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46
Evaluation of Definite Integrals via the Residue Theorem ECE 6382 Notes are from D. R. Wilton, Dept. of ECE 1 David R. Jackson Fall 2020 Notes 11

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Page 1: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Evaluation of Definite Integrals via the Residue Theorem

ECE 6382

Notes are from D. R. Wilton, Dept. of ECE

1

David R. Jackson

Fall 2020

Notes 11

Page 2: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Review of Singular Integrals

Logarithmic singularities are examples of integrable singularities:

( )1 1 1

0 00 0Ln Ln Ln 1 lim Ln 0lim lim

x xx dx x dx x x x x x

εεε ε = →→ →= = − = − =∫ ∫ since

1x

Ln x

2

Note: There might be numerical trouble if one integrates this function numerically!

π θ π− < <( ) ( )Ln lnz r iθ= +

Page 3: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Singularities like 1/x are non-integrable.

3

Review of Singular Integrals (cont.)

x

1x

( )1 1 1

0 0 0

1 1 Lnlim limx

dx dx xx x εεε ε =→ →

= = = −∞∫ ∫

Page 4: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Review of Cauchy Principal Value Integrals

2 0 2 0 2

1 01 1 0Ln Ln

x x

dx dx dxI x xx x x =− =− −

= = + = +∫ ∫ ∫

Consider the following integral:

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 xx x x

ε ε

εεε ε

εε

− −

=− =+− −→ →

= = + = +

=

∫ ∫ ∫Ln1 Ln2 Lnε− + −( ) Ln2=

The infinite contributions from the two symmetrical shadedparts shown exactly cancel in this limit. Integrals evaluated in this way are said to be (Cauchy) principal value (PV) integrals:

2 2

1 1PV dx dxI

x x− −= =∫ ∫

or

4

x

1/xε−

ε1−

2

Notation:

Page 5: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

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 them to permit cancellation of infinities

x

1x

5

Cauchy Principal Value Integrals (cont.)

Page 6: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Singularities like 1/x2 are non- integrable (even in the PV sense).

2 2

2

2

2

1

1

1 1 1 1 1 1

, 0sgn( ) , 0

b

a

x

x

dx dxx x a b

xxx x

ε

ε ε ε− + = + + − → ∞

> = − <

∫ ∫

but note that has a PV integral

6

x

2

1x

a bεε−

Cauchy Principal Value Integrals (cont.)

Page 7: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Summary of some results:

• Ln x is integrable at x = 0

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

• 1/xα is non-integrable at x = 0 for α ≥ 1

• f (x)sgn(x)/|x|α has a PV integral at x = 0 for α < 2 if f(x) is smooth

(The above results translate to singularities at a point a via the transformation x → x-a.)

7

Singular Integral Examples

Page 8: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Integrals Along the Real Axis

( )f x dx∞

−∞∫

On the large semicircle we then have

0

lim ( ) lim ( ) 0R

i i

R RC

f z dz f Re iRe dπ

θ θ θ→∞ →∞

= =∫ ∫

( )( ) 2 ( )C

I f z dz f x dx i f zπ∞

−∞

= = = ∑∫ ∫

residues of in the UHP

f is analytic in the UHP except for a finite number of poles (can be extended to handle poles on the real axis via PV integrals).

f is , i.e. in the UHP.( )1 /o z ( )lim 0z

z f z→∞

=

8

Assumptions:

Hence

R−

: ,iR

i

C z R edz iR e d

θ

θ θ

=

=

Rx

××

×

Note:If the function is analytic in the LHP except for poles, then we would close the

contour in the LHP.

