D. R. WiltonECE Dept.

ECE 6382 ECE 6382

Evaluation of Definite Integrals Via the Residue

Theorem

8/24/10

Review of Cauchy Principal Review of Cauchy Principal Value IntegralsValue Integrals

2 0 2 0 2

1 01 1 0ln ln

x x

dx dx dxI x x

x x x

Recall for real integrals,

but a finite result is obtained if the integral interpreted as

2 2 2

11 10 0

0

lim lim ln ln

lim ln

x x

dx dx dxI x x

x x x

ln1 ln 2 ln ln 2

because the infinite contributions from the two symmetrical shadedparts shown exactly cancel. Integrals evaluated in this way are said to be (Cauchy) principal value integrals (or “deleted” integrals) and are often written as 2 2

1 1

dx dxI PV

x x

or

x

1/x

Cauchy Principal Value IntegralsCauchy Principal Value Integrals

2 1 1

dxI

x x

1

2 210

2 00

lim1 1 1 1

lim lim1 1

R

R

R

RC R C C

i

RC C

dz dz

z z z z

i edzI I

z z

21 1i i

d

e e

2

2 20

2 2 2

2 20 0

lim1 1

lim lim 02 2 2

2 Res ( ) Res ( 1) 2 lim

i

i iR

i i

R R

z i

iR e d

R e R e

id ie d i ie d iI I I

R R

z ii f z i f z i

z i

1

1lim

1 z

z

z i z

2 1 1z z

2

1 12

4 2

1 12

4 2 2 21 1

ii

dx i iI PV i

x x

To evaluate

consider the integral

Note: Principal value integrals have either symmetric limits extending to infinity or a vanishing, symmetric deleted interval at a singularity. Both types appear in this problem!

z i

z i

: ,iR

i

C z R e

dz iR e d

:1 ,i

i

Cz edz i e d

RR1z

Brief Review of SingularBrief Review of Singular IntegralsIntegrals

• Logarithmic singularities are examples of integrable singularities:

1 1

00 0ln ln 1 lim ln 0since

x xx dx x x x x x

1 x

ln x

SingularSingular Integrals,Integrals, cont’dcont’d• 1/x singularities are examples

of singularities integrable only in the principal value (PV) sense.

• Principal value integrals must not start or end at the singularity, but must pass through to permit cancellation of infinities

1 1

00 0

2 2 2

11 10 0

0

1ln lim ln ,

1 1lim ( ) lim ln ln

lim ln

since

but

x x

x x

dx x xx

PV dx dx x xx x

ln 2 ln ln 2

x

1

x

Singular Integrals, cont’dSingular Integrals, cont’d

• Singularities like 1/x2 are non- integrable:

11

20 00

2

2

2

1

1

1 1 1 lim

, 0sgn( )

, 0

since

but note that has a PV integral

xx

x

x

dxx x x

xx

x x

x

2

1

x

Singular Integrals, cont’dSingular Integrals, cont’d

Summary:

• ln x is integrable at x = 0

• 1/x is integrable at x = 0 for 0 < < 1

• 1/x is non-integrable at x = 0 for or

• f(x)sgn(x)/|x| has a PV integral at x = 0 if f(x) is continuous

• Above results translate to singularities at a point a via the transformation x x-a

Dispersion RelationsDispersion Relations

0

0

0

,0 00

0,

limix R

iRC R x

f z u iv f z z

f x ef z f xdz dx

z x x x e

analytic, in UHP :

ii e

2

0

0 0 00

0 0 00 0 0

00

00

2

2

1 1( ) ( )

1( ) ,

1( )

( ), ( )

d i f x

f xPV dx i f x i f x i f x

x x

f x v x u xif x u x iv x PV dx PV dx PV dx

i x x x x x x

v xu x PV dx

x x

u xv x PV dx

x x

u x v x

are Hilbert Transfo of one anotherrms

y

RRx0x

Dispersion Relation, Example 1Dispersion Relation, Example 1

in in

in in in in in in

inin in

,0

( ) ( ) 0,

* ;

( ) ( )lim

iR

iRC R

Z Z

Z Z R R X X

Z eZ Zd d

e

Assume analytic in LHP, in LHP,

:

ii e

in

0

in

in inin in in

in inin 2 2

0

in inin 2 2

2

1

21 1,

21 1

i Z

d i Z

X RiZ R iX PV d PV d

X XR PV d PV d

R RX PV d PV

0

d

in

in in

( )

( ) ( )

