Page 761 - Advanced_Engineering_Mathematics o'neil

22.3 Evaluation of Real Integrals 741

Take the limit in this equation as R →∞. In this limit, the semicircle γ R expands over the entire
upper half-plane, and the interval [−R, R] expands over the real line. Furthermore, because q(z)
2
has degrees at least 2 more than the degree of p(z), the degree of z p(z) does not exceed that
2
2
of q(z). This means that, for large R, z p(z)/q(z) is bounded. If |z p(z)/q(z)|≤ M for |z|≥ R,
then
M M

p(z)

≤ ≤ for |z|≥ R.

2 2
q(z) |z| R
But then, because γ R has length π R,
p(z) M

dz
≤ (π R) → 0as R →∞.
q(z) R

2
S R
Therefore, in the limit as R →∞ in equation (22.2), the integral over γ R tends to zero and we
are left with
m
p(x)
∞

dx = 2πi Res(p(z)/q(z), z j ). (22.3)
−∞ q(x)
j=1
∞
In sum, under the stated assumptions, p(x)/q(x)dx can be evaluated as 2πi times the
−∞
sum of the residues of p(z)/q(z) at the zeros of q(z) occurring in the upper half-plane.
EXAMPLE 22.17
We will evaluate

∞
1
dx.
∞ x + 64
6
6
The conditions of the method are met with p(z)=1 and q(z)= z +64. The zeros of q(z) are the
sixth roots of −64. To find these, put −64 in polar form as −64 = 64e i(π+2nπ) in which n can be
any integer. The sixth roots are 2e i(π+2nπ) for n = 0,1,2,3,4,5. The three sixth roots in the upper
half-plane correspond to n = 0,1,2 and are
z 1 = 2e πi/6 , z 2 = 2e πi/2 = 2i, and z 3 = 2e 5πi/6 .
We need the residue of p(z)/q(z) at each of these simple poles:
1 1
Res(p(z)/q(z), z 1 ) = = e −5πi/6 ,
6(2e πi/6 5 192
)
1 i
Res(p(z)/q(z), z 2 ) = =− ,
6(2i) 5 192
and
1 1 −25πi/6 1 −πi/6
Res(p(z)/q(z), z 3 ) = = e = e .
6(2e 5πi/6 5 192 192
)
Then
∞

p(x) 2πi
dx = [e −5πi/6 − i + e −πi/6 ]
−∞ q(x) 192
π
π
πi 5π 5π
= cos − i sin − i + cos − i sin .
96 6 6 6 6
Now
5π
5π
1
π

π
cos + cos = 0 and sin = sin = ,
6 6 6 6 2
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October 14, 2010 15:37 THM/NEIL Page-741 27410_22_ch22_p729-750