Complex Analysis - Maths KU

–R
y
–C r
c
C R
Figure 6.13 Indented contour
x
R
6.6 Some Consequences of the Residue Theorem 359
from (4) of Section 6.5. Then, from Theorem 6.18 we conclude
�
CR f(z)eiz dz → 0asR →∞, and so
But by (16),
� ∞
−∞
� ∞
P.V.
−∞
x
x 2 +9 eix dx =2πi
� � −3 e
2
= π
i.
e3 x
x2 +9 eix � ∞
x cos x
dx =
−∞ x2 � ∞
x sin x
dx + i
+9 −∞ x2 +9
π
dx = i.
e3 Equating real and imaginary parts in the last line gives the bonus result
� ∞
x cos x
P.V.
−∞ x2 � ∞
x sin x
dx = 0 along with P.V.
+9 −∞ x2 +9
π
dx = . (18)
e3 Finally, in view of the fact that the integrand is an even function, we obtain
the value of the prescribed integral:
� ∞
0
x sin x
x 2 +9
� ∞
1 x sin x
dx =
2 −∞ x2 +9
π
dx = .
2e3 Indented Contours The improper integrals of forms (2) and (3) that
we have considered up to this point were continuous on the interval (−∞, ∞).
In other words, the complex function f(z) =p(z)/q(z) did not have poles on
the real axis.In the situation where f has poles on the real axis, we must
modify � the procedure illustrated in Examples 2–4.For example, to evaluate
∞
−∞
f(x) dx by residues when f(z) has a pole at z = c, where c is a real
number, we use an indented contour as illustrated in Figure 6.13. The
symbol Cr denotes a semicircular contour centered at z = c and oriented in
the positive direction.The next theorem is important to this discussion.
Theorem 6.19 Behavior of Integral as r → 0
Suppose f has a simple pole z = c on the real axis.If Cr is the contour
defined by z = c + reiθ , 0 ≤ θ ≤ π, then
�
lim f(z) dz = πi Res(f(z),c).
r→0
Cr
Proof Since f has a simple pole at z = c, its Laurent series is
f(z) = a−1
+ g(z),
z − c