# How to prove this integral problem: $\int_0^{\infty}\frac{dx}{1+x^n}=\frac{\pi}{n}\csc\frac \pi n$? [duplicate]

How to prove this integral $$\int_0^{\infty}\frac{dx}{1+x^n}=\frac{\pi}{n}\csc\frac \pi n\ {?}$$ $n\ge 2$ is a positive integer

Frankly speaking i have no clue how to start someone please explain me.

• math.stackexchange.com/questions/1999869/… might be of help – bigfocalchord Jan 11 '17 at 8:36
• Use complex analysis tools namely Theorem of residue. start by computing the nth roots of -1. And there residues – Guy Fsone Jan 11 '17 at 8:37
• I haven't thought about this, but maybe you can use $$\frac{1}{1+x^n}=\frac{1}{n}\,\sum_{k=1}^n\,\frac{1}{1-\omega^{2k-1}x}\,,$$ where $\omega:=\exp\left(\frac{\pi}{n}\right)$. – Batominovski Jan 11 '17 at 9:44
• – user82588 Jan 11 '17 at 14:02

Method 1: Let $x=\sqrt[n]{\tan^{2}\theta }$, then \begin{align*} \int_{0}^{\infty }\frac{1}{1+x^{n}}\, \mathrm{d}x&=\frac{2}{n}\int_{0}^{\pi /2}\cos^{1-2/n}\theta \sin^{2/n-1}\theta \, \mathrm{d}\theta \\ &=\frac{1}{n}\mathrm{B}\left ( 1-\frac{1}{n},\frac{1}{n} \right )\\ &=\frac{1}{n}\Gamma \left ( 1-\frac{1}{n} \right )\Gamma \left ( \frac{1}{n} \right )\\ &=\frac{\pi }{n}\mathrm{csc}\frac{\pi }{n} \end{align*} where $\mathrm{B}\left ( \cdot \right )$ is the Beta function and $\Gamma\left ( \cdot \right )$ is the Gamma function.
More general, using the same way as Method 1 mentioned,we get $$\int_{0}^{\infty} \frac{x^{\mu-1}}{1+x^{\nu}} \; \mathrm{d}x=\frac{\pi}{\nu} \csc \left( \frac{\pi \mu}{\nu} \right)$$ where $0< \mu < \nu$
• now, the question is : $\Gamma\left(1-\frac1n\right)\Gamma\left(\frac1n\right) =\pi\mathrm{csc}\frac{\pi }{n}$ ? how did you get this in last step? – jeanne clement Jan 12 '17 at 10:57