Show that $\int_0^ \infty \frac{1}{1+x^n} dx= \frac{ \pi /n}{\sin(\pi /n)}$ , where $n$ is a positive integer. Using residues, try the contour below with $R \rightarrow  \infty$ and  $$\lim_{R \rightarrow  \infty }  \int_0^R  \frac{1}{1+r^n}  dr \rightarrow \int_0^\infty  \frac{1}{1+x^n}  dx$$

I've attempted the residue summation, but my sum did not converge.
 A: The integral of
$$
\int_\gamma\frac1{1+z^n}\mathrm{d}z\tag{1}
$$
on the outgoing ray on the real axis tends to
$$
\int_0^\infty\frac1{1+x^n}\mathrm{d}x\tag{2}
$$
On the incoming ray parallel to $e^{2\pi i/n}$, the integral tends to
$$
-e^{2\pi i/n}\int_0^\infty\frac1{1+x^n}\mathrm{d}x\tag{3}
$$
For $n\ge2$, the integral on the circular arc vanishes. Therefore,
$$
\int_\gamma\frac1{1+z^n}\mathrm{d}z
=\left(1-e^{2\pi i/n}\right)\int_0^\infty\frac1{1+x^n}\mathrm{d}x\tag{4}
$$
There is one singularity contained in $\gamma$ at $z_0=e^{\pi i/n}$. The residue of $\frac1{1+x^n}$ at $z_0$ is $\frac1{nz_0^{n-1}}=-\frac{z_0}{n}$. Thus,
$$
2\pi i\left(-\frac{e^{\pi i/n}}{n}\right)
=\left(1-e^{2\pi i/n}\right)\int_0^\infty\frac1{1+x^n}\mathrm{d}x\tag{5}
$$
which resolves by division to
$$
\int_0^\infty\frac1{1+x^n}\mathrm{d}x=\frac{\pi/n}{\sin(\pi/n)}\tag{6}
$$
For $n=1$, the integral diverges and $\frac{\pi}{\sin(\pi)}=\frac\pi0$.
A: Supposed that you have no idea of how to do integration over the complex, this problem can also be solved neatly.
Denote $t:=x^n$, and we have
\begin{align}
\int_0^\infty\frac{\mathrm{d} x}{1+x^n} &= \frac1n\int_0^\infty\frac{t^{\frac1n-1}}{1+t}\mathrm{d} t\\
&=\frac1n\:\mathrm{B}(\frac1n, 1-\frac1n) \\
&=\frac1n\:\Gamma(\frac1n)\cdot\Gamma(1-\frac1n)\\
&=\frac{\pi/n}{\sin \pi/n}
\end{align}
Here we used a famous result, $\displaystyle\Gamma(z)\cdot\Gamma(1-z)=\frac{\pi}{\sin{\pi z}} \quad (0<\mathrm{Re}(z)<1)$
