Complex Analysis - Maths KU

6.6 Some Consequences of the Residue Theorem 361
Therefore,
� ∞
P.V.
−∞
eix x(x2 dx = πi
− 2x +2)
� �
1
2
+ 2πi
�
− e−1+i
�
(1 + i) .
4
Using e −1+i = e −1 (cos 1 + i sin 1), simplifying, and then equating real and
imaginary parts, we get from the last equality
and
� ∞
P.V.
−∞
� ∞
P.V.
−∞
cos x
x(x2 π
dx =
− 2x +2) 2 e−1 (sin 1 + cos 1)
sin x
x(x2 π
dx =
− 2x +2) 2 [1 + e−1 (sin 1 − cos 1)].
6.6.3 Integration along a Branch Cut
Branch Point at z =0 In the next discussion we examine integrals
of the form � ∞
f(x) dx, where the integrand f(x) is algebraic.But similar
0
to Example 5, these integrals require a special type of contour because when
f(x) is converted to a complex function, the resulting integrand f(z) has, in
addition to poles, a nonisolated singularity at z = 0.Before proceeding, the
reader is encouraged to review the discussion on branch cuts in Sections 2.6
and 4.1.
In the example that follows we consider a special case of the real integral
� ∞
xα−1 dx, (21)
x +1
0
where α is a real constant restricted to the interval 0