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

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#### Chris L T521

##### Well-known member
Staff member
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

##### Well-known member
Staff member
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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