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