102,103,107,108 Sommer, F. (1912) 114 Stegun, I. A. (1919) 82 Stewart, I. 38 Stieltjes, T. J. (1856-1894) 177 Stirling, 1. (1696-1770) 21 Sveshnikov, A. (1911) 185 Szego, G. (1895-1985) 34, 148, 160, 161 Tall,D. 38 Taylor,B. (1685-1731) 88,91,103,142

Tchebychef, see Ceby§ev Tichonov, A. (1906) 185 Tisserand, F. (1845-1896) 12, 13,31, 53, 55 Titchmarsh, E. C. (1899-1963) 25, 95 Toii~, D. D. (1932) 1, 133, 138, 139 Toii~, D. V. 139

Turin. P. (1910-1976) 79 Vietoris, L. (1891) 79 Vitali, G. (1875-1932) 177

Volkovysky, L. (1913) 45,60

Wang, E. T. H 37 Watson, G. N. (1886-1962) 14 Weber, H R. 48 Wei~,lC.(1815-1897) 18,94,95,177 Whittaker, E. T. (1873-1956) 14 Wong,R. 139 Young, R. L. 37 Zakian, V. 139

Zuser, lC. 49 Zwier,P. 75

Subject Index division of discs 141

Additionofdiscs 141 lIII8lytic continuation 88, 90 - of Cauchy type integrals 26 - of Mittag-leIDer's function 27 - principle 164 asymptotic formula for Mittag-leIDer's function 29

Elliptic integral 59 essential singularity, order of 14 Euler-Maclaurin's formula 21 Euler's constant 57 Euler's formula 82 evaluation of residues - at a pole 4, 5, 8, 9 - at a regular point 4 - at a removable singularity 4 - at infinity 4

Basic functions 167 Bernoulli's numbers 71,73,74 Bessel's function 9 Bonnet's formula 104 BOrmann's series 22, 23

F~~-]~sinequality

Carleman's theorem 23, 24 Carleman's formula 25 Casorati-Sohocki's theorem 94,95 Cauchy type integral 26 - analytic continuation 26 Cauchy's formula (principle of the argument) 18 Cauchy's integral formula 163, 181 Cauchy's integral of a distribution 171, 173, 174,176,177,178,179,180,181,183 Cauchy's residue theorem 5, 19,50,56,57, 75,77,81,134,136 Cauchy's theorem 3, 150, 157 circular inclusive extension 142 circular complex function 141 compact 167 compact support 147,167 continuous fractions 64,65, 101, 102, 108

76

finite functions 167 Fourier's coefficient 138 Fresnel's integrals 36, 37 Frullani's formula 38 functional 167, 168 Gamma function 57, 81, 82, 83, 86 Gauss' sums 68 GaUS&-Legendre's method 133, 135, 136, 137 Gegenbauer polynomials 158 Gegenbauer weight 158 generalized Liouville's theorem 165, 176 generalized value of an improper integral 118, 119, 120, 121, 122, 123, 128, 130

Heaviside's function 169 Hermite's formula 73 Hermitian symmetry 147 highly oscillating function 133, 137 hyperelliptic integral 59,62

Dirac's distribution 169,170,171,180 disc arithmetic 140 discontinuous factor 39 distribution 167, 168

190

191

Subject IDdex

Improper integral. generalized value 118. 119.120.121.122.123.124.126.127. 131 -principal value 110.112.114.115.116. 117.119.120.121.123.124.126.127. 131 -value 117 including approximation 142 inclusive calculus of residues 140 inclusive isotone function 142 inclusive isotonicity 141 infinite support 147 inner product 147 interval arithmetic 140 inversion of a disc 141 Lagrange's expansion 23 Lagrange's formulas for inverse functions 100.101 Lagrange's interpolation formula 19 left factorial function 83.84 Legendre's polynomials 103. 107. 108 Liouville's theorem 164 - generalized 165. 176 locally analytic function 162. 173 locally integrable function 167

