Is it true that $\int_{0}^{\infty}\frac{x^{m}}{x^n+1}dx=\frac{\pi}{n\sin{(\frac{m+1}{n}\pi})}$?

Question

Is it true that $$\int_{0}^{\infty}\frac{x^{m}}{x^n+1}dx=\frac{\pi}{n\sin{(\frac{m+1}{n}\pi})}$$ where $1\leq m<n$ ?

If so, how can I prove it?

Answer 1

Let $x=\tan^{2/n} t \implies dx= \frac{2}{n} \tan^{2/n-1}t \sec^2 t dt$, then $$I=\int_{0}^{\infty}\frac{x^m}{1+x^n} dx= \frac{2}{n} \int _{0}^{\pi/2}\sin ^{2m/n+2/n-1} t ~ \cos^{1-2m/n-2/n}~ t dt =\frac{1}{n}\Gamma((m+1)/n)\Gamma(1-(m+1)/n)$$ By using $\Gamma(z) \Gamma(1-z)=\pi \csc z,$ we get $$ I=\frac{\pi}{n \sin (\frac{(m+1)\pi}{n})}$$

Answer 2

Thanks to @Lord Shark the Unknown, @Ali Shather, and @Dr Zafar Ahmed DSc, I was able to prove it.

Let $x^n=y$, then $dx=\frac{y^{1/n-1}}{n}dy$ , $$\int_{0}^{\infty}\frac{x^{m}}{x^n+1}dx =\frac{1}{n}\int_{0}^{\infty}\frac{y^{m/n+1/n-1}}{y+1}dy =\frac{1}{n}\mathrm{B}(m/n+1/n,1-(m/n+1/n)) =\frac{\pi}{n\sin{\pi(m/n+1/n)}}$$ For $0<m/n+1/n<1\implies-1<m<n-1$