TITLE: Improper integration involving complex analytic arguments
QUESTION [9 upvotes]: I am trying to evaluate the following:
$\displaystyle \int_{0}^{\infty} \frac{1}{1+x^a}dx$, where $a>1$ and $a \in \mathbb{R}$
Any help will be much appreciated.

REPLY [6 votes]: Here's another approach using contour integrals. 
There are some similarities with @ChristopherA.Wong's answer, though the transformation and contour are completely different.
Let $x = e^{2t/a}$. Then 
$$\begin{align*}
I_1 &= \int_0^\infty \frac{dx}{1+x^a} \\
    &= \frac{1}{a} \int_{-\infty}^\infty dt\, \frac{e^{(2/a-1)t}}{\cosh t} \\
    &= \lim_{R\to\infty} \frac{1}{a} \int_{\gamma_1} dt\, \frac{e^{(2/a-1)t}}{\cosh t}.
\end{align*}$$
The curve $\gamma_1$ is shown below.
The integrand has poles at $(2n+1)i \pi/2$ for $n\in\mathbb{Z}$. 
Consider the integral along the contour, 
$I_\gamma = \sum_{k=1}^4 I_k.$
By Cauchy's residue theorem
$$\begin{align*}
I_\gamma &= 2\pi i \, \mathrm{Res}_{t=i\pi/2}
    \frac{1}{a} \frac{e^{(2/a-1)t}}{\cosh t} \\
    &=  \frac{2\pi i}{a} \frac{e^{(2/a-1)i\pi/2}}{\sinh i\pi/2} \\
    &= -\frac{2\pi i}{a} e^{i\pi/a}.
\end{align*}$$
The integrals $I_2$ and $I_4$ are exponentially suppressed in the limit $R\to\infty$ since $2/a-1 < 1$.
The integral $I_3$ can be written in terms of $I_1$,
$$\begin{align*}
I_3 &= \frac{1}{a}\int_{\gamma_3} dt\, \frac{e^{(2/a-1)t}}{\cosh t} \\
    &= -e^{2\pi i/a} I_1.
\end{align*}$$
Therefore, in the limit $I_\gamma = I_1 + I_3 = I_1(1-e^{2\pi i/a}) = -2\pi i e^{i\pi/a}/a$ and so
$$I_1 = \frac{\pi \csc \pi/a}{a}.$$
Figure 1. The contour for $I_\gamma$.