1

Chapter 7 Applications of Residues - evaluation of definite and improper integrals occurring in real analysis and applied math - finding inverse Laplace transform by the methods of summing residues.

60. Evaluation of Improper Integrals In calculus

R

0 0( ) ( )lim

R

f x dx f x dx

when the limit on the right exists, the improper integral is said to converge to that limit.

2

2

11 2

0

0( ) ( ) ( ) (2)lim lim

R

RR R

f x dx f x dx f x dx

• If f(x) is continuous for all x, its improper integral over the

x is defined by

When both of the limits here exist, integral (2) converges to their sum.

• There is another value that is assigned to integral (2). i.e., The Cauchy principal value (P.V.) of integral (2).

R

-R. . ( ) ( ) (3)lim

R

P V f x dx f x dx

provided this single limit exists

3

If integral (2) converges its Cauchy principal value (3) exists.

If is not, however, always true that integral (2) converges when its Cauchy P.V. exists.

Example. (ex8, sec. 60)

2

11 2 1 2

1 2

2 20

0

2 21 2

R

0 2

01

( )

2 2

= ( ) ( )2 2

neither of these limits exist

On the other hand,

P.V.

lim lim lim lim

lim lim

R

RR R R R

R

R

R

f x x

x xxdx xdx xdx

R R

xdx

2 2 2

( ) 02 2 2 lim0lim lim lim

R

RR R R R

R

R

x R Rxdx

4

0

R

0 0

But suppose that ( ) (- < < ) is an even function,

( ) ( ) for each

1( ) ( )

21

( ) ( ) ( )2lim lim

R R

R

R

RR R

f x x

f x f x x

f x dx f x dx

f x dx f x dx f x dx

converge

1to P.V.

2

if exist

(1) (3)

1

1

2

1

0

0

0

0

Moreover, ( ) ( )

( ) lim lim lim ( )

R

R

R R

R R

f x dx f x dx

f x dx f x dx

5

• To evaluate improper integral of

( )( )

( )

p xf x

q x

p, q are polynomials with no factors in common.

and q(x) has no real zeros.

0

when ( ) is even

2 ( ) ( ) . . ( )

f x

f x dx f x dx PV f x dx

See example

6

Example2

60

1

xdx

x

2

6( )

1

zf z

z

has isolated singularities at 6th roots of –1.

and is analytic everywhere else.

Those roots are

56 6

0 1 2

R

0 1 2-R

2exp ( ) 0,1,2,3,4,5

6 6

, C lie in the upper half plane

when R>1

( ) ( ) 2 ( )R

k

i i

C

kC i k

C e C i e

f x dx f z dz i B B B

Residues of ( )

at ( =0,1,2)k

f z

C k

RC

-R R

2C 1C

0C

7

22

k 5 36

0

1 0 1 2

2

are simple poles,

1B ( =0,1,2)Re 1 6 6

1B

61

2 ( )6 3

1

6

K

k

k

Z C k k

C

Czks z C C

i

B i B B Bi

Bi

R

-R ( ) ( )

3 RCf x dx f z dz

8

R

-R

2 2

66 6

2 2

66

2

6

2

6R

( ) ( )3

when

1 1 1

for on

( )11

( ) 0 as 1

lim or1 3

R

R

C

R

C

R

R

f x dx f z dz

z R

z R

z z R

z C

z Rf z

Rz

Rf z dz R R

R

xdx

x

2 2

6 6- 0 P.V.

1 3 1 6

x xdx dx

x x

9

61. Improper Integrals Involving sines and cosines

To

evaluate

( ) sin or ( ) cos ( 0)f x ax dx f x ax dx a

Previous method does not apply since

sin , cos as az az y

sinhay (See p.70)

However, we note that

R

-R( ) cos dx+ ( ) sin dx= ( )

is bounded in 0

R R iax

R R

iaz iax ay

f x ax i f x ax f x e dx

e e e y

10

Ex1.

