Gamma function identities

Question

In the book Stochastic Geometry and Wireless Network vol.1 i found some equation that state : using the fact of

\begin{equation} \int_{0}^{\infty}\frac{1}{1+x^u}\mathrm{d}x = 1/u\Gamma[1/u]\Gamma[1-1/u] \end{equation} where $u>0$

1. The question is how to prove this fact? or where can i start to prove this integral form equal to this Gamma function?
2. Is this only applied to interval $0 ~to~ \infty$ only? what if we define some interval from A to B (i.e A does not equal 0 and $B = \infty$)?

In This Paper i found a formula that almost similar

\begin{align} I &=\int_b^{\infty} \frac{1}{1+w^{\alpha/2}}\mathrm{d}w\\ &=\frac{2\pi}{\alpha}\mathrm{csc}(2\pi/\alpha)-b_2F_1(1,2/\alpha;(2+\alpha)/\alpha;-b^{\alpha/2}) \end{align}

From this formula i try to expand the answer as follows

\begin{align} I&=\frac{2\pi}{\alpha}\frac{1}{sin(2\pi/\alpha)}-b_2F_1(1,2/\alpha;(2+\alpha)/\alpha;-b^{\alpha/2})\\ &=\frac{2}{\alpha}\Gamma(2/\alpha)\Gamma(1-2/\alpha)-b_2F_1(1,2/\alpha;(2+\alpha)/\alpha;-b^{\alpha/2})\\ &=\Gamma(1+2/\alpha)\Gamma(1-2/\alpha)-b_2F_1(1,2/\alpha;(2+\alpha)/\alpha;-b^{\alpha/2})\\ \end{align}

However from here i still cannot see how can i prove it from the original integral

Any comments are appreciated

Answer 1

May be an answer for the second part.

If you consider $$I=\int\frac{dx}{1+x^u}$$ the result is given in terms of the gaussian or ordinary hypergeometric function $$I=x \, _2F_1\left(1,\frac{1}{u};1+\frac{1}{u};-x^u\right)$$ which can be integrated without any problem between any non negative bounds.

The particular case of $$J=\int_0^\infty\frac{dx}{1+x^u}=\frac{\Gamma \left(1-\frac{1}{u}\right) \Gamma \left(\frac{1}{u}\right)}{u}=\frac{\pi }{u}\,\csc \left(\frac{\pi }{u}\right)$$ by Euler reflection formula (provided $u>1$).