Evaluation of Integral(ebooksforgate.blogspot.in)

Embed Size (px)

Citation preview

  • 8/13/2019 Evaluation of Integral(ebooksforgate.blogspot.in)

    1/5

  • 8/13/2019 Evaluation of Integral(ebooksforgate.blogspot.in)

    2/5

    178

    ba

    f(x) dx= lim0

    a

    f(x) dx+

    b+

    f(x) dx (7.2)

    provided the limits exist. The limits in equations (7.1) and (7.2) do not always produce

    the same results. Whenever the integral given by equation (7.2) exists, then the value of

    the integral is called the Cauchy principal value. Note that the Cauchy principal value issymmetric with the limits decreasing at the same rate, while in comparison the limits in

    equation (7.1) can decrease at different rates.

    The same idea applies to integration involving infinite limits. The integration of rational

    functions f(x) over the interval [0, ) are called improper integrals and are defined by a

    limiting process

    0

    f(x) dx= limR

    R0

    f(x) dx

    if this limit exists. Similarly, the integration of real rational functions f(x), defined for all

    real values of x, over the interval (, )are calculated from the limiting process

    f(x) dx= limR1

    0R1

    f(x) dx+ limR2

    R20

    f(x) dx (7.3)

    provided that both limits exist. The limiting process given by equation (7.3) is not the same

    as the limiting process

    f(x) dx= limR

    RR

    f(x) dx (7.4)

    because in one case the limiting process is symmetric and nonsymmetric in the other case.

    If the limit in equation (7.3) exists, then the same value can be obtained from the limiting

    process given by equation (7.4). However, if the limit given by equation (7.4) exists, thenthe limit given by equation (7.3) may or may not exist. This can be illustrated with a simple

    example. Consider the limiting process given by equation (7.4) in the special case f(x) =x

    limR

    RR

    x dx= limR

    x2

    2

    R

    R

    = limR

    R2

    2

    (R)2

    2

    = 0 (7.5)

    However, the limiting process defined by equation (7.3) for the function f(x) =x produces

    limR1

    0R1

    x dx + limR2

    R20

    x dx= limR1

    x2

    2

    0

    R1

    + limR2

    x2

    2

    R2

    0

    = limR1

    R212

    + limR2

    R222

    (7.6)

    which becomes an indeterminate form in the limit asR1and R2 increase without bound. The

    limiting process defined by equation (7.4) is used to define the Cauchy principal value of an

    improper integral.

    The Cauchy principal value of the improper integral

    f(x) dx, of a real rational function

    f(x) is defined by the limiting process

    Cauchyprincipal

    value

    f(x) dx= limR

    RR

    f(x) dx (7.7)

    ebooksforgate.blogspot.in

    ebooksforgate.blogspot.in

  • 8/13/2019 Evaluation of Integral(ebooksforgate.blogspot.in)

    3/5

    179

    provided that the limit exists. If the limit exists, then the integral is said to be a convergent

    integral in the Cauchy sense, otherwise it is called divergent.

    Consider the contour integral of a complex rational function f(z)taken around the special

    contour illustrated in the figure 7-1. Assume that the rational function f(z) has poles at the

    points z1

    , z2

    , . . . , z n

    lying in the upper half plane. The contour of integration consists of two

    parts. The first part is the upper semicircle CR centered at the origin and having a radius

    R large enough to enclose all the poles of f(z) in the upper half plane. The second part of

    the contour is the straight line segment from (R, 0)to (R, 0). The integration off(z)around

    this special contour Cis performed in the positive sense. The semicircular part of the path

    of integration is given the notation CR to emphasize that this portion of the path depends

    upon the radius R.

    Figure 7-1. Contour path Cconsisting of semicircleCR and straight line path

    from R to Ron the x-axis. It is assumed that f(z) has poles at points

    z1, z2, . . . , z n in the upper half-plane which lie inside the contour C.

    Consider the contour integral off(z) around the path Cof figure 7-1 in the limit as the

    radius R increases without bound. This type of integral can be evaluated using the residue

    theorem by writing

    limR...........................................................................................C f(z) dz= limR

    RR f(x) dx+

    CR f(Re

    i

    )iR ei

    d

    = 2i

    nj=1 Res [ f, z

    j] (7.8)

    where z = x is the value of z on the line segment and z = Rei is the value of z on the

    semicircular path CR. If we can show that the line integral along CR approaches zero as R

    increases without bound, then one can write

    limR

    CR

    f(z) dz= limR

    0

    f(Rei) Rei i d = 0 (7.9)

    ebooksforgate.blogspot.in

    ebooksforgate.blogspot.in

  • 8/13/2019 Evaluation of Integral(ebooksforgate.blogspot.in)

    4/5

    180

    and consequently the integral in equation (7.8) reduces to the Cauchy principal value of an

    improper integral

    limR

    RR

    f(x) dz=

    f(x) dx= 2in

    j=1

    Res [ f, zj] (7.10)

    If the integrand f(z), for z= Rei

    on the semicircle CR, satisfies |f(z)|

    M

    Rk where M and kare constants with k >1, then integrals of the type given by equations (7.9) and (7.10) can

    be shown to be valid. Iff(z) satisfies the above boundedness property, then one can employ

    the magnitude of a line integral property, equation (3.9), to obtain the resultCR

    f(z) dz

    MRk R= MRk1 , k >1 (7.11)where R is the length of the semicircle arc CR. It then follows that limR

    CR

    f(z) dz= 0.