UHP: Upper Half Plane

Page 9: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Example:

( )( ) ( )( )2 2

2 22 2 2 20

4( )

1 12 29 4 9 4

f z

z dx zI dxz z z z

z

∞ ∞

−∞

−=

= = =+ + + +

∫ ∫

2 ( ) ( )

( )3

Res 3 + Res 2

3limz i

i f i f i

z ii

π

π→

−=

( ) ( )

2

3 3

z

z i z i+ − ( )( )2

2 22

2 + lim

4 z i

z iddzz →

+ ( ) ( )

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 + lim50 9 2

3 13 50 200 200

z i

z i

z z i z z z z i z i ziiz z i

i ii

π

ππ

+ + + − + + + + = + + = − =

9

( ) ( )2

22 20 9 4

x dxIx x

=+ +

( )1

0 01

0

1Res ( ) ( ) ( )1 !

mm

mdf z z z f z

m dz z z

− = − − =

( ) ( )2

22 20 2009 4

x dxIx x

π∞

= =+ +

z = 2iz = 3i

x

y

2z i=(Note the double pole at !)

×

×

Integrals Along the Real Axis (cont.)

Page 10: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

( ) ( ) ( ) ( )

( ) ( )

1

2 2,10

2 00

lim1 1 1 1

lim lim1 1

R

R

R

C RC R C C

i

RC C

dz dzIz z z z

i edzI Iz z

ρ

ρ

ρ

ρρ

θ

ρρ

ρ

− −

→∞− − +→

→∞ →→

= = + + + + + + +

= + + = + + +

∫ ∫ ∫ ∫ ∫

∫ ∫

( )21 1i i

d

e eθ θ

θ

ρ ρ − + + ( ) ( )

2

2 20

2 2 2

2 20 0

lim1 1

lim lim 02 2 2

i

i iR

i i

R R

iR e dR e R e

id ie d i ie d iI I IR R

π π φ

φ φπ

π π πφ φ

π

φ

θ φ π φ π

→∞

− −

→∞ →∞

++ +

= + + = + + = + +

∫ ∫

∫ ∫ ∫

Consider the following integral:I

10

Cauchy Principal Value IntegralsExample

( ) ( )2 1 1dxI

x x∞

−∞=

+ +∫

2CiI I π

= +

Hence,

Evaluate:z i=

z i= −

: ,iR

i

C z R edz iR e d

φ

φ φ

=

=

: 1 ,i

i

C z edz i e d

θρ

θ

ρρ θ

+ ==

RR−1z = −

×

×

×

C

We could have also used the formula from Notes 10 for going

halfway around a simple pole on a small semicircle.

Page 11: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

[ ]( )

2 Res ( ) Res ( 1) 2 limC z i

z iI i f z i f z iπ π

−= = + = − =

( )z i− ( )( )( )

1

1lim

1 z

z

z i z → −

++

+ + ( ) ( )2 1 1z z+ +

( )1 1 1 1 1 12 2 2

2 1 2 4 2 4 2 2 2i ii i i i

i i iπ ππ π π

− − − = + = + = + = + +

Next, evaluate the integral ICusing the residue theorem:

11

( ) ( )2 1 1dxI

x x∞

−∞=

+ +∫

2I π

=

Hence:

Cauchy Principal Value Integrals (cont.)

2 2CI iπ π = + 2C

iI I π= +

Recall:Thus:

z i=

z i= −

: ,iR

i

C z R edz iR e d

φ

φ φ

=

=

: 1 ,i

i

C z edz i e d

θρ

θ

ρρ θ

+ ==

RR−1z = −

×

×

×

C

Page 12: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Choosing the contour shown, the contribution from the semicircular arc vanishes as R → ∞ by Jordan’s lemma.

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 via PV integrals),

lim ( ) 0, 0 arg ( )z

f z z zπ→∞

= ≤ ≤ in UHP

0a >Assume (close the path in the UHP)

12

Assumptions:

(See next slide.)Fourier

R−

: ,iR

i

C z R edz iR e d

θ

θ θ

=

=

Rx

y

××

×

Page 13: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Jordan’s lemma

( )

( )

/2cos sin sin

0 0/2 2

0

( ) ( ) 2 max ( )

2 max ( ) 2

R

iaz i iaR aR i i aR

CaR

i

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

R f Re e d R

π πθ θ θ θ θ θ

π θθ π

θ θ

θ

− −

= ≤

≤ =

∫ ∫ ∫

∫ ( ) max ( )2

if Rea R

θ π ( )1 0aRe−− →

2sin20 / 2 : sin

aRaRe e

θθ πθθ π θ

π−−⇒≤ ≤ ≥ ≤

13θ

1

sinθ

2θπ

Fourier Integrals (cont.)

lim ( ) 0, 0 arg ( )z

f z z zπ→∞

= ≤ ≤ in UHP

Assume

( ) 0R

iaz

C

f z e dz →∫Then

Proof:

R−

: ,iR

i

C z R edz iR e d

θ

θ θ

=

=

Rx

y

××

×

Page 14: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Fourier Integrals (cont.)

