Complex Analysis - Maths KU

6.6 Some Consequences of the Residue Theorem 357
It is often tedious to have to show that the contour integral along CR
approaches zero as R →∞.Sufficient conditions under which this behavior
is always true are summarized in the next theorem.
Theorem 6.17 Behavior of Integral as R →∞
Suppose f(z) = p(z)
is a rational function, where the degree of p(z) is
q(z)
n and the degree of q(z) ism≥ n +2.IfCR is a semicircular contour
z = Reiθ , 0 ≤ θ ≤ π, then �
f(z) dz → 0asR→∞. CR
In other words, the integral along CR approaches zero as R →∞when
the denominator of f is of a power at least 2 more than its numerator.The
proof of this fact follows in the same manner as in Example 2.Notice in that
example, the conditions stipulated in Theorem 6.17 are satisfied, since degree
of p(z) = 1 is 0 and the degree of q(z) =(z 2 + 1)(z 2 +9)is4.
EXAMPLE 3 Cauchy P.V. of an Improper Integral
� ∞
1
Evaluate the Cauchy principal value of
x4 +1 dx.
Solution By inspection of the integrand we see that the conditions given in
Theorem 6.17 are satisfied. Moreover, we know from Example 3 of Section
6.5 that f(z) =1/(z 4 + 1) has simple poles in the upper half-plane at z1 =
e πi/4 and z2 = e 3πi/4 . We also saw in that example that the residues at these
poles are
Res(f(z),z1) =− 1
4 √ 1
−
2 4 √ 2 i and Res(f(z),z2) = 1
4 √ 1
−
2 4 √ 2 i.
Thus, by (14),
� ∞
P.V.
−∞
−∞
1
x 4 +1 dx =2πi [Res(f(z),z1)+Res(f(z),z2)] = π √ 2 .
Since the integrand is an even function, the original integral converges to
π �√ 2.
Integrals of the Form � ∞
f(x) cos αxdx and
−∞ � ∞
f(x) sinαxdx Because improper integrals of the form
� −∞
∞
f(x) sinαxdx are encountered in applications of Fourier analysis, they
−∞
often are referred to as Fourier integrals. Fourier integrals appear as the
real and imaginary parts in the improper integral � ∞
−∞ f(x)eiαx dx. ‡ In view
‡ See Section 6.7.