Slides 12 Handout

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Transcript of Slides 12 Handout

  • Evaluation of integrals

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type I

    Type I: Integrals of the form 2pi0

    F (cos , sin ) d

    If we take z = e i, then cos = 12(z + 1

    z), sin = 1

    2i(z 1

    z) and d =

    dz

    iz.

    Substituting for sin , cos and d the definite integral transforms intothe following contour integral 2pi

    0

    F (cos , sin ) d =

    |z|=1

    f (z) dz

    where f (z) = 1iz

    [F ( 12(z + 1

    z), 1

    2i(z 1

    z))]

    Apply Residue theorem to evaluate|z|=1

    f (z) dz .

    Lecture 18 Evaluation of integrals

  • Example of Type I

    Consider 2pi0

    1

    1 + 3(cos t)2dt.

    2pi0

    1

    1 + 3(cos t)2dt =

    |z|=1

    1

    1 + 3( 12(z + 1

    z))2

    dz

    iz

    = 4i|z|=1

    z

    3z4 + 10z2 + 3dz

    = 4i|z|=1

    z

    3(z +

    3i)(z 3i

    )(z + i

    3

    )(z i

    3

    ) dz= 4

    3i

    |z|=1

    z

    (z +

    3i)(z 3i)(z + i

    3

    )(z i

    3

    ) dz= 4

    3i 2pii{Res(f , i

    3) + Res(f , i

    3)}.

    Lecture 18 Evaluation of integrals

  • Improper Integrals of Rational Functions

    The improper integral of a continuous function f over [0,) is defined by 0

    f (x)dx = limb

    b0

    f (x)dx

    provided the limit exists.

    If f is defined for all real x , then the integral of f over (,) isdefined by

    f (x)dx = lim

    a

    0a

    f (x)dx + limb

    b0

    f (x)dx

    provided both limits exists.

    There is another value associated with the improper integral f (x)dx

    namely the Cauchy Principal value(P.V.) and it is given by

    P. V.

    f (x)dx := limR

    RR

    f (x)dx

    provided the limit exists.

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type II

    If the improper integral f (x)dx converges, then P. V.

    f (x)dx

    exists and

    f (x)dx = P. V.

    f (x)dx .

    The P. V. f (x)dx exists 6= the improper integral

    f (x)dx

    exists. Take f (x) = x .

    However if f is an even function (i.e. f (x) = f (x) for all x R) thenP. V.

    f (x)dx exists = the improper integral

    f (x)dx exists and

    their values are equal.

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type II

    Consider the rational function f (z) =P(z)

    Q(z)where P(z) and Q(z) are

    polynomials with real coefficients such that

    Q(z) has no zeros in the real line

    degree of Q(z) > 1+ degree of P(z)

    then P. V.

    f (x)dx can be evaluated using Cauchy residue theorem.

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type II

    Type II Consider the integral

    1

    (x2 + 1)2dx ,

    To evaluate this integral, we look at the complex-valued function

    f (z) =1

    (z2 + 1)2

    which has singularities at i and i . Consider the contour C like semicircle, theone shown below.

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type II

    Note that: C

    f (z) dz =

    aa

    f (z) dz +

    Arc

    f (z) dz aa

    f (z) dz =

    C

    f (z) dz

    Arc

    f (z) dz

    Furthermore observe that

    f (z) =1

    (z2 + 1)2=

    1

    (z + i)2(z i)2 .

    Then, by using Residue Theorem,C

    f (z) dz =

    C

    1(z+i)2

    (z i)2 dz = 2piid

    dz

    (1

    (z + i)2

    ) z=i

    =pi

    2

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type II

    Again, Arc

    f (z) dz

    api(a2 1)2 0 as a.So

    1

    (x2 + 1)2dx =

    f (z) dz = lima+

    aa

    f (z) dz =pi

    2.

    Theorem

    Suppose f is analytic in C except at a finite number ofsingular points in C \R. Let R denote the semi-circle |z | = Rin the upper half plane =(z) 0. If lim

    zzf (z) = 0 then

    limR

    Rf (z)dz = 0.

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type II

    Exercise.

    Show that P.V.

    (x2 x + 2)dx(x4 + 10x2 + 9)

    =5pi

    12. [Poles are i ,3i .]

    Show that

    0

    x2dx

    (x2 + 9)(x2 + 4)2=

    pi

    200.

    Show that

    0

    dx

    x3 + 1=

    2pi

    3

    3.

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type III

    Type III Integrals of the form

    P. V.

    P(x)

    Q(x)cosmx dx or P. V.

    P(x)

    Q(x)sinmx dx ,

    where

    P(x),Q(x) are real polynomials and m > 0

    Q(x) has no zeros in the real line

    degree of Q(x) > degree of P(x)

    then

    P. V.

    P(x)

    Q(x)cosmx dx or P. V.

    P(x)

    Q(x)sinmx dx

    can be evaluated using Cauchy residue theorem.

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type III

    Evaluate:

    cosx

    x2 + 1dx or

    sinx

    x2 + 1dx

    Consider the integral

    e ix

    x2 + 1dx

    We will evaluate it by expressing it as a limit of contour integrals along the

    contour C that goes along the real line from a to a and thencounterclockwise along a semicircle centered at 0 from a to a. Take a > 1 sothat i is enclosed within the curve.

