calculus - Evaluating $intlimits_0^infty ! frac{x^{1/n}}{1+x^2} mathrm{d}x$

I've been trying to evaluate the following integral from the 2011 Harvard PhD Qualifying Exam. For all $n\in\mathbb{N}^+$ in general:

$$\int\limits_0^\infty \! \frac{x^{1/n}}{1+x^2} \ \mathrm{d}x$$

However, I'm not quite sure where to begin, even. There is a possibility that it is related to complex analysis, so I tried going at it with Cauchy's and also with residues, but I still haven't managed to get any further in solving it.

Answer

Beta function

I see @robjohn beat me to it.
The substitution is slightly different, so here it is.

Here's a simple approach using the beta function.

First, notice the integral diverges logarithmically for $n=1$, since the integrand goes like $1/x$ for large $x$.

Let $t=x^2$.

Then
$$\begin{eqnarray*} \int_0^\infty dx\, \frac{x^{1/n}}{1+x^2} &=& \frac{1}{2}\int_0^\infty dt\, \frac{t^{(1-n)/(2n)}}{1+t} \\ &=& \frac{1}{2} \textstyle B\left(\frac{1}{2} + \frac{1}{2n}, \frac{1}{2} - \frac{1}{2n}\right) \\ &=& \frac{1}{2} \textstyle\Gamma\left(\frac{1}{2} + \frac{1}{2n}\right) \Gamma\left(\frac{1}{2} - \frac{1}{2n}\right) \\ &=& \frac{\pi}{2 \sin\left(\frac{\pi}{2} + \frac{\pi}{2n}\right)} \\ &=& \frac{\pi}{2} \sec \frac{\pi}{2n}. \end{eqnarray*}$$

Some details

Above we use the integral representation for the beta function
$$B(x,y) = \int_0^\infty dt\, \frac{t^{x-1}}{(1+t)^{x+y}}$$
for $\mathrm{Re}(x)>0$, $\mathrm{Re}(y)>0$.
We also use Euler's reflection formula,
$$\Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin\pi z}.$$

Addendum: A method with residue calculus

Let $t = x^{1/n}$. Then
$$\begin{eqnarray*} I &=& \int_0^\infty dx\, \frac{x^{1/n}}{1+x^2} \\ &=& n\int_0^\infty dt\, \frac{t^n}{t^{2n}+1} \end{eqnarray*}$$
Notice the last integral has no cuts for integral $n$.
The residues are at the roots of $t^{2n}+1=0$.

Consider the pie-shaped contour with one edge along the positive real axis, another edge along the line $e^{i\pi/n}t$ with $t$ real and positive, and the "crust" at infinity.

$\hspace{4.5cm}$

There is one residue in the contour at $t_0 = e^{i\pi/(2n)}$.
The integral along the real axis is just $I$.
The integral along the other edge of the pie is
$$\begin{eqnarray*} I' &=& n\int_\gamma dz\,\frac{z^n}{z^{2n}+1} \\ &=& n \int_\infty^0 dt e^{i\pi/n} \frac{t^n e^{i\pi}}{t^{2n}+1} \\ &=& -e^{i(n+1)\pi/n} I. \end{eqnarray*}$$
The integral along the crust goes like $1/R^{2n-1}$ as the radius of the pie goes to infinity, and so vanishes in the limit.
Therefore,
$$\begin{eqnarray*} I + I' &=& 2\pi i \,\mathrm{Res}_{t=t_0}\, \frac{t^n}{t^{2n}+1} \\ &=& 2\pi i \frac{t_0^n}{2n t_0^{2n-1}}. \end{eqnarray*}$$
Using this and the formula for $I'$ in terms of $I$ above we find

$$I = \frac{\pi}{2} \sec \frac{\pi}{2n},$$
as before.