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

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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