2 2 3

cos3 2 =? ( )

( 1)

xdx

x e

An even function

2 2

1Let ( )

( 1)f z

z

And note that is analytic everywhere on and above the real axis except at

3( ) i zf z e.z i

33

12 2 2 ( )

( 1) R

i xR i z

R C

edx iB f z e dz

x

, ( 1)z R R

iR

iRC

11

Take real part

32 2 3

3 32 2

3 32 2

cos3 2 Re ( )

( 1)

1but ( ) , 1

( 1)

1 Re ( ) ( ) 0 as

( 1)

R

R R

R i z

R C

i z y

i z i z

C C

xdx f z e dz

x e

f z e eR

f z e dz f z e dz R RR

31

3

2

2 2

33

1 3 2

33 3

Re ( )

( ) ( ) ( ) pole of order 2

( ) ( )

2 3 '( )

( ) ( )

2 3 ( ) 1 = e

( )

i z

z i

i z

i zi z

z i

z i

i z

z i

B s f z e

ez z i

f zz i z i

i eB z e

z i z i

i z i

z i ie

12

It is sometimes necessary to use a result based on

Jordan’s inequality to evaluate ( )cos ,..........f x ax dx

sin

0 < (R>0)

RRe d

2 sin

2 sin2 2

0 0

2pf: sin when 0

2 for 0

, 02

So 12

<2

since sin is symmetric w.

RR

RR R

R

e e

e d e d eR

R

y

sin

0

r.t. =2

( 0)Re d RR

y

siny

2

y

2

13

Suppose f is analytic at all points

00 and above

(0 )

iy z R e

0R

RC

R

0R

0

Let

denote

e (0 ),

R

i

C

z R R R

0

sin cos

sin

0

( ) ( e ) exp ( e ) e

Since we also have ex

If ( e ) , and 0 as

p( e )

( ) 0 as

lim ( )

R

R

R

iaz i i i

C

i aR iaR

iaz aR RRC

ia

C

iR

R

R

f z e dz f R iaR iR d

iaR e e

Mf z e dz M R e d R

a

f z

f

e

R M M R

0 (a>0) Jordan's m a Le mzdz

14

Example 2.

2

sin

2 2

x x dx

x x

Sol:

1

21 1

1

1

( 1 ) 11

1

1 1

( )2 2 ( )( )

1

is a simple pole of ( )

with residue

( 1 ) ( 1 ) (cos1 sin1)

( 1 ) ( 1 ) 2

1 = ( 1)(cos1 sin1)

2

iz

iz i i

z zf z

z z z z z z

z i

z f z e

z e i e i e iB

i i iz z

i ie

1 = cos1 sin1 (cos1 sin1)

2ei

RC

R

1Z

1Z 2

15

12

12

when > 2

2 ( )

2 2 sin

Im(2 B ) Im ( )2 2

R

R

ixR iz

R C

R iz

R C

R

x e dxi B f z e dz

x xx x dx

i f z e dzx x

21 1

21

1

But Im ( ) ( )

( 2) ( )

2( 2)

2

and 1

R R

iz iz

C C

iz y

f z e dz f z e dz

z R

Rz z z zR

f zz z RR

z z R

e e

16

But from Jordan’s Lemma

12-R

( ) 0 as

sin lim ( ) 0 P.V. Im(2 )

2 2

= (sin1 cos1)e

R

iz

C

f z R

x x dxf z e dz i B

x x

2

2( ) does not tend to zero as

( 2)R

iz

C

Rf z e dz R

R

17

62. Definite Integrals Involving Sines and Cosines

2

0

-1 -1

1 1

C

To evaluate (sin ,cos )

: 0 2 consider an unit circle : , 0 2 ,

- sin cos

2 2

( , ) parametric form2 2

i

F d

C z e

z z z z dzd

i iz

z z z z dzF

i iz

2

20

2Example : (-1< <1)

1 sin 1 0, trivial

suppose 0

da

a aa

a

18

2

2

0 2

22

1 2

2 2

2

2

1( ) 2

1 sin 12 2

2 /

21 sin 1

42 4

2 2 1poles ,

2 2

1 1 1 ( )

1 1 >11

<1

C

a z iz az azai iz

d adz

ia z za

ii aa az z

a

i i a ai

a a

az

a

C

1Z 1

19

1 2 1

1 2

1 1 21 2

12 2C

but 1, 1

( ) 2 / ( ) , where ( )

-

2 / 1 ( )

12 / 2

2 2( ) -1 1

z z z

z af z z

z z z z

aB z

z z i aa

dz i Biz z aa

20

63. Indented Paths

0

0 2 0

If a function ( ) has a simple pole at a point

on the real axis, with a Laureut series representation

in a punctured disk 0 - and with residue .