Mittag-Leffier's function 26 - analytic continuation 27 - asymptotic formula 29 multiplication of discs 141 Newton-Coates' formulas 133 numerical evaluation of definite integrals 133 Order of essential singularity 18 orthogonal polynomials 147 orthogonality 147 - on a curve 148 - on the semicircle 149 - on the unit circle 148

Plemelj's formulas 174.175 polygamma functions 79 polylogarithm of order n 90

practical computation of residues 9. 10 principal value distribution 169 principal value of an improper integral 110. 112.114.115.116.117.119.120.121. 123.124.126.127.131.158 principle of the argument. generalization 18. 19 Regular distribution 169 residue. at a pole 4. 5 - at a regular point 4 - at a removable singularity 4 - at infinity 4 - definition 4 - of Jacobi's elliptic functions 14 - practical computation 9. 10 - with respect to a line 100 residue theorem 20. 22. 34. 48. 86. 180 Riemann-Roch theorem 62 Riemann sum 177.179.181.184 Riemann surface 60. 61 Riemann zeta function 56. 71 Rodrigues' formula 106 Romberg's method 133 Roucbe's theorem 22 Runge's phenomenon 19

Sectionally analytic functions 162. 173 singular distribution 169 Sohocki's theorem 94 Sohocki-Weierstrass'theorem 94 Stirling's formula 21 subtraction of discs 141 support of a function 162. 167 symmetry 147 - Hermitian 147 Test functions 167 Weierstrass' theorem 94 weight function 149 - distribution of zeros 160 - of Gegenbauer 158 - of Jacobi 156. 157 Zeta function 56

Other Mathematics and Its Applications titles of interest J. Chaillou: Hyperbolic Differential Polynomials and their Singular Perturbations. 1979, 184 pp. ISBN 90-277-1032-5 S. Fucik: Solvability 0/ Nonlinear Equations and Boundllry Value Problems. 1981, 404 pp. ISBN 90-277-lOTI-5 V.I. Istratescu: Fixed Point Theory. An Introduction. 1981,488 pp. out o/print. ISBN 90-277-1224-7 F. Langouche, D. Roekaerts and E. Tirapegui: Functional Integration and Semiclassical Expansions. 1982,328 pp. ISBN 90-277-1472-X N.E. Hurt Geometric Quantization in Action. Applications of Harmonic Analysis in Quantum Statistical Mechanics and Quantum Field Theory. 1982,352 pp. ISBN 90-277-1426-6 F.H. Vasilescu: Analytic Functional Calculus and Spectral Decompositions. 1983,

392 pp.

ISBN 90-277-1376-6

W. Kecs: The Convolution Product and Some Applications. 1983,352 pp. ISBN 90-277-1409-6 C.P. Bruter, A. Aragnol and A. Lichnerowicz (eds.): Bifurcation Theory. Mechanics and Physics. Mathematical Developments and Applications. 1983, 400 out o/print. ISBN 90-277-1631-5 pp. J. Aczel (ed.): Functional Equations: History. Applications and Theory. 1984,256 pp. out o/print. ISBN 90-277-1706-0 P.E.T. Jorgensen andR.T. Moore: Operator Commutation Relations. 1984,512 pp. ISBN 90-277-1710-9

D.S. Mitrinovic and JD. Keckic: The Cauchy Method 0/ Residues. Theory and ISBN 90-277-1623-4

Applications. 1984,376 pp.

R.A. Askey, T.H. Koomwinder and W. Schempp (eds.): Special Functions: Group Theoretical Aspects and Applications. 1984,352 pp. ISBN 90-277-1822-9 R. Belbnan and G. Adomian: Partial Differential Equations. New Methods/or their Treatment and Solution. 1984,308 pp. ISBN 90-277-1681-1 S. Rolewicz: Metric linear Spaces. 1985,472 pp.

ISBN 9O-2TI-1480-O

Y. Cherruault Mathematical Modelling in Biomedicine. 1986,276 pp. ISBN 90-277-2149-1 R.E. Bellman and R.S. Roth: Methods in Approximation. Techniques/or Mathematical Modelling. 1986,240 pp. ISBN 90-277-2188-2 R. Belbnan and R. Vasudevan: Wave Propagation. An Invariant Imbedding Approach. 1986,384 pp. ISBN 90-277-1766-4 ISBN 9O-2TI-2103-3 A.G. Ramm: Scattering by Obstacles. 1986,440 pp.