( ) i axI f x e dx∞

−∞

= ∫

( ) ( ) 2 ( )iax iaz iaz

C

I f x dx f z dz i f ze e eπ∞

−∞

= = = ∑∫ ∫ residues of in the UHP

14

We then have

0a >

Question: What would change if a < 0 ?

R−

: ,iR

i

C z R edz iR e d

θ

θ θ

=

=

Rx

y

××

×

Page 15: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Use the symmetries of cosλx and sinλx and the Euler formula, we have:

Example:

( )2 20

cos , , 0.xI dx ax a

λ λ∞

= >+∫

cos sini xe x i xλ λ λ= +Note :

2 212

i xeI dxx a

λ∞

−∞

=+∫ (imaginary part vanishes by symmetry!)

( )2 2 2 2

1 1 12 Res = 2 lim2 2 2

i z i z

z iaz ia

z iae eI dz i iz a z a

λ λ

π π∞

→−∞ =

− = = + +

∫ ( ) ( )

i ze

z ia z ia

λ

+ −

= 2 iπ 12 2

aei

λ−

2

aeaa

λπ −

=

15

Fourier Integrals (cont.)

2 20

cos2

ax eIx a a

λλ π∞ −

= =+∫

z = ia

x

y

×

Page 16: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Rational Functions of sin and cos

( )2

0

sin ,cosf dπ

θ θ θ∫ f is finite within the interval. f is a rational function (ratio of polynomials) of sinθ, cosθ.

1 1

,

sin , cos2 2

i i dzz e dz ie d d iz

z z z zi

θ θ θ θ

θ θ− −

⇒= = = −

− += =

unit circle

Let

( )1 1

1

1 1

2

0

,2 2

12 ,2 2

sin ,cosz

z z z z dzI fi z

z z z zfz i

f d iπ

π

θ θ θ− −

=

− −

− +=

− +=

= − ∫

Residues of inside the unit circle

1 11 ,2 2

z z z zf zz i

− − − +

will be a rational function ofNote : 16

Assumptions:

Page 17: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Example:

( ) ( )( )

( )( )( )

1

12

2

5 504 4 21

2 121 1

12

121

1sin

4 42 5 2 2 2

2 2 lim2

z ziz

z z

z iz

d dzI iz

i dzidzz iz z i z i

z idz iz i z i

π θθ

π

−−=

= =

→ −=

= = − + +

= − =+ − + +

+= =

+ +

∫ ∫

∫ ∫

( ) ( )12

2

2z i z i+ +83π

=

17

2

504 sin

dIπ θ

θ=

+∫

Multiply top and bottom by 4i.

2

504

8sin 3

dIπ θ π

θ= =

+∫

12z i= −

2z i= −

1z =

×

×

x

y

Rational Functions of sin and cos (cont.)

Page 18: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

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.

, 0 11

ax

xeI dx a

e

−∞

= < <+∫

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

1 2 3 41 1

az az

C z zC

e eI dz dze eγ γ γ γ

= = + + + + +

∫ ∫ ∫ ∫ ∫

The contribution from each contour segment in the limit must be separately evaluated (next slide).

R → ∞

(2 1) , 0, 1, 2,z n i nπ= + = ± ±

The integrand has simple poles at

18

Example:

2γ1γ

R− R

2z iπ=

z iπ=x

y

×

× 3z iπ=

Page 19: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Exponential Integrals (cont.)