If denotes the upper half of a circle

f z z x

z x R B

C z

0

2

- ,

where < , and negative oriented, then

x

R

00

00

lim ( )

lim ( ) 2

C

C

f z dz B i

f z dz i B

y

x0

C

0X

2R C

21

00

0

0 2

pf: ( ) can be written as

( ) ( ) , 0

(0 - )

f z

Bf z g z B

z x

z x R

0 2continuous in - , not sure if it is analytic.z x R

00

0 2

( ) ( )

since ( ) is continuous, there is a positive

such that ( ) for all -

C C C

dzf z dz g z dz B

z x

g z M

g z M z x R

0( )

0

00 0

00

( ) 0 as 0

-C : (0 )

lim ( )

C

i

i

iC C

C

g z dz M

z x e

dz dz ie di

z x z x e

f z dz B i

22

2L

C

01L R

RC

Ex1.

0

sinTo show

2

xdx

x

Consider a simple closed contour

1 2

1 2

1

2

01

2

,

:

:

From Cauchy-Goursat theorem

0

or

Since : ( r )

:

R

R

iz iz iz iz

L C L C

iz iz iz iz

L L C C

i

R

L x R

L R x

e e e edz dz dz dz

z z z z

e e e edz dz dz dz

z z z z

L z re r R

L z re

( r )i r R

23

1 2

( )( )

sin 2

sin2

R

iz iz ir irR R

L L

ir irR R

iz izR

C C

e e e edz dz dr dr

z z r r

e e rdr i dr

r r

r e ei dr dz dz

r z z

2

2 32

1 ( )Now 1 ....

1! 2!

1 ...... 0

1! 2! 3!

ize iz iz

z z

i i iz z z

z

0

0

has a simple pole at 0, with residue 1

lim

1 1 1also 0 as

1 sinFrom (9) in sec. 61, lim 0

2R

iz

C

R

iz

CR

z

edz i

z

M Rz z R

re dz dr

z r

Jordan’s Lemma

24

64. Integrating Along a Branch Cut

ln

0

Consider , where 0, and 0 1.

Let it denote the principal vabue of

i.e.,

shall evaluate (0 1)1

important in gamma function study

a

a

a a x

a

x x a

x

x e

xdx a

x

log

ln Log

is multivalueda xe

(P.81, complex exponent)

-

-

- -1

( has an infinite discontinuity at =0.1

The integral converges when 0<a<1 since the

integrand behaves like near 0, and like

as tends to infinity)

a

a

a

xx

x

x x

x x

25

Begin by :

:

consider the branch

( ) ( 0, 0 arg 2 )1

of the multiple-valued function

, with branch cut arg =01

Since it is piecewise continuous o

R

a

a

C z

C z R

zf z z z

z

zz

z

R

C

n C and C ,

then integrals

( ) and ( ) exist

exp (ln )exp( log )Write ( ) , ( )

1 1and use =0 and =2 along the upper and lower "edges", respectively.

RC

ii

f z dz f z dz

a r ia zf z z re

z re

RC

C

1

R

1 R

26

Then

0

2i2

2

1

exp (ln 0 for z , ( )

1 1exp (ln 2

for =re , ( )1 1

from residue thm,

( ) ( )1 1

2 Re ( )

( )

R

ai

a i a

a a i aR R

C C

z

a r i rre f z

r ra r i r e

z f zr r

r r edr f z dz dr f z dz

r ri s f z

f z

is not defined on the branch cut involved, but ......see Ex.9

Note that ( ) exp( log ) ( 0, 0 arg 2 )

is analytic at -1 and

( 1) exp (ln1 )

a

ia

z z a z z z

z

a i e

27

1

2

1

This shows that -1 is a simple pole of the function ( )

and Re ( )

( ) ( ) 2 ( 1) 1

2but ( ) 2 0 as 0

1 1

and ( )

R

ia

z

aRia i a

C C

aa

C

z f z

s f z e

rf z dz f z dz ie e dr

r

f z dz

f z d

-R

20

-

0

2 12 0 as

1 1

2 2also lim

1 1

1 sin

R

a

aC

a ia

i a ia ia

R

a

R Rz R R

R R R

r ie idr

r e e e

xdx

z a

28

65. Argument Principle and Rouche’s Theorem

A function f is said to be meromorphic in a domain D if it is analytic throughout D - except possibly for poles.

Suppose f is meromorphic inside a positively oriented simple close contour C, and analytic and nonzero on C.

The image of C under the transformation w = f(z),

is a closed contour, not necessarily simple, in the w plane.

y z

0z

C

x

v

u

0w

w

29

As a point traverses in the positive direction, its image

traverses in a particular direction that determines the

orientation of .

Note, since has no zeros on , the contour does not

pass

z C w

f C

0 0 0 0

0

0

through the origin in the plane.

Let and be points on , where is fixed and arg .

Let arg vary continuously, starting with , as begins

at and traverses

w

w w w w

w w w

w

0

0 1

1 0

once.