Z

R i X

in ( )V

in ( )I

Re

RR

Im

C

Dispersion Relation, Example 2Dispersion Relation, Example 2

( ) ( ) 1,

Im2Re ( ) 1

r r

r PV

The , is analytic in the LHP and

in LHP; hence, similar to the input impedance analysis, one obtains

the

relative permittivity

Kronig - Kramers dispersion relations

2 20

2 20

( ) 1

Re ( ) 12Im ( ) 1

r

rr

d

PV d

( ) ( ) 1

Note :

relative permittivity

electric susceptibilityr

e

e r

Kramers

Integrals of the form Integrals of the form 2

0

sin ,cosf d

• f is finite• f is a rational function of• Let

so the above integral becomes

sin ,cos

1 1

,

sin , cos2 2

i i dzz e dz ie d d i

z

z z z z

i

Unit circle

1 1

1

1 1

2

0

,2 2

12 ,

2 2

sin ,cosz

z z z z dzI f

i z

z z z zf

z i

f d i

Residues of inside the unit circle

1 11,

2 2

z z z zf z

z i

Note : will be a rational polynomial in

Integrals of the form Integrals of the form (cont.) (cont.)

2

0

sin ,cosf d

Example:

1

12

2

5 504 4 21

2 121 1

12

121

1

sin

4 4

2 5 2 2 2

22 lim

2

z ziz

z z

z iz

d dzI i

z

i dzidz

z iz z i z i

z idzi

z i z i

12

2

2z i z i 8

3

x

iy|z|=1

12z i

2z i

Integrals of the form Integrals of the form f x dx

Since

0 0

lim ( ) lim ( ) lim 0R

i i

R R RC

Kf z dz f Re iRe d d

R

( ) 2 ( )C

f z dz f x dx i f z

residues of in the UHP

• f is analytic in the UHP except for a finite number of poles (can easily be extended to handle poles on the real axis)

• f is , i.e. , a

constant, in the UHP 2z O 2lim

zz f z K

R

: ,iR

i

C z R e

dz iR e d

R

Ques: What changes to the problem conditions and result must be made if f is only analytic in the lower half plane (LHP)?

Integrals of the form Integrals of the form (cont.) (cont.)

f x dx

Example: z = 2iz = 3i

2 2

2 22 2 2 20

4( )

1 1

2 29 4 9 4

x zf x

x dx xI dx

x x x x

z

O

2

3

Res 3 + Res 2

3limz i

i f i f i

z ii

2

3 3

z

z i z i 2

2 22

2 + lim

4 z i

z id

dzz

2

22 9 2

z

z z i

2

2 22 2 2

2 42 2

2

9 2 2 2 2 2 2 93 + lim

50 9 2

3 13 2

50 200 200

z i

z i

z z i z z z z i z i zii

z z i

i ii z i

(Note the double pole at !)

Integrals of the form Integrals of the form (Fourier Integrals) (Fourier Integrals)

i axf x e dx

• f is analytic in the UHP except for a finite number of poles (can easily be extended to handle poles on the real axis),

•

(i.e., z in UHP)

lim ( ) 0, 0 argz

f z z

R

: ,iR

i

C z R e

dz iR e d

R

/2cos sin sin

0 0/2 2

0

lim ( ) lim ( ) lim 2 max ( )

lim 2 max ( ) lim 2

R

iaz i iaR aR i i aR

R R RC

aRi

R R

f z e dz f Re e e iRe d R f Re e d

R f Re e d R

max ( )2

if Rea R

1 0aRe

• Choosing the contour shown, the contribution from the semicircular arc vanishes by Jordan’s lemma:

since for

2sin2

sin aR

aRe e

02

( ) 2 residues of ( ) in the UHPiaz iax iaz

C

f z dz f x dx i f ze e e

0a

1

2

Using the symmetries of and and the Euler formula, , we write

and identify .

Hence

Integrals of the form Integrals of the form (Fourier Integrals) (cont.) (Fourier Integrals) (cont.)

iaxf x e dx

( ) 2 residues of ( ) in the UHPiaz iax iaz

C

f z dz f x dx i f ze e e

Example: 2 20

cos, , 0.

xI dx a

x a

cos x sin xcos sini xe x i x

2 2

1

2

i xeI dx

x a

( imag. part vanishes by symmetry!)