1

1 : , ,

lim1

az

zR

z x dz dxe dz I

γ

→∞

= =

=+∫

3

3

2 2

: 2 , ,

lim lim1 1

Raz axia ia

z xR RR

z x i dz dx

e edz e dx e Ie e

π π

γ

γ π−

→∞ →∞

= + =

= = −+ +∫ ∫

( )

2

2

2

0

2 2 1

0 0

1

: ,

1 1

0,1 1

az aR iay

z R iy

a RaR

R R

z R iy dz idy

e e edz i dye e e

e edy dye e

a

π

γ

π π

γ

= + =

=+ +

≤− −

<= →

∫ ∫

∫ ∫19

R → ∞

y π=Smallest for :

1 1R iy Re e e+ ≥ −

1Re y

1 R iye e+

2γ1γ

R− R

2z iπ=

z iπ=x

y

×

× 3z iπ=

Page 20: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Exponential Integrals (cont.)

4

4

0

2

2

0

: ,

0

,

1 1

0,1

az aR iay

z R iy

aR

R

z R iy dz idy

e e edz i dye e e

e d aye

γ π

π

γ−

= − + =

=+ +

≤ →−

>

∫ ∫

Hence

( ) ( )

( ) ( ) ( ) ( )

21 2 Res

Res lim lim lim1 1 1

ia

azaz az

z z iz i z i z i

e I i f z i

z i ee ef z i z i z ie e

π

ππ π π

π π

ππ π π −→ → →

− = =

−= = − = − =

+ − 1− ( ) ( )

( )

212

12

lim1

azia

z i

z i z i

e ez i

π

π

π π

π→

+ − + − +

= = −− − − −

( )( )( ) ( )0

0

0

0

( )( ) , 0( )

Res ( ) lim lim1

1

az

z z z i z

a i a ia i

i

g zf z h zh z

g z ef z i dh z edz

e e ee

π

π ππ

π

π→ →

= =

⇒ = = =′ +

= = = −−

Alternatively,

20

2γ1γ

R− R

2z iπ=

z iπ=x

y

×

Page 21: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Exponential Integrals (cont.)

Therefore

22 2

1 1

ax ia ia

x iae ie i eI dx

e e

π π

π

π π∞

−∞

− −= = =

+ −∫ iae π ( ), 0 1

sinia iaa

ae eπ π

ππ−

= < <−

( )21 2ia iae I ieπ ππ− = −

21

We thus have

, 0 11 sin

ax

xeI dx a

e aπ

π

−∞

= = < <+∫

Hence

Page 22: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Integration Around a Branch Cut A given contour of integration is chosen: usually problem specific, usually must

not cross a branch cut.

We take advantage of the fact that the integrand changes across the branch cut.

Usually an evaluation of the contribution from the branch point is required.

0

, 0 11

kxI dx kx

∞ −

= < <+∫

22

Example:

Assume the branch 0 ≤ θ < 2π

( kx− = positive real)

1z = −x

y

×

Note: We choose the branch cut on the positive real axis (the axis of integration).

Page 23: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Integration Around a Branch Cut (cont.)

0

, 0 11

kxI dx kx

∞ −

= < <+∫

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

x x x kxx x x kx

−−

−− −

→∞

→ = <+

→ = ∞ >+

which is integrable at

which is integrable at

23

Page 24: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Integration Around a Branch Cut

0

, 0 11

kxI dx kx

∞ −

= < <+∫

We’ll evaluate the integral using the contour shown.

24

( ) ( ) ( )lnln ln ln ln , 0 2k kk r ik z k z k r ik r ik k ikz e e e e e e e r eθ θ θ θ θ π

− −− +− − − − − − −= = = = = = < <

0 2θ π< <

iz re θ=

RC

1z = −

0C

R

εε

1L

2Lx

y

0ε →

×0r

Page 25: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

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 zz

π+ + +

= −

=+

where

0 0 0 0: , ,i i k k ikC z r e dz ir e d z r eθ θ θθ − − −= = =

( )0 0

0

010 0

00, 002 20

lim lim 01

k ik ik i ik

ir rC

r e ir e df z dz ir e dr e

ε θ θθ θ

θπ ε πε

θ θ− −

− −

→ →−→

= = →+∫ ∫ ∫

: , ,i i k k ikRC z Re dz iRe d z R eθ θ θφ − − −= = =

( )2 2

,00

lim lim 01

R

k ik ik ik

iR RC

R e iRe df z dz iR e dRe

π ε πθ θθ

θεε

φ φ− − −

− −

→∞ →∞→

= = →+∫ ∫ ∫

25

Integration Around a Branch Cut (cont.)