Other MathemLltics and Its Applications tides of interest

c.w. Kilmister (ed.): Disequilibrium and Self-Organisation. 1986,320 pp.

ISBN 90-277-2300-1

A.M. Krall: Applied Analysis. 1986, 576 pp. ISBN 90-277-2328-1 (hb), ISBN 90-277-2342-7 (Pb) J.A. Dubinskij: Sobolev Spaces of Infmite Order and Differential Equations. 1986, 164 pp. ISBN 90-277-2147-5 H. Triebel: Analysis and MathemLltical Physics. 1987,484 pp. ISBN 90-277-2077-0 B.A. Kupershmidt Elements of Superintegrable Systems. Basic Techniques and Results. 1987,206 pp. ISBN 90-277-2434-2 M. Gregus: Third Order Linear Differential Equations. 1987, 288 pp. ISBN 90-277-2193-9 M.S. Birman and M.Z. Solomjak: Spectral Theory of Self-AdjOint Operators in Hilbert Space. 1987,320 pp. ISBN 90-277-2179-3 V.L Isttatescu: Inner Product Structures. Theory and Applications. 1987,912 pp. ISBN 90-277-2182-3 R. Vich: ZTransform Theory and Applications. 1987,260 pp.

ISBN 90-277-1917-9 N.V. Krylov: Nonlinear Elliptic and Parabolic Equations of the Second -Order. 1987, 480.pp. ISBN 90-277-2289-7 W.I. Fushchich and A.G. N'ddtin: Symmetries of Maxwell's Equations. 1987.228 pp. ISBN 90-277-2320-6 P.S. Bullen, D.S. Mitrinovic and P.M. Vasic (eds.): Means and their Inequalities. 1987,480 pp. ISBN 90-277-2629-9 V.A. Marchenko: Nonlinear Equations and Operator Algebras. 1987, 176 pp. ISBN 9O-277-2654-X YuL. Rodin: The RiemLlnn Boundary Problem on RiemLlnn SUrfaces. 1988,216 pp. ISBN 90-277-2653-1 A. Cuyt (ed.): Nonlinear Numerical Methods and Rational Approximation. 1988, 480 pp. ISBN 90-277-2669-8 D. Przeworska-Rolewicz: Algebraic Analysis. 1988,640 pp. ISBN 90-277-2443-1 V.S. Vladimirov. YU.N. Drozzinov and B.I. Zavialov: Tauberian Theorems for Generalized Functions. 1988.312 pp. ISBN 90-277-2383-4 G. Morosanu: Nonlinear Evolution Equations and Applications. 1988. 352 pp. ISBN 90-277-2486-5

Other Mathematics and Its Applications titles of interest: A.F. Filippov: Differential Equations with Discontinuous Righthand Sides. 1988. 320 pp. ISBN 9O-277-2699-X A.T. Fomenko: Integrability and Nonintegrability in Geometry and Mechanics. 1988.360 pp. ISBN 90-277-2818-6 G. Adomian: Nonlinear Stochastic Systems Theory and Applications to Physics. 1988.244 pp. ISBN 90-277-2525-X A. Tesar and Ludovt Fillo: Transfer Matrix Method. 1988.260 pp. ISBN 90-277-2590-X A. Kaneko: Introduction to the Theory of Hyperfunctions. 1989.472 pp. ISBN 90-277-2837-2 O.S. Mitrinovic. J.E. Pecaric and V. Volenec: Recent Advances in Geometric ISBN 90-277-2565-9

Inequalities. 1989. 734 pp.

