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Problem of the Week #13 - June 25th, 2012

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Chris L T521

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Jan 26, 2012
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Thanks to those who participated in last week's POTW!! Here's this week's problem.

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Problem: Let $0<a<1$ be a real number. Compute

\[\int_{-\infty}^{\infty}\frac{e^{ax}}{1+e^x}\,dx.\]

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
This week's problem was correctly answered by sbhatnagar and Sudharaka.

Sudharaka's solution can be seen below.

\(\mbox{Let, }u=\dfrac{1}{1+e^x}.\mbox{ Then the given integral becomes,}\).

\begin{eqnarray}


\int_{-\infty}^{\infty}\frac{e^{ax}}{1+e^x}\,dx&=&\int_{0}^{1}(1-u)^{a-1}u^{2-a}\,du\\


&=&\int_{0}^{1}(1-u)^{a-1}u^{(3-a)-1}\,du


\end{eqnarray}


\(\mbox{Note that, }0<a<1\Rightarrow 2<3-a<3.\mbox{ Therefore, both \(a\) and \(3-a\) are positive. Hence,}\)


\[\int_{-\infty}^{\infty}\frac{e^{ax}}{1+e^x}\,dx=\beta(a,3-a)\]


\(\mbox{where }\beta\mbox{ denotes the Beta function.}\)


\(\mbox{Using the relation between the Beta function and the Gamma function}(\Gamma)\mbox{ we get,}\)


\begin{eqnarray}


\int_{-\infty}^{\infty}\frac{e^{ax}}{1+e^x}\,dx=\beta(a,3-a)&=&\frac{\Gamma(a)\,\Gamma(3-a)}{\Gamma(3)}\\


&=&\frac{2!\,\Gamma(a)\,\Gamma(1-a)}{2!}\\


&=&\Gamma(a)\,\Gamma(1-a)


\end{eqnarray}


\(\mbox{Using Euler's reflection formula we can simplify this into,}\)


\[\int_{-\infty}^{\infty}\frac{e^{ax}}{1+e^x}\,dx=\frac{\pi}{\sin\pi a}\]
 
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