# 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.

• Here's the details of proving this+ It's an even functions so $\int_{-\infty}^\infty=2\int_0^\infty$ – Mason Dec 11 '18 at 23:32
• Essentially because $B(s,1-s)=\Gamma(s)\Gamma(1-s)=\frac{\pi}{\color{red}{\sin(\pi s)}}$. – Jack D'Aurizio Dec 11 '18 at 23:40

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 $$\alpha\in\mathbb{C}$$.
$$\int_{-\infty}^{\infty} \frac{1}{x^n+1}dx = 2\int_0^\infty \frac{1}{x^n+1}dx.$$
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.