    Wheneverf(z) =P(z)

    Q(z)is a rational function with P(z)a polynomial of degree m with real

    coefficients a0, a1, . . . , am and Q(z) is a polynomial of degree n m+ 2 with real coefficients

    b0, b1, . . . , bn, then one can writeP(z)

    Q(z) =

    a0zm +a1z

    m1 + +am1z+amb0zn +b1zn1 + +bm1z+bm

    = zmp(z)

    zn q(z)

    where

    p(z) =a0+a1z1 + +am1z

    m+1 +amzm and q(z) =b0+b1z1 + +bn1zn+1 +bnzn

    In the special case z= Rei lies on the path CR in figure 7-1 one finds that

    |zf(z)| =

    zP(z)

    Q(z)

    Rm+1

    Rn

    |p(z)|

    |q(z)|

    and in the limit as Rincreases without bound we can write

    limR

    Rm+1Rn |p(z)||q(z)| limR R

    m+1

    Rna0b0

    = 0

    because n m + 2. Therefore, one can select a radius Rso large that zP(z)Q(z) which implies

    the inequalityP(z)Q(z)

    < R whenever z= Rei is on the semicircle CR.It then follows from themagnitude of an integral property, equation (3.9), that

    CR

    P(z)

    Q(z)dz

    CR

    R|dz| =

    RR= (7.12)

    In the limit as R increases without bound the integral on the left-hand side of equation (7.12)

    is some constant which is less than which can be made arbitrarily small by making small.This implies that the constant on the left-hand side must be zero. Hence, for P(z) and Q(z)

    polynomials of degree mand n respectively, with n m+ 2, one can write

    limR

    CR

    P(z)

    Q(z)dx = 0 (7.13)

    ebooksforgate.blogspot.in

    ebooksforgate.blogspot.in

  • 8/13/2019 Evaluation of Integral(ebooksforgate.blogspot.in)

    5/5

    181

    Example 7-1. (Evaluation of improper integral)

    Evaluate the Cauchy principal value of

    dx

    2 +x2, where >0 is a real constant.

    Solution: Use the path of integration illustrated in the figure 7-1 and replace the integrand

    by the complex function f(z) = 1

    2 +z2. The residue theorem gives

    ....

    ...........................................................

    ............................

    C

    f(z) dz= 2iRes [ f(z), i ]

    since only the simple pole at z= i lies within the contour Cof figure 7-1. The residue of

    f(z) at z= i is found to be 1/2i. In the limit as R we write

    limR

    ...............................................................

    ............................

    C

    f(z) dz= limR

    CR

    f(z) dz+

    RR

    dx

    2 +x2

    = 2i

    1

    2i=

    If we can show that limR

    CRf(z) dz

    = 0, then we are left with the Cauchy principal

    value

    dx

    2 +x2 =

    . We can use the previous discussion to show limRCR f(z) dz = 0.

    Alternatively, for z on CR we can use the inequality |z1+z2| |z1| |z2| to show for zon CR

    that |f(z)| = 1z2+2

    1R22. One can then show limR CR f(z) dz

    limR RR22 = 0.Here f(z) satisfies the boundedness property required for the integral along CR to approach

    zero as Rincreases without bound. One can use the rational function P(z)/Q(z)approach or

    some other inequality to show a boundedness property on a curve.

    Example 7-2. (Evaluation of improper integral)

    Evaluate the improper integral 0

    dx

    x6 + 1

    Solution: Replace the integrand by the complex function f=f(z) = 1

    z6 + 1and use the residue

    theorem to evaluate the integral off(z) around the path Cillustrated in the figure 7-1 in the

    limit as R increases without bound. One can then write

    limR

    .....

    ..........................................................

    ............................

    C

    f(z) dz= limR

    RR

    f(x) dx+

    CR

    f(Rei)Rei id

    = 2i

    nk=1

    Res [ f, zk] (7.14)

    where n represents the number of singularities of f(z) in the upper half plane. The line

    integral along the semicircular path CR approaches zero as Rincreases without bound. Note

    that if|z1+z2| |z1| |z2|, then 1|z1+z2| 1|z1| |z2| consequently one can show

    limR

    CR

    f(z) dz

    limRCR

    1z6 + 1 dz limR RR6 1= 0 (7.15)

    The singularities off(z) occur where z6 + 1 = 0. We can write

    z6 = 1 =ei (+2k), or z= ei(/6+2k/6) where for k= 0, 1, 2, 3, 4, 5we obtain

    ebooksforgate.blogspot.in

    ebooksforgate.blogspot.in