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

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...

• Are you sure that it's indefinite integral? – Jaideep Khare Jun 8 '17 at 9:25
• Yes, sure. I edited the integral. This is the final form of it which needs to be calculated. In my book the result is given: pi/(sin(pi * a)) , but I can't solve it – Nfff3 Jun 8 '17 at 9:28
• Do you know about residue theorem? – Shashi Jun 8 '17 at 9:32
• @KelemenNorbi I want to say, that this in fact isn't indefinite integral, it's definite integral. Whether a integral is definite or not is determined by whether it has upper and lower limits of not, respectively. – Jaideep Khare Jun 8 '17 at 9:34
• Okay, can you at least add to your question in what course you have seen that integral and what kind of methods you know, your attempts etc – Shashi Jun 8 '17 at 9:47

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 $x \to \infty$

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

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

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

Euler's Reflection Formula

• Thanks. This solution is great. – Nfff3 Jun 8 '17 at 11:53
• You are welcome :) – W.R.P.S Jun 8 '17 at 15:53