A.W. Leung: Systems of Nonlinear PDEs: Applications to Biology and EngineerISBN 0-7923-0138-2 ing. 1989.424 pp. N.E. Hurt: Phase Retrieval and Zero Crossings: Mathematical Methods in Image ISBN 0-7923-0210-9 Reconstruction. 1989.320 pp. V.L Fabrikant: Applications of Potential Theory in Mechanics. A Selection of New Results.1989.484pp. ISBN 0-7923-0173-0 R. Feistel and W. Ebeling: Evolution of Complex Systems. Sel!organization. Entropy at:ld Development. 1989.248 pp. ISBN 90-277-2666-3

S.M. Ermakov. V.V. Nekrutkin and A.S. Sipin: Random Processes for Classical Equations of Mathematical PhySics. 1989.304 pp. ISBN 0-7923-0036-X B.A. Plamenevskii: Algebras of Pseudodifferential Operators. 1989. 304 pp. ISBN 0-7923-0231-1 N. Bakhvalov and G. Panasenko: Homogenisation: Averaging Processes in Periodic Media. Mathematical Problems in the Mechanics of Composite MaterialS.

1989.404 pp.

ISBN 0-7923-0049-1

A.Ya. Helemskii: The Homology of Banach and Topological Algebras. 1989.356 pp. ISBN 0-7923-0217-6 M. Toda: Nonlinear Waves and Solitons. 1989.386 pp.

ISBN 0-7923-0442-X

M.I. Rabinovich and 0.1. Trubetskov: Oscillations and Waves in Linear and ISBN 0-7923-0445-4

Nonlinear Systems. 1989.600 pp.

A. CrumeyroUe: Orthogonal and Symplectic Clifford Algebras. Spinor Structures. 1990. 364 pp. ISBN 0-7923-0541-8 V. Goldshtein and Yu. Reshetnyak: Quasiconformal Mappings and Sobolev ISBN 0-7923-0543-4

Spaces. 1990.392 pp.

Other Mathematics and Its Applications titles of interest

I.H. Dimovski: Convolutional Calculus. 1990, 208 pp. ISBN 0-7923-0623-6 Y.M. Svirezhev and V.P. Pasekov: Fundamentals of Mathematical Evolutionary Genetics. 1990, 384 pp. ISBN 90-277-2772-4 S. Levendorskii: Asymptotic Distribution of Eigenvalues of Differential Operators. 1991, 297 pp. ISBN 0-7923-0539-6 V.G. Makhankov: Soliton Phenomenology. 1990,461 pp. ISBN 90-277-2830-5 I. Cioranescu: Geometry of Banach Spaces, Duality Mappings and Nonlinear Problems. 1990,274 pp. ISBN 0-7923-0910-3 Bl. Sendov: Hausdorff Approximation. 1990, 384 pp. ISBN 0-7923-0901-4 A.B. Venkov: Spectral Theory of Automorphic Functions and Its Applications. 1991,280 pp. ISBN 0-7923-0487-X V.I. Arnold: Singularities of Caustics and Wave Fronts. 1990,274 pp. ISBN 0-7923-1038-1 A.A. Pankov: Bounded and Almost Periodic Solutions of Nonlinear Operator Differential Equations. 1990,232 pp. ISBN 0-7923-0585-X A.S. Davydov: Solitons in Molecular Systems. Second Edition. 1991,428 pp. ISBN 0-7923-1029-2 B.M. Levitan and I.S. Sargsjan: Sturm-Uouville and Dirac Operators. 1991,362 pp. ISBN 0-7923-0992-8 V.L Gorbachuk and M.L. Gorbachuk: Boundary Value Problems for Operator Differential Equations. 1991,376 pp. ISBN 0-7923-0381-4 Y.S. Samoilenko: Spectral Theory of Families of Self-Adjoint Operators. 1991,309 pp. ISBN 0-7923-0703-8 Bl. Golubov A.V. Efimovand V.A. Scvortsov: Walsh Series and Transforms. 1991,382 pp. ISBN 0-7923-1100-0 V. Laksmikhantam, V.M. Mattosov and S. Sivasundaram: Vector Lyapunov Functions and Stability Analysis ofNonlinear Systems. 1991,250 pp. ISBN 0-7923-1152-3 F.A. Berezin and M.A. Shubin: The SchriJdinger Equation. 1991, 556 pp. ISBN 0-7923-1218-X D.S. Mibinovic, J.E. Pecaric and A.M. Fink: Inequalities Involving Functions and their Integrals and Derivatives. 1991, 588 pp. ISBN 0-7923-1330-5 Julii A. Dubinskii: Analytic Pseudo-Differential Operators and their Applications. 1991,252 pp. ISBN 0-7923-1296-1 V.I. Fabrikant: Mixed Boundary Value Problems in Potential Theory and their Applications. 1991,452 pp. ISBN 0-7923-1157-4

