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...

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    $\begingroup$ Are you sure that it's indefinite integral? $\endgroup$ – Jaideep Khare Jun 8 '17 at 9:25
  • $\begingroup$ 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 $\endgroup$ – Nfff3 Jun 8 '17 at 9:28
  • $\begingroup$ Do you know about residue theorem? $\endgroup$ – Shashi Jun 8 '17 at 9:32
  • $\begingroup$ @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. $\endgroup$ – Jaideep Khare Jun 8 '17 at 9:34
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    $\begingroup$ 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 $\endgroup$ – 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

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

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