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

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• Show that $\int_0^ \infty \frac{1}{1+x^n} dx= \frac{ \pi /n}{\sin(\pi /n)}$ , where $n$ is a positive integer. 1 answer
• Prove $\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}$ using real analysis techniques only 4 answers

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.

Answer 1

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*}

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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 $