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- Jan 26, 2012

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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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- Thread starter
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- #1

- Jan 26, 2012

- 995

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

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

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- Jan 26, 2012

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Sudharaka's solution can be seen below.

\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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