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 November 16th, 2012, 05:58 PM #1 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions A challenging integral $\int^{\infty}_{0}\, \frac{x^m}{1+x^n}\,\,\text{ n,m are positive reals such }n-m>2$
 November 17th, 2012, 04:43 AM #2 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: A challenging integral $\int_{0}^{\infty}\frac{x^{m}}{1+x^{n}}dx$ This can be evaluated with the Beta: $B(p,q)=\int_{0}^{\infty}\frac{y^{p-1}}{(1+y)^{p+q}}dy$ Let $x=y^{1/n}, \;\ dx=\frac{1}{n}y^{\frac{1}{n}-1}dy$ This gives: $\frac{1}{n}\int_{0}^{\infty}\frac{y^{\frac{m+1}{n}-1}}{(1+y)}dy$ This means that $p=\frac{m+1}{n}, \;\ q=1-\frac{m+1}{n}$ So, $\frac{1}{n}\cdot B\left(\frac{m+1}{n}, \;\ 1-\frac{m+1}{n}\right)=\frac{1}{n}\int_{0}^{\infty}\ frac{y^{\frac{m+1}{n}-1}}{(1+y)}dy$ But, $B(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$ So, we have $B\left(\frac{m+1}{n}, \;\ 1-\frac{m+1}{n}\right)=\frac{\Gamma(\frac{m+1}{n})\G amma(1-\frac{m+1}{n})}{\Gamma(1)}$ Now, using the identity: $\Gamma(p)\Gamma(1-p)=\frac{\pi}{\sin(\pi p)}$, we happily arrive at: $\fbox{\frac{1}{n}\cdot \frac{\pi}{\sin(\pi(\frac{m+1}{n}))}}$
 November 17th, 2012, 04:56 AM #3 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: A challenging integral Really , I thought that can't be solved without using residues. Nice Approach [color=#0000BF]galactus[/color].

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