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I'm self studying complex analysis. I've encountered the following integral:

$$\int_{- \infty}^{+ \infty} \frac{e^{ax}}{1+e^x} dx \text{ with } a \in \mathbb{R},\ 0 \lt a \lt 1. $$

I've done the substitution $e^x = y$. What kind of contour can I use in this case ?

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It is possible to calculate the original integral by taking the usual contour of the semicircle in the upper half-plane. Note that the integrand has a simple pole in this region at $z=i\pi$. –  Antonio Vargas Apr 11 '12 at 16:32
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@Antonio Vargas, it also has poles at $z = i(\pi + 2\pi n)$ for $n\in \mathbb{N}$ –  Chris Janjigian Apr 11 '12 at 16:53
    
@Chris, quite right. I was too hasty. –  Antonio Vargas Apr 11 '12 at 18:00
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@stef You might find this useful. –  Pedro Tamaroff Apr 23 '12 at 1:54
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1 Answer

up vote 8 down vote accepted

The substitution $e^x=y$ leads to the integral $$ \int_0^\infty\frac{y^{a-1}}{y+1}\,dy. $$ This can be computed integrating the function $f(z)=z^{a-1}/(z+1)$ along the keyhole contour. We consider the branch of $z^{a-1}$ defined on $\mathbb{C}\setminus[0,\infty)$ with $f(-1)=e^{\pi i}$. For small $\epsilon>0$ and large $R>0$, he contour is made up of the interval $[\epsilon,R]$, the circle $C_r=\{|z|=R\}$ counterclockwise, the interval $[R,\epsilon]$ and the circle $C_\epsilon=\{|z|=\epsilon\}$ clockwise. The function $f$ has a simple pole at $z=-1$ with residue $(-1)^{a-1}=e^{\pi(a-1)i}$. It is easy to see that $$ \lim_{\epsilon\to0}\int_{C_\epsilon}f(z)\,dz=\lim_{R\to\infty}\int_{C_R}f(z)\,dz=0. $$ Then $$ \int_0^\infty\frac{y^{a-1}}{y+1}\,dy+\int_\infty^0\frac{(e^{2\pi i}y)^{a-1}}{y+1}\,dy=2\,\pi\,i\operatorname{Res}(f,-1), $$ from where $$ \bigl(1-e^{2\pi(a-1)i}\bigr)\int_0^\infty\frac{y^{a-1}}{y+1}\,dy=2\,\pi\,i\,e^{\pi(a-1)i} $$ and $$ \int_0^\infty\frac{y^{a-1}}{y+1}\,dy=\frac{\pi}{\sin((1-a)\pi)}. $$

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Is it a bad idea to use a rectangular contour here rather than the keyhole contour? –  tacos_tacos_tacos Nov 2 '12 at 4:27
    
@jshin47 It is a bad idea because $z^{a-1}$ has a branching point at $z=0$. –  Julián Aguirre Nov 3 '12 at 21:43
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