solutions to $\int_{-\infty}^\infty \frac{1}{x^n+1}dx$ for even $n$ I was playing around with glasser's master theorem and integrals of the form $$\int_{-\infty}^\infty \frac{1}{x^n+1}dx$$ I observed that for positive, even values of n, the solution to the integral followed the form $$\frac{\pi}{\frac{n}{2}\sin(\frac{\pi}{n})}$$, the obvious example being at n = 2, the integral equals $\pi$. I've yet to prove this pattern, but I recognized it after some fiddling around. Is this integral part of some larger concept that is already documented? For example, the way that $\int_0^\infty \frac{x^{p-1}}{e^x-1}dx = \zeta(p)\Gamma(p), p>1$, I can't help but think about why the general solution for even values has a sine function in it. 
 A: For any complex number $\alpha$ such that $0<\text{Re}(\alpha)<1$, let
$$I(\alpha):=\int_0^\infty\,\frac{t^{\alpha-1}}{t+1}\,\text{d}t\,,$$
where the branch cut of the map $z\mapsto z^{\alpha-1}$ is taken to be the positive-real axis.
For a real number $\epsilon\in(0,1)$, consider the positively oriented keyhole contour $\Gamma_\epsilon$ given by 
$$\begin{align}\left[\epsilon\,\exp(\text{i}\epsilon),\frac{1}{\epsilon}\,\exp(\text{i}\epsilon)\right]&\cup\left\{\frac{1}{\epsilon}\,\exp(\text{i}t)\,\Big|\,t\in[\epsilon,2\pi-\epsilon]\right\}
\\&\cup\left[\frac{1}{\epsilon}\,\exp\big(\text{i}(2\pi-\epsilon)\big),\epsilon\,\exp\big(\text{i}(2\pi-\epsilon)\big)\right]\cup\Big\{\epsilon\,\exp(\text{i}t)\,\Big|\,t\in[2\pi-\epsilon,\epsilon]\Big\}\,.\end{align}$$
Then,
$$\lim_{\epsilon\to0^+}\,\oint_{\Gamma_\epsilon}\,\frac{z^{\alpha-1}}{z+1}\,\text{d}z=I(\alpha)-\exp\big(2\pi\text{i}(\alpha-1)\big)\,I(\alpha)=-2\text{i}\,\exp(\pi\text{i}\alpha)\,\sin(\pi\alpha)\,I(\alpha)\,.$$
Via the Residue Theorem,
$$\lim_{\epsilon\to0^+}\,\oint_{\Gamma_\epsilon}\,\frac{z^{\alpha-1}}{z+1}\,\text{d}z=2\pi\text{i}\,\text{Res}_{z=-1}\left(\frac{z^{\alpha-1}}{z+1}\right)=2\pi\text{i}\,(-1)^{\alpha-1}=-2\pi\text{i}\,\exp(\pi\text{i}\alpha)\,.$$
Hence, $$\int_0^\infty\,\frac{t^{\alpha-1}}{t+1}\,\text{d}t=I(\alpha)=\frac{\pi}{\sin(\pi\alpha)}\,.$$
Now, take $\alpha:=\dfrac1n$ for some integer $n\geq 2$.  Then,
$$\frac{\pi}{\sin\left(\frac{\pi}{n}\right)}=\int_0^\infty\,\frac{t^{\frac1n-1}}{t+1}\,\text{d}t=\int_0^\infty\,\frac{n}{x^n+1}\,\text{d}x\,,$$
by setting $x:=t^{\frac1n}$.  This proves the equality
$$\int_0^\infty\,\frac{1}{x^n+1}\,\text{d}x=\frac{\pi}{n\,\sin\left(\frac{\pi}{n}\right)}\,.$$  If $n$ is even, then
$$\int_{-\infty}^{+\infty}\,\frac{1}{x^n+1}\,\text{d}x=2\,\int_0^\infty\,\frac{1}{x^n+1}\,\text{d}x=\frac{2\pi}{n\,\sin\left(\frac{\pi}{n}\right)}=\frac{\pi}{\frac{n}{2}\,\sin\left(\frac{\pi}{n}\right)}\,.$$ 
In fact, it can be seen that $I(\alpha)=\text{B}(\alpha,1-\alpha)=\Gamma(\alpha)\,\Gamma(1-\alpha)$, where $\text{B}$ and $\Gamma$ are the usual beta and gamma functions, respectively.  Therefore, this gives a proof of the Reflection Formula for complex numbers $\alpha$ such that $0<\text{Re}(\alpha)<1$.  Then, using analytic continuation, we prove the Reflection Formula for all $\alpha\in\mathbb{C}$.
A: Using Mason's tip in the comments you can say
$$ \int_{-\infty}^{\infty} \frac{1}{x^n+1}dx = 2\int_0^\infty \frac{1}{x^n+1}dx. $$
And now substitute $x^n + 1 \mapsto 1/u$, and you should get something in the form of the beta function, which has a known representation in terms of the gamma function.
