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### Transcript of Notes 11 Evaluation of Definite Integrals via the Residue ... Notes/Notes 11... There is no general

• 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

• Review of Singular Integrals

 Logarithmic singularities are examples of integrable singularities:

( )1 1 1 0 00 0 Ln Ln Ln 1 lim Ln 0lim lim

x x x dx x dx x x x x x

εεε ε = →→ → = = − = − =∫ ∫ since

1 x

Ln x

2

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

π θ π− < < ( ) ( )Ln lnz r iθ= +

• Singularities like 1/x are non-integrable.

3

Review of Singular Integrals (cont.)

x

1 x

( )1 1 1 0 0 0

1 1 Lnlim lim x

dx dx x x x εεε ε =→ →

= = = −∞∫ ∫

• Review of Cauchy Principal Value Integrals

2 0 2 0 2

1 01 1 0 Ln Ln

x x

dx dx dxI x x x x x =− =− −

= = + = +∫ ∫ ∫

Consider the following integral:

A finite result is obtained if the integral interpreted as

( )2 2 211 10 0 0

lim lim Ln Ln

lim Ln

x x

dx dx dxI x x x x x

ε ε

εεε ε

ε ε

− −

=− =+− −→ →

= = + = +

=

∫ ∫ ∫ Ln1 Ln2 Lnε− + −( ) Ln2=

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

2 2

1 1 PV dx dxI

x x− − = =∫ ∫

or

4

x

1/x ε−

ε 1−

2

Notation:

•  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

1 x

5

Cauchy Principal Value Integrals (cont.)

• 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 dx x x a b

xx x x

ε

ε ε ε −    + = + + − → ∞   

   

 > =  −

• 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

• Integrals Along the Real Axis

( )f x dx ∞

−∞ ∫

On the large semicircle we then have

0

lim ( ) lim ( ) 0 R

i i

R R C

f z dz f Re iRe d π

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 0 z

z f z →∞

=

8

Assumptions:

Hence

R−

: ,iR i

C z R e dz iR e d

θ

θ θ

=

=

R x

y ×

× ×

×

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

• Example:

( )( ) ( )( ) 2 2

2 22 2 2 2 0

4( )

1 1 2 29 4 9 4

f z

z dx zI dx z z z z

z

∞ ∞

−∞

      

−=

= = = + + + +

∫ ∫ 

2 ( ) ( )

( ) 3

Res 3 + Res 2

3 lim z i

i f i f i

z i i

π

π →

  

− =

( ) ( )

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 50 200 200

z i

z i

z z i z z z z i z i zii z z i

i ii

π

ππ

         +      + + − + + + +  =    + +     = − =  

9

( ) ( ) 2

22 2 0 9 4

x dxI x x

= + +

( ) 1

0 01

0

1Res ( ) ( ) ( ) 1 !

m m

m df z z z f z

m dz z z

−  = − − =

( ) ( ) 2

22 2 0 2009 4

x dxI x x

π∞ = =

+ + ∫

z = 2i z = 3i

x

y

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

×

×

Integrals Along the Real Axis (cont.)

• ( ) ( ) ( ) ( )

( ) ( )

1

2 2, 10

2 0 0

lim 1 1 1 1

lim lim 1 1

R

R

R

C R C R C C

i

R C C

dz dzI z z z z

i edzI I z z

ρ

ρ

ρ

ρρ

θ

ρ ρ

ρ

− −

→∞ − − +→

→∞ → →

   = = + + +  + + + + 

   = + + = +   + + 

∫ ∫ ∫ ∫ ∫

∫ ∫

( )21 1i i d

e eθ θ θ

ρ ρ − + +   ( ) ( )

2

2 2 0

2 2 2

2 2 0 0

lim 1 1

lim lim 0 2 2 2

i

i iR

i i

R R

iR e d R e R e

id ie d i ie d iI I I R R

π π φ

φ φ π

π π πφ φ

π

φ

θ φ π φ π

→∞

− −

→∞ →∞

+ + +

= + + = + + = + +

∫ ∫

∫ ∫ ∫

Consider the following integral: I



10

Cauchy Principal Value Integrals Example

( ) ( )2 1 1 dxI

x x ∞

−∞ =

+ +∫

2C iI I π= +

Hence,

Evaluate: z i=

z i= −

: ,iR i

C z R e dz iR e d

φ

φ φ

=

=

: 1 ,i i

C z e dz i e d

θ ρ

θ

ρ ρ 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.

• [ ] ( )

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

I i f z i f z iπ π →

− = = + = − =

( )z i− ( )( ) ( )

1

1 lim

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 2 i ii i i i

i i i π ππ π π

      

  − − −     = + = + = + = +      +       

Next, evaluate the integral IC using the residue theorem:

11

( ) ( )2 1 1 dxI

x x ∞

−∞ =

+ +∫

2 I π=

Hence:

Cauchy Principal Value Integrals (cont.)

2 2C I iπ π = +  

  2 C

iI I π= +

Recall:Thus:

z i=

z i= −

: ,iR i

C z R e dz iR e d

φ

φ φ

=

=

: 1 ,i i

C z e dz i e d

θ ρ

θ

ρ ρ θ

+ = =

RR− 1z = −

×

×

×

C

• 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 e dz iR e d

θ

θ θ

=

=

R x

y

× ×

×

• Jordan’s lemma

( )

( )

/2 cos sin sin

0 0 /2 2

0

( ) ( ) 2 max ( )

2 max ( ) 2

R

iaz i iaR aR i i aR

C aR

i

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

R f Re e d R

π π θ θ θ θ θ θ

π θ θ π

θ θ

θ

− −

= ≤

≤ =

∫ ∫ ∫

∫ ( ) max ( ) 2 if Re

a R θ π ( )1 0aRe−− →

2 sin20 / 2 : sin