When returns to the starting point , arg assumes

a particular value of arg , .

Thus the change in arg as describes once is - .

w w w

w

w w

30

0

1 0

0

1 0

This change is independent of .

is in fact the change in arg ( ) as

describes once, starting at .

arg ( ) .

a multiple of 2

1The integer arg (

2

C

C

f z z

C z

f z

f

) represents the

number of times the point winds around the

origin in the plane.

called winding number of w.r.t. the origin =0.

z

Positive:

Negative:

winding number=0 when does not enclose =0

31

The winding number can be determined from the number of

zeros and poles of f interior to C.

Number of poles

zerosare finite

(Ex 15, sec. 57)

(Ex 4)

Argument principle

Thm1. is meromorphic inside a simple closed, positively oriented

Contour , and is analytic and nonzero on .

If, counting multiplicities, is the number of

zeros

f

C C

Z

and is the number of poles inside ,

1 then arg ( ) .

2 C

P C

f z Z P

32

Pf.

( )

( )

Let be ( ) (a t b)

' ( ) '( )'( )

( ) ( )

The image of ( ) can be expressed as

= (t) ( ( ) 0)

Thus ( ) ( ) (0 )

and

' (

b

C a

i t

i t

C z z t

f z t z tf zdz dt

f z f z t

z t

e f z

f z t t e t b

f z t

( )

( ) ( )

) '( ) ( ) ( )

'( ) ( ) '( )

'( ) '( )'( )

( ) ( )

ln ( ) ( )

i t

i t i t

b b

C a a

d dz t f z t t e

dt dt

t e i t e t

f z tdz dt i t dt

f z t

b bt i t

a a

33

But ( ) ( ) and ( ) ( ) arg ( )

'( ) arg ( )

( )

'( )Another way to evaluate

( )

'( )Since is analytic inside and on except at

( )

points inside at which zeros

C

CC

C

b a b a f z

f zdz i f z

f z

f zdz

f z

f zC

f z

C

0

0 0

0 0

0

10 0 0

0

0

and poles of occur.

If ( ) has a zero of order at , then

( ) ( ) ( )

Hence '( ) ( ) ( ) ( ) '( )

'( ) '( ) or

( ) ( )

m

m m

f

f z m z

f z z z g z

f z m z z g z z z g z

mf z g z

f z z z g z

0analytic, nonzero at .z

34

0

0 0

'( )Since is analytic at , it has Taylor series representation

( )

about that point.

'( ) has a simple pole at , with residue .

( )

On the other hand, if has a pole of order at p

g zz

g z

f zz m

f z

f m

0,

0

10 0

0

( ) ( )

( )

( ) '( ) '( )

( ) ( )

'( ) '( )

( ) ( ) ( )

p

p p

m

p

m m

p

z

zf z

z z

m z zf z

z z z z

mf z z

f z z z z

0analytic and nonzero at .z

0has a simple pole at , with residue - .pz m

35

,

2

Applying the residue theorem, then

'( ) '( ) 2 Re 2 ( )

( ) ( )

1 arg ( )

2

Example 1.

1 ( ) , 0 pole of order 2.

Let ,(0 2 )

Cpoles zeros

C

i

f z f zdz i s i Z P

f z f z

f z Z P

f z zz

z e

2

22

1 1arg( ) 2 by theorem.

21

or = (0 2 )

C

i

z

ez

36

Rouche’s theorem

Thm 2. Let two functions ( ) and ( ) be analytic inside

and on a simple closed contour , and suppose

( ) ( ) at each point on . Then ( )

and ( ) ( ) have the same number of zeros,

counting multiplicitie

f z g z

C

f z g z C f z

f z g z

s, inside .C

Pf.

Since ( ) ( ) on

( ) ( ) ( ) ( ) 0 on .

( )arg ( ) ( ) arg ( ) 1

( )

( ) = arg ( ) arg 1

( )

C C

C C

f z g z C

f z g z f z g z C

g zf z g z f z

f z

g zf z

f z

37

( )Let F( ) 1

( )

( )then ( ) 1 1,

( )

g zz

f z

g zF z

f z

C

Therefore under the transformation ( ),

the image of lies in 1 1, which does not

enclose 0.