Other Mathematics and Its Applications titles of interest A.M. Samoilenko: Elements of the Mathematical Theory of Multi-Frequency Oscillations. 1991,314 pp. ISBN 0-7923-1438-7 YuL. Dalecky and S.V. Fomin: Measures and Differential Equations in InfiniteDimensional Space. 1991,338 pp. ISBN 0-7923-1517-0 W. Mlak: Hilbert Space and Operator Theory. 1991,296 pp. ISBN 0-7923-1042-X NJ. Vilenkin and A.U. KJimyk: Representations of Ue Groups and Special Functions. Volume 1: Simplest Ue Groups, Special Functions, and Integral Transforms. 1991, 608 pp. ISBN 0-7923-1466-2 K. Gopalsamy: Stability and Oscillations in Delay Differential Equations of PopUlation Dynamics. 1992, 502 pp. ISBN 0-7923-1594-4 N.M. Korobov: Exponential Sums and their Applications. 1992,210 pp. ISBN 0-7923-1647-9 Chuang-Gan Hu and Chung-Chun Yang: Vector-Valued Functions and their Applications. 1991, 172 pp. ISBN 0-7923-1605-3 Z. Szmydt and B. Ziemian: The Mellin Transformation and Fuchsian Type Partial Differential Equations. 1992,224 pp. ISBN 0-7923-1683-5 L.I. Ronkin: Functions of Completely Regular Growth. 1992, 394 pp. ISBN 0-7923-1677-0 R. Delanghe, F. Sommen and V. Soucek: Clifford Algebra and Spinor-valued Functions, A Function Theory of the Dirac Operator. 1992, 486 pp. ISBN 0-7923-0229-X A. Tempelman: Ergodic Theoremsfor Group Actions. 1992,400 pp. ISBN 0-7923-1717-3 D. Bainov and P. Simenov: Integral Inequalities and Applications. 1992, 426 pp. ISBN 0-7923-1714-9 I. Imai: Applied Hyper/unction Theory. 1992,460 pp. ISBN 0-7923-1507-3 Yu.I. Neimark and P.S. Landa: Stochastic and Chaotic Oscillations. 1992,502 pp. ISBN 0-7923-1530-8 H.M. Srivastava and R.G. Buschman: Theory and Applications of Convolution Integral Equations. 1992,240 pp. ISBN 0-7923-1891-9 A. van der Burgh and J. Simonis (eds.): Topics in Engineering Mathematics. 1992, 266 pp. ISBN 0-7923-2005-3 F. Neuman: Global Properties of Unear Ordinary Differential Equations. 1992, 320 pp. ISBN 0-7923-1269-4 A. Dvurecenskij: Gleason's Theorem and its Applications. 1992,334 pp. ISBN 0-7923-1990-7

Other Mathematics and Its Applications titles of interest:

D.S. Mittinovic. I.E. Pecaric and A.M. Fink: Classical and New Inequalities in Analysis. 1992. 740 pp. ISBN 0-7923-2064-6 H.M. Hapaev: Averaging in Stability Theory. 1992.280 pp. ISBN 0-7923-1581-2 S. Gindinkin and L.R. Volevich: The Method

Theory of PDE' s. 1992.276 pp.

of Newton's Polyhedron in the

ISBN 0-7923-2037-9 Yu.A. Mitri>polsky. AM Samoilenko and DJ. Martinyuk: Systems of Evolution Equations with Periodic and Quasiperiodic Coefficients. 1992. 280 pp. ISBN 0-7923-2054-9 LT. Kigmadze and T.A. Chantmia: Asymptotic Properties of Solutions of Nonautonomous Ordinary Differential Equations. 1992.332 pp. ISBN 0-7923-2059-X

The Cauchy Method of Residues (Vol. 2)

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