RC

1z = −

0C

R

εε

1L

2Lx

y

0ε →

×0r

Page 26: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

1 : , ,i i k k ikL z re dz e dr z r eε ε ε− − −= = =

( )1 00

( )

000

lim1 1

R k ki k

iRL rr

r dr r drf z dz e Ire r

ε εε

ε

∞− −−

→∞→→

= = =+ +∫ ∫ ∫

( ) ( )2 22 : , ,i iki i k kL z re re dz e dr z r eπ ε π εε ε− − −− − − −= = = =

( ) [ ]0

2 0

(2 ) 2 2

000

lim1 1

r k ki k i k i k

iRL Rr

r dr r drf z dz e e e Ire r

ε π ε π πε

ε

∞− −− + − − −

−→∞→→

= = − = −+ +∫ ∫ ∫

26

Integration Around a Branch Cut (cont.)

For the path L2:

For the path L1:( )

1

kzf zz

=+

RC

1z = −

0C

R

εε

1L

2Lx

y

0ε →

×0r

Page 27: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

( ) ( )

( )

2

1

1 2 Res 1

2 lim 1

i k

z

e I i f z

i z

π π

π

→−

− = = −

= +1

kzz

+

( )

( )2 1

2

2

k

ki

ik

i

i e

i e

π

π

π

π

π

= −

=

=

( )2

2 21

ik ik

i k

i e i eIe

π π

π

π π− −

−= =

− ike π− ( ), 0 1

sinik i kk

ke eπ π

ππ−

= < <−

Hence

Therefore, we have

27

Integration Around a Branch Cut (cont.)

Note: Arg (-1) = π here.

0

, 0 11 sin

kxI dx kx k

ππ

∞ −

= = < <+∫

Hence

Page 28: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

“Dispersion” Relations

28

Assumptions:

( )( ) 0 , z

f z u iv

f z

= +

(including the real axis)is analytic in the UHP

in the UHP

0iz x e θε− =Let

( ) ( ) ( )0

0

0

0 00

limix R

iRC R x

f x ef z f xdz dx

z x x x e

θε

θεε

ε

ε

→∞− +→

+= + + − −

∫ ∫ ∫

ii e θε ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2

0

0 0 00

0 0 00 0 0

2

2

1 1( ) ( )

d i f x

f xdx i f x i f x i f x

x x

f x v x u xif x u x iv x dx dx dxi x x x x x x

π

π

θ π

π π π

π π π

−∞

∞ ∞ ∞

−∞ −∞ −∞

=

= − =−

= + = = −− −

−⇒

∫ ∫ ∫

(From the residue theorem or the Cauchy integral theorem.)

CPV ( )0i f xπ

UHP: Upper Half Plane

y

RR−x0x

ε×

0Rε

→ ∞→

C

Page 29: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

“Dispersion” Relations (cont.)

29

( ) ( ) ( )0 0 0

0 0

1( ) ( )v x u xif x u x iv x dx dxx x x xπ π

∞ ∞

−∞ −∞

= + = −− −∫ ∫

Equate the real and imaginary parts:

( )

( )

00

00

1( )

1( )

v xu x dx

x x

u xv x dx

x x

π

π

−∞

−∞

=−

= −−

Next, relabel: 0,x x x x′→ →

Page 30: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

“Dispersion” Relations (cont.)

