What is the improper integral of $\int_0^{\infty}\frac{x^{a-1}}{1+x} \,dx$?

Question

What is the improper integral of $$\int_0^{\infty}\frac{x^{a-1}}{1+x} \,dx$$ a is between 0 and 1.

The result has to be pi/(sin(pi * a)) and is calculated somehow using Fourier series...

Answer 1

we consider : $y=\frac{x}{1+x}$ => $ x=\frac{y}{1-y} \to dx=\frac{1}{(1-y)^2}dy$

and $y\to 0 $ as $x\to 0 $ and $y \to 1$ as $x \to \infty $

$\int^{\infty}_{0}\frac{x^{a-1}}{1+x}dx=\int^{\infty}_{0}\frac{x}{1+x}.x^{a-2}dx=\int ^{1}_{0}y.(\frac{y}{1-y})^{a-2}\frac{1}{(1-y)^2}dy$

$=\int^{1}_{0}y^{a-1}.(1-y)^{(1-a)-1}dy=\beta(a,1-a) $

$=\frac{\Gamma (a).\Gamma (1-a)}{\Gamma (a+1-a)}=\Gamma(a).\Gamma(1-a)=\frac{\pi}{\sin(a\pi)}$

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Euler's Reflection Formula