2 2

1 1

2f z

z a

2 20

cos 12 Res = 2 lim

2i z

z ia

z iaxI dx i e f ia i

x a

i ze

z ia z ia

= 2 i 1

2 2

ae

i

2

ae

aa

Exponential Integrals Exponential Integrals • There is no general rule for choosing the contour of integration; if the

integral can be done by contour integration and the residue theorem, the contour is usually specific to the problem.

Example: , 0 11

ax

x

eI dx a

e

Consider the contour integral over the path shown in the figure:

1 2 3 41 1

az az

z z

e edz dz

e e

The contribution from each contour segment in the limit

must be separately evaluated:

R

43

21

R R

2z i

z i

Exponential Integrals (cont.) Exponential Integrals (cont.)

1

1 : , ,

lim lim1 1

Raz ax

z xR RR

z x dz dx

e edz dx I

e e

3

3

2 2

: 2 , ,

lim lim1 1

Raz axia ia

z xR RR

z x i dz dx

e edz e dx e I

e e

2

2

2

0

2 2 1

0 0

: ,

lim lim1 1

lim lim 0, 11 1

az aR iay

z R iyR R

a RaR

R RR R

z R iy dz idy

e e edz i dy

e e e

e edy dy a

e e

1 1R iy Re e e

1

Rey

43

21

R R

2z i

z i

Exponential Integrals (cont.) Exponential Integrals (cont.)

4

4

0

2

2

0

: , ,

lim lim1 1

lim 0, 01

az aR iay

z R iyR R

aR

RR

z R iy dz idy

e e edz i dy

e e e

edy a

e

Hence

21 2 Res 2 lim1

2 lim 2 lim1 1

azia

zz i

azaz

i z iz i z i

ee I i f z i i z i

e

z i eei z i i

e e

1

212

2 limz i

z i z i

z ii

aze

z i 212 z i

2 iaie

43

21

R R

2z i

z i

Exponential Integrals (cont.) Exponential Integrals (cont.)

Finally,

2

2 2

1 1

ax ia ia

x ia

e ie i eI dx

e e

iae , 0 1

sinia iaa

ae e

21 2ia iae I ie

Integration around a Branch Cut Integration around a Branch Cut • A given contour of integration, usually problem specific, must not cross

a branch cut.

• Care must be taken to evaluate all quantities on the chosen branch. Integrand discontinuities are often used to relate integrals on either side of the cut.

• Usually a separate evaluation of the contribution from the branch point is required.

Example:

RC

1z 0C

R

1L

2L

0

, 0 1.1

kxI dx k

x

• We’ll evaluate the integral using the contour shown

Define the branch of such that

Integration around a Branch Integration around a Branch Cut (cont.) Cut (cont.)

RC

1z 0C

R

1L

2L

0

, 0 1.1

kxI dx k

x

First, note the integral exists since the integral of the asymptotic forms of the integrand at both limits exists:

0

1

0, 11

, 01

kk

x

kk

x

xx x k

x

xx x k

x

which is integrable at (must check!)

which is integrable at (must check!)

kz

ln arg lnln , 0 2k z i z k r ik k z k ikz e e e r e

Integration around a Integration around a Branch Cut (cont.) Branch Cut (cont.)

RC

1z 0C

R

1L

2L

Now consider the various contributions to the contour integral

1 2 0

2 Res 1 ,

1

RL L C Ck

f z dz i f

zf z

z

where

0 0 0 0: , ,i i k k ikC z r e dz ir e d z r e

0 0

0

010 0

00, 0

02 20

lim lim 01

k ik ik i ik

ir rC

r e ir e df z dz ir e d

r e

: , ,i i k k ikRC z Re dz iRe d z R e

2 2

,00

lim lim 01

R

k ik ik ik

iR RC

R e iRe df z dz iR e d

Re

Integration around a Integration around a Branch Cut (cont.) Branch Cut (cont.)

RC

1z 0C

R

1L

2L1 : , ,i i k k ikL z re dz e dr z r e

1 0

0

( )

,00,

0

lim1 1

R k ki k

iRL r

r

r dr r drf z dz e I

re r

2 22 : , ,i iki i k kL z re re dz e dr z r e

0

2

0

(2 ) 2 2

,00,

0

lim1 1

r k ki k i k i k

iRL R

r

r dr r drf z dz e e e I

re r

21 2 Res 1 2 lim 1i

i k i

z ee I i f z e i z

1

kz

z

2 iki e

20

2 2

1 1

k ik ik

i k

x i e i eI dx

x e

ike , 0 1.

sinik i kk

ke e

Hence

and finally,