( ), ( )u x v x⇒ are of one anotherHilbert Transforms

30

Hence, we have

( )

( )

1( )

1( )

v xu x dx

x x

u xv x dx

x x

π

π

−∞

−∞

′′=

′ −

′′= −

′ −

∫ Hilbert

( )( ) 0 ,

f z u iv

f z z

= +

(including the real axis)

Assumptions :is analytic in the UHP

in the UHP

Page 31: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

“Dispersion” Relations: Circuit Theory

31

Note 1: A pole in the LHP would corresponds to a nonphysical growing response:ir ti ti t

r ii e e e ωωωω ω ω −= + ⇒ =

Note 2: The system is assumed to be unable to respond to a signal at very high frequency.

( ) ( ) ( ) ( ) ( ) ( )* ;R R I IH H H H H Hω ω ω ω ω ω− = ⇒ − = − = −

(see Appendix)

Symmetry property:

Reω

Imω

LHP: Lower Half Plane

( )( )

( )

1

2

exp(

( )

( ) 0

)

,

i

H

t

H ω

ω

ω

ω→ → ∞

time convention

is an

Assumptio

alytic in the LHP

for in the

ns

LHP

:

( ) ( ) ( )R IH H i Hω ω ω= +( )inV ω+

−( )outV ω

+

( ) ( )( ) /out inH V Vω ω ω≡

Transfer function

Page 32: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

“Dispersion” Relations: Circuit Theory (cont.)

32

Use the path shown below:

( )0

( ) ( )limiR

iRC R

H eH Hd de

θω ε

θω εε

ω εω ωω ωω ω ω ω ε

→∞− +→

+ ′ ′′ ′= + + ′ ′− −

∫ ∫ ∫

CPV

ii e θε

( )

( )

( ) ( )

( ) ( )

0

2

1

i H

d i H

Hd i H

HH d

i

π ω

π

θ π ω

ωω π ω

ω ω

ωω ω

π ω ω

−∞

−∞

= −

′′ = −

′ −

⇒′

′= −−

ReRR−

Im

ε

ω

ω′

C

×

(From the residue theorem or the Cauchy integral theorem.)

Page 33: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

33

( ) ( )1 HH d

ω ωπ ω ω

−∞

′′= −

′ −∫

( ) ( ) ( ) ( ) ( )1 I RR I

H HiH H iH d dω ω

ω ω ω ω ωπ ω ω π ω ω

∞ ∞

−∞ −∞

′ ′′ ′= + = − +

′ ′ −⇒

−∫ ∫

( ) ( )

( ) ( )

1

1

IR

RI

HH d

HH d

ωω ω

π ω ω

ωω ω

π ω ω

−∞

−∞

′′= −

′ −

′′=

′ −

We thus have

“Dispersion” Relations: Circuit Theory (cont.)

The real and imaginary parts of the transfer functionare Hilbert transforms of each other.

Page 34: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

34

We can also derive an alternative form:

( ) ( ) ( )( )( )2 2

0

21 1R RI

H HH d d

ω ω ωω ω ω

π ω ω π ω ω

∞ ∞

−∞

′ ′′ ′= = −

′ − ′ −∫ ∫

Similarly, we have

“Dispersion” Relations: Circuit Theory (cont.)

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( )( )( )( )

( )( )( )

0

00

0

0 0

0

2 20

1 1 1

1 1

1 1

1

21

I I IR

I I

I I

I

I

H H HH d d d

H Hd d

H Hd d

Hd

Hd

ω ω ωω ω ω ω

π ω ω π ω ω π ω ω

ω ωω ω

π ω ω π ω ω

ω ωω ω

π ω ω π ω ω

ω ω ω ω ωω

π ω ω ω ω

ω ωω

π ω ω

∞ ∞

−∞ −∞

∞ ∞

′ ′ ′′ ′ ′= − = − −

′ ′ ′− − −

′ ′−′ ′= −

′ ′− − −

′ ′′ ′= − −

′ ′+ −

′ ′ ′− + +′= −

′ ′+ −

′ ′′= −

′ −

∫ ∫ ∫

∫ ∫

∫ ∫

ω ω′ ′→ −First one :Use

This integral starts at zero.

