How to compute the improper integral

Question

Suppose $n$ is an integer and $\alpha$ is a real number satisfying $0 < \alpha <n$. Compute the integral $$I=\int\limits_{0}^{\infty}\frac{t^{\alpha -1}}{t^n+1}dt.$$

Take $\lambda=n/\alpha >1$ and set $u=t^{\alpha}$ then $$I=\frac{1}{\alpha}\int\limits_{0}^{\infty} \frac{du}{u^\lambda+1}.$$ I have a trouble when compute this integral. I don't know how to compute this for real $\lambda$ (it may be easier if $\lambda$ is an integer). Help me, please. Thank you for your help.

Answer 1

$$I(\lambda) = \int\limits_{0}^{\infty} \frac{du}{u^\lambda+1}$$

Let $u^\lambda = x$

$$I(\lambda) =\frac{1}{\lambda} \int\limits_{0}^{\infty} \frac{x^{1/\lambda-1}}{x+1}\,dx = \frac{\Gamma(1/\lambda)\Gamma(1-1/\lambda)}{\lambda\Gamma(2)} =\frac{\pi}{\lambda \sin(\pi/\lambda)}$$

Note that

$$\int^\infty_0 \frac{t^{x-1}}{(1+t)^{x+y}}\,dt = B(x,y)$$

And

$$\Gamma(x)\Gamma(1-x) = \pi \csc(\pi x)$$

Answer 2

It can be computed using the Residue Theorem. Use the contour $$ \gamma_R=\gamma_{R,1}+\gamma_{R,2}-\gamma_{R,3} $$ where $$ \gamma_{R,1}(t)=t,\,\, t\in[0,R], \quad \gamma_{R,2}(t)=R\mathrm{e}^{it}, \,\, t\in [0,2\pi/\lambda],\quad \gamma_{R,3}(t)=t\mathrm{e}^{2\pi/\lambda}, \,\, t\in [0,R]. $$ Then $$ \lim_{R\to\infty}\int_{\gamma_R} f=\big(1-\mathrm{e}^{2\pi i/\lambda}\big)\int_0^\infty\frac{dx}{1+x^\lambda}, $$ while, for $R>1$, $$ \int_{\gamma_R} f=2\pi i\,\mathrm{Res}\big(\,f,z=\mathrm{e}^{\pi i/\lambda}\big)=\frac{2\pi i}{\lambda \mathrm{e}^{(\lambda-1)\pi i/\lambda}}. $$ Therefore, $$ \int_0^\infty\frac{dx}{1+x^\lambda}=\frac{2\pi i}{\lambda \mathrm{e}^{(\lambda-1)\pi i/\lambda}(1-\mathrm{e}^{2\pi i/\lambda})}=\frac{\pi/\lambda}.{\sin(\pi/\lambda)}. $$