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.


marked as duplicate by tired, Daniel Fischer Jan 11 '17 at 16:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


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 $

  • $\begingroup$ Nice Hoe do you move from the next to the last row? I think it is residue tools right? $\endgroup$ – Guy Fsone Jan 11 '17 at 9:13
  • $\begingroup$ @Guy Fabrice which row did you mean,bro? $\endgroup$ – Renascence_5. Jan 11 '17 at 9:15
  • $\begingroup$ @Guy Fabriceoh I see,the last row follows from Euler reflexion formula. $\endgroup$ – Renascence_5. Jan 11 '17 at 9:20
  • $\begingroup$ 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? $\endgroup$ – jeanne clement Jan 12 '17 at 10:57

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