What is the integral of e^(ax)/(1+e^x) for a real number 0<a<1?

[Chris L T521]

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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[Chris L T521]

This week's problem was correctly answered by sbhatnagar and Sudharaka.

Sudharaka's solution can be seen below.

Spoiler

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

\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}\(\mbox{Using Euler's reflection formula we can simplify this into,}\)\[\int_{-\infty}^{\infty}\frac{e^{ax}}{1+e^x}\,dx=\frac{\pi}{\sin\pi a}\]