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I dont know if this has already been asked.

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.

Renascence_5.
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jeanne clement
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    http://math.stackexchange.com/questions/1999869/evaluate-int-frac11xndx-for-n-in-mathbb-r might be of help – bigfocalchord Jan 11 '17 at 08:36
  • 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 08:37
  • 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 09:44
  • This is duplicate of http://math.stackexchange.com/questions/247866/show-that-int-0-infty-frac11xn-dx-frac-pi-n-sin-pi-n-wh http://math.stackexchange.com/questions/270118/a-few-improper-integral http://math.stackexchange.com/questions/255875/improper-integration-involving-complex-analytic-arguments – Martin Nicholson Jan 11 '17 at 14:02

1 Answers1

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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.

Method 2: \begin{align*} &{\int_0^\infty \frac{1}{x^n+1} \:{\rm{d}}x}=\int_0^\infty\int_0^\infty e^{-(x^n+1)t} \:{\rm{d}}t\:{\rm{d}}x \\&=\int_0^\infty e^{-t}\int_0^\infty e^{-x^n t} \:{\rm{d}}t\:{\rm{d}}x =\int_0^\infty e^{-t}\left(\int_0^\infty e^{-x^n t}\:{\rm{d}}x\right)\:{\rm{d}}t \\&=\frac1n\int_0^\infty t^{-\frac1n}e^{-t}\left(\int_0^\infty u^{\frac1n-1} e^{-u}{\rm{d}}u\right)\:{\rm{d}}t =\frac1n \Gamma\left(1-\frac1n\right)\Gamma\left(\frac1n\right) \\&=\frac{\pi }{n}\mathrm{csc}\frac{\pi }{n} \end{align*}

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 $

Renascence_5.
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