( ) ( )I IH Hω ω− = −First one :Use

Page 35: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

35

Summarizing, we have

( ) ( ) ( )( )( )

( ) ( ) ( )( )( )

2 20

2 20

21 1

21 1

I IR

R RI

H HH d d

H HH d d

ω ω ωω ω ω

π ω ω π ω ω

ω ω ωω ω ω

π ω ω π ω ω

∞ ∞

−∞

∞ ∞

−∞

′ ′ ′′ ′= − = −

′ − ′ −

′ ′′ ′= =

′ − ′ −

∫ ∫

∫ ∫

Dispersion Relation: Circuit Theory (cont.)

( )( ) 0,

HH

ωω ω→ → ∞

is analytic in the LHPfo

Assumption

r in t

s:

he LHP

Page 36: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Kramers-Kronig Relations

( ) ( ) 1,r rε ω ε ω ω→The is analytic in the LHP and in LHPrelative permittivity

( ) ( ) 1

r

e

e r

εχχ ω ε ω

≡≡

≡ −

relative permittivityelectric susceptibility

Material parameters :

36

( ) ( )

( ) ( )

2 20

2 20

Im ( ) 12Re ( ) 1

Re ( ) 12Im ( ) 1

rr

rr

d

d

ω ε ωε ω ω

π ω ω

ω ε ωε ω ω

π ω ω

′ ′ −′− = −

′ −

′ −′− =

′ −

( ) 0eχ ω ω→ → ∞as in LHP

Assumption:

Similar to the transfer - function analysis, one obtains the disperson relations :Kronig - Kramers

0 eP Eε χ= (polarization per unit volume)

Page 37: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Kramers-Kronig Relations (cont.)

Kramers

37

( ) ( )

( ) ( )( )

2 20

2 20

21

12

rr

rr

d

d

ω ε ωε ω ω

π ω ω

ω ε ωε ω ω

π ω ω

′ ′′ ′′ ′= +

′ −

′ ′ −′′ ′= −

′ −

The final form of the Kramers-Kronig relations is then:

( )( ) ( ) ( )r r rj j iε ω ε ω ε ω′ ′′= −Denote using instead of

Note: This shows that if there is no loss (εr′′ = 0), then the relative permittivity must be 1. Hence, practical materials will always have some loss.

Page 38: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Laplace Transform

38

( ) ( )0

stF s f t e dt∞

−≡ ∫

The Laplace transform is defined as:

Assume

( ) 0Re s γ>

( ) 0tf t Aeγ<

Then F(s) is analytic in the region ( )Re s

( )Im s

Analytic××

×

( ) ( ) ( ) ( )00

Re >stF s t f t e dt s γ∞

−′ = −∫ valid forNote :

Page 39: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Laplace Transform (cont.)

39

( ) ( )12

i tg t g e dωω ωπ

−∞

= ∫

Define a function g(t) (using any γ > γ0):

We then have, from the inverse Fourier transform,

( ) ( ), 00, 0

te f t tg t

t

γ− >≡

<

Then, from the relation between f and g:

The Fourier transform of g(t) exists.(The function g(t) stays finite along the entire real axis

and tends to zero.)

( ) ( )1 , 02

t i tf t g e e d tγ ωω ωπ

−∞

= >∫

s iγ ω≡ +

( ) ( )( )1 , 02

ist

i

f t g i s e ds ti

γ

γ

γπ

+ ∞

− ∞

= − − >∫

Let

( )Re s

( )Im s

γ

C××

×

( ) ( ) i tg g t e dωω ω∞

−∞

Page 40: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

40

( ) ( )( )1 , 02

ist

i

f t g i s e ds ti

γ

γ

γπ

+ ∞

− ∞

= − − >∫

( )( ) ( ) ( )( )

( ) ( )( )

( ) ( )

( )

( )

0

0

0

i i s t

i i s t

s t

st

g i s g t e dt

g t e dt

g t e dt

f t e dt

F s

γ

γ

γ

γ∞

− − −

−∞

∞− − −

∞− −

∞−

− − =

=

=

=

=

We have for the integrand term:

( )Re s

( )Im s

γ

C××

×

Laplace Transform (cont.)

