# Improper Integral via Residues

I want to evaulate $\displaystyle{ \int_{-\infty}^{\infty} \frac{e^{kx}}{1+e^x} dx, \ k \in \mathbb{R} }$ via the Residue Theorem over the contour $[x=\pm a, y=0, y=2\pi]$ oriented counterclockwise.

1. Lets define the complex function, $\displaystyle{ f(z) = \frac{e^{kz}}{1+e^z} }$.

2. The function has singularities at $z_n = i\pi(2n+1), \ n \in \mathbb{Z}$ and thus we have a single singularity within the desired domain at $n=0$, $z_0 = i\pi$.

Any tips on how to proceed?

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This and this are very related. –  Pedro Tamaroff Jul 6 '12 at 4:58
Note the integral converges $\iff a \in (0,1)$ –  Pedro Tamaroff Jul 6 '12 at 5:00

Let $$I = \int_{-\infty}^\infty dz\, \frac{e^{kz}}{1+e^z}.$$ Consider the integral $I'$ along the top of the rectangular contour. We find
$$\begin{eqnarray*} I' &=& \int_{\infty + 2\pi i}^{-\infty + 2\pi i} dz\, \frac{e^{kz}}{1+e^z} \\ &=& \int_{\infty}^{-\infty} d\zeta\, \frac{e^{k(\zeta+2\pi i)}}{1+e^\zeta} \hspace{10ex} (\textrm{let } z = \zeta+2\pi i) \\ &=& -e^{2\pi k i} I. \end{eqnarray*}$$ The argument of the integral along the right side of the rectangle goes like $e^{(k-1)x}$ for $x\gg 0$. This must be suppressed, so $\mathrm{Re}\, k < 1$. Likewise, to the left the argument goes like $e^{k x}$ for $x\ll 0$, so $\mathrm{Re}\, k > 0$. For $0<\mathrm{Re}\, k < 1$, the integrals along the sides of the rectangle vanish. The integral along the closed rectangular contour is then $I + I' = (1-e^{2\pi k i})I = 2\pi i \mathrm{Res}_{z=\pi i},$ and so
$$\begin{eqnarray*} I &=& \frac{1}{1-e^{2\pi k i}} 2\pi i \,\mathrm{Res}_{z=\pi i} \, \frac{e^{kz}}{1+e^z} \\ &=& \frac{1}{1-e^{2\pi k i}} 2\pi i \frac{e^{k\pi i}}{e^{\pi i}}. \end{eqnarray*}$$ To calculate the residue we use the fact that if $f(z)$ is analytic and $f(z)/g(z)$ has a simple pole at $z_0$ that $\mathrm{Res}_{z=z_0}f(z)/g(z) = f(z_0)/g'(z_0).$ Therefore, for $0<\mathrm{Re}\, k < 1$, $$\begin{equation*} \int_{-\infty}^\infty dz\, \frac{e^{kz}}{1+e^z} = \frac{\pi}{\sin k\pi}. \end{equation*}$$