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type III

    Res

    (e iz

    z2 + 1, i

    )= lim

    zi(z i) e

    iz

    z2 + 1= lim

    zie iz

    z + i=

    e

    2i.

    So by residue theoremC

    f (z) dz = (2pii)Res(f , i) = 2piie

    2i= pie.

    The contour C may be split into a straight part and a curved arc, so thatstraight

    +

    arc

    = pie

    and thus aa

    e ix

    x2 + 1dx = pie

    arc

    e iz

    z2 + 1dz .

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type III

    arc

    e iz

    z2 + 1dz

    pi0

    a e ia(cos+i sin )a2 1 d

    aa2 1

    pi0

    ea sin d.

    Hence, arc

    e iz

    z2 + 1dz

    0 as aand

    P.V.

    cosx

    x2 + 1dx = lim

    a

    aa

    e ix

    x2 + 1dx

    = lima

    [pie

    arc

    e iz

    z2 + 1dz

    ]= pie.

    Lecture 18 Evaluation of integrals

  • Jordans Lemma

    Lemma

    Let f be analytic in C except for finite number of singular points inC \ R. Let R denote the semicircle |z | = R in the upper halfplane. If lim

    R

    (max|z|=R

    |f (z)|)

    = 0, then limR

    R

    f (z)e iazdz = 0.

    The result follows from Jordans inequality: pi0

    eR sin d 0).

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type IV

    Type IV Integrals of the form 0

    sin x

    xdx

    can be evaluated using Cauchy residue theorem.Before we discuss integrals of Type IV we need the following result.

    Lemma: Suppose f has a simple pole at z = a on the real axis. If c is thecontour defined by c(t) = a + e

    i(pit), t (0, pi) then

    lim0

    c

    f (z)dz = ipiRes(f , a).

    Proof: Since f has a simple pole at z = a, the Laurent series expansion of fabout z = a is of the form

    f (z) =Res(f , a)

    z a + g(z).

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type IV

    Nowc

    f (z)dz =

    c

    Res(f , a)

    z a dz +c

    g(z)dz

    = Res(f , a) pi

    0

    ie i(pit)

    e i(pit)dt

    pi0

    g(a + e i(pit))ie i(pit)dt

    = ipiRes(f , a) pi

    0

    g(a + e i(pit))ie i(pit)dt.

    Note that f has Laurent series expansion in 0 < |z a| < R for some R > 0.The function g is continuous on |z a| 0 for every < 0 < R. So|g(z)| < M on |z a| 0. So pi

    0

    g(a + e i(pit))ie i(pit)dt

    Mpi 0 as 0.Hence

    lim0

    c

    f (z)dz = ipiRes(f , a).

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type IV

    Consider the integral 0

    sin x

    xdx .

    Define f (z) = eiz

    z, (z = 0 is a simple pole on the real axis).

    Consider the contour C = [R,] [,R] .

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type IV

    By Cauchys theoremC

    e iz

    zdz =

    [R,]

    e iz

    zdz +

    e iz

    zdz +

    [,R]

    e iz

    zdz +

    e iz

    zdz = 0.

    But [R,]

    e iz

    zdz +

    [,R]

    e iz

    zdz =

    [,R]

    e ix eixx

    dx

    So [,R]

    e ix eixx

    dx =

    e iz

    zdz

    e iz

    zdz = ipi

    as 0 (by the previous Lemma ) and R (by Jordans inequality) andhence,

    0

    sin x

    xdx =

    pi

    2.

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type V

    Integration along a branch cut: Consider the improper integral 0

    xa

    1 + xdx (0 < a < 1).

    Define

    f (z) =za

    1 + z(|z | > 0, 0 < arg z < 2pi).

    The functionza

    1 + zis a multiple valued function with branch cut

    arg z = 0 (positive real axis).

    Consider the contour C = [+ i,R + i] R [R i, i] {}.

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type V

    By residue theorem([+i,R+i]

    +

    R

    +

    [Ri,i]

    +

    )f (z)dz = 2piiRes(f ,1) = 2piieiapi.

    Since

    f (z) =exp(a log z)

    z + 1=

    exp(a(ln r + i))re i + 1

    ,

    where z = re i, it follows thatOn [+ i,R + i], 0 as 0,

    f (z) =exp(a(ln r + i .0))

    re i.0 + 1 r

    a

    1 + ras 0.

    On [R i, i], 2pi as 0,

    f (z) =exp(a(ln r + i .2pi))

    re i.2pi + 1 r

    a

    1 + re2apii as 0.

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type V

    But R

    za

    1 + zdz

    RaR 1 2piR = 2piRR 1 1Ra 0 as R and

    za

    1 + zdz

    a 1 2pi = 2pi1 1a 0 as 0.So

    limR,0

    ( R

    ra

    1 + rdr +

    R

    ra

    1 + re2apiidr

    )= 2piieiapi

    That is

    (1 e2apii )

    0

    ra

    1 + rdr = 2piieiapi

    and hence 0

    ra

    1 + rdr =

    2piieiapi

    (1 e2apii ) =pi

    sin api(0 < a < 1).

    Lecture 18 Evaluation of integrals

  • Evaluation of certain contour integrals: Type VI

    Integration around a branch cut:Consider the improper integral

    0

    log x

    1 + x2dx .

    Define

    f (z) =log z

    1 + z2(|z | > 0, pi

    2< arg z