Page 41: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

41

( ) ( )1 , 02

ist

i

f t F s e ds ti

γ

γπ

+ ∞

− ∞

= >∫

“Bromwich integral”

( )Re s

( )Im s

γ

C××

×

Laplace Transform (cont.)

The Bromwich contour C is chosen to the right of all singularities of the function F(s).

“Inverse Laplace transform”

Hence we have:

Page 42: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

42

( ) ( )12

R

st

C

f t F s e dsiπ

= ∫

0t <

Close the contour to the right.

( ) 0f t =

Laplace Transform (cont.)

Consider the case:

( )Re s

( )Im s

γ

RC×

×

×

∞C

( ) ( )12

ist

i

f t F s e dsi

γ

γπ

+ ∞

− ∞

= ∫ RCThe integrand is analytic inside .

( )

:

00 ( )

R

st

C C

eF s

→ ∞

→ assumption

Only the vertical path contributes as

Right :Top and bottom :

(by Cauchy’s theorem)

Page 43: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

43

Laplace Transform (cont.)

( ) ( ) , 012 0, 0

ist

i

f t tF s e ds

i t

γ

γπ

+ ∞

− ∞

>=

<∫

Summary

( )Re s

( )Im s

γ

C××

×

Note: This inverse Laplace transform integral gives us zero for t < 0, no matter how the original

function f(t) was defined for t < 0.

Page 44: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

44

Close the contour to the left.

( ) ( )12

L

st

C

f t F s e dsiπ

= ∫

( ) Cf t = ∑ residues at poles to the left of

(This assumes that there are only pole singularities.)

Laplace Transform (cont.)Evaluation of the Bromwich integral for the case of poles:

( )Re s

( )Im s

γ

××

×

LC

−∞C

( ) ( )12

ist

i

f t F s e dsi

γ

γπ

+ ∞

− ∞

= ∫0t >

( )

:

00 ( )

L

st

C C

eF s

→ ∞

→ assumption

Only the vertical path contributes as

Left :Top and bottom :

Poles only

Page 45: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Laplace Transform (cont.)

45

Example ( ) atf t e=

( ) ( ) ( )

00 0

1s a t s a tat stF s e e dt e dt es a

∞∞ ∞− − − −− −

= = =−∫ ∫

( ) ( )1 , ReF s s as a

= >−

( ) 1 1 ,2

ist

i

f t e ds ai s a

γ

γ

γπ

+ ∞

− ∞

= >−∫

( )Re s

( )Im s

×

LC−∞

γa

( ) 1 12 Res2

st

s a

f t i ei s a

ππ =

= −

( ) , 0atf t e t⇒ = >

For t > 0:

Page 46: Notes 11 Evaluation of Definite Integrals via the Residue ...courses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes 11...There is no general rule for choosing the contour of integration;

Appendix

46

( ) ( )*H Hω ω− =Proof of symmetry property:

( ) ( )12

i th t H e dωω ωπ

−∞

= =∫ real

( ) ( ) ( ) ( ) ( ) ( )* * * * *1 1 1 12 2 2 2

i t i t i t i th t h t H e d H e d H e d H e dω ω ω ωω ω ω ω ω ω ω ωπ π π π

∞ −∞ ∞ ∞′ ′−

−∞ ∞ −∞ −∞

′ ′ ′ ′= = = − − = − = −∫ ∫ ∫ ∫

ω ω′ = −Use

( ) ( )*H Hω ω− =

so

( ) ( )*1 12 2

i t i tH e d H e dω ωω ω ω ωπ π

∞ ∞

−∞ −∞

= −∫ ∫

Hence:

Impulse response:

( ) ( ) i tH h t e dtωω∞

−∞

= ∫

ω ω′ →Relabel

( ) ( )1 1 *F H F Hω ω− − = −

( ) ( ) ( )R IH H i Hω ω ω= +( )inV ω+

−( )outV ω

+

( ) ( )( ) /out inH V Vω ω ω≡

Transfer function