( ) arg 1 0

( )

arg ( ) ( ) arg ( )

They have the same numbC C

w F z

C w

w

g z

f z

f z g z f z

er of zeros. (No poles)

38

7 3

3 7

73

Ex2. To determine the number of roots

4 1 0

inside 1,

write ( ) 4 and ( ) 1

( ) 4 4, and ( ) 1 3 when 1.

since (

z z z

z

f z z g z z z

f z z g z z z z

f

7 3

) has three zeros, inside 1,

4 1 has three roots inside 1.

z z

z z z z

39

66. Inverse Laplace Transforms

Suppose that a function F of complex variable s is analytic throughout the finite s plane except for a finite number of isolated singularities.

Let denote a vertical line segment from

to ,

where is positive and large enough

that singularities of all lie to the

left of the segment.

RL

s iR s iR

F

RC

NS

2S

RL

1S

iRr

r0

y

define

1 ( ) lim ( ) ( 0)

2 provided the limit exists

1 or ( ) P.V. ( ) ( 0)

2

R

st

LR

st

f t e F s ds ti

if t e F s ds t

ii

Bromwich integral

40

0

1

0

It can be shown that ( ) is the inverse Laplace

transform of ( ). That is, if ( ) is

( ) ( )

then ( ) ( )

Let (n=1,2,...N) denote the singularities of ( ).

Let deno

st

n

f t

F s F s

F s e f t dt

f t L F s

s F s

R

0

te the largest of their moduli and consider

a semicircle

3 e ( )

2 2 where

R

i

C

s R

R R

41

0

n=1

Note that for each ,

Hence all singularities are inside the semicircle

and ( ) 2 Re ( )

- ( )

Suppose that (

R n

R

n

n n

Nst st

L s s

st

C

s

s s R R

e F s ds i s e F s

e F s ds

F s

32

2

32 cos

2

sin

0

) , where 0 as ,

( ) exp( ) ( Re )

( )

=

R

R

R R

st i i i

C

st t RtRC

Rt

M M R

e F s ds t Rte F Rie d

e F s ds e M R e d

e dRt

( )2

Jordan’s inequality

iRr

0R

tRe M

t

42

1

1

lim ( ) 0

( ) Re ( )

can be exteded to the case with infinite singular points

( ) Re ( ) ( 0)

R

n

n

st

CR

Nst

s sn

st

s sn

e F s ds

f t s e F s

f t s e F s t

43

67. Example

Exercise 12

0

0

0

1 20 2

0 0 0 0

121

0

Suppose that ( ) has a pole of order at ,

( ) ( ) .... ( 0)( ) ( )

Then Re ( ) .......1! ( 1)!

when ( 0) an

n mn mm

n

s tst mm

s s

F s m s

bb bF s a s s b

s s s s s s

bbs e F s e b t t

m

s i

0 0

0

121

d ( ) ( ),

is also a pole of order .

Moreover,

Re ( ) Re ( )

=2 .....1! ( 1)!

st st

s s s s

t i t mme

F s F s

s i m

s e F s s e F s

bbe R e b t t

m

When t is real

44

0

0 0

0 00

0If is a simple pole

Re ( ) Re ( )

and Re ( ) Re ( ) 2 Re Re ( )

s tst

s s s s

st st t i t

s s s ss s

s

s e F s e s F s

s e F s s e F s e e s F s

2 2 2

0 0

2 2

0 2 2 2 2 22 2

( )( )

singularities and

( )( ) , ( )

( ) ( )

is a pole of order 2. ( )( 2 )( )

Furthermore ( ) ( ). ( )(

sF s

s a

s ia s ia

s sF s s

s ia s ia

s ss F s

x y a ixys a

sF s F s F s

2 2 2 22 2 2 ( 2 ))

s

x y a ixys a

Ex1.

45

0 01 2

2 2

2

Re ( ) Re ( ) 2Re ( )

1 1 '( )But ( ) ( ) ( ) ( ) ....

( ) ( ) 1!

( ) '( ) = ........ (0 2 )

( )

st st iat

s s s ss e F s s e F s e b b t

aiF s s ai s ai

s ia s ia

ai ias ia a

s ia s ia

2

4

( ) 2( )'( )

( )

s ia s s ias

s ia

1 2

( ) , '( ) 04

0, b4

1 Re ( ) 2Re ( ) sin

4 2

1 ( ) sin ( 0)

2 provided ( ) satisfies the boundedness condition

n

st iat

s s

iia ia

ai

ba

is e F s e t t at

a a

f t t at ta

F s

(in sec. 66) p.236.

46

22 2 2 2 2

2 22 2 2 2

3Let s e ( )

2 2where 0,

e

e

Since ( ) 0

( ) 0 as .( )

i

i

i

R

R a

s R R

s R R R a

s a s a R a

s RF s R

s a R a