Arnold's Trivium problem 51

We divide the integral into two parts:

\begin{align*} f(k) &= \int_{\Bbb{R}} \mathrm{sgn}(x) e^{-ikx} \, dx + \int_{\Bbb{R}} e^{-ikx} \left( \frac{1-e^{-x}}{1+e^{-x}} - \mathrm{sgn}(x) \right) \, dx \end{align*}

The first one can be easily computed by integration by parts as follows:

\begin{align*} \int_{\Bbb{R}} \mathrm{sgn}(x) e^{-ikx} \, dx &= -\int_{\Bbb{R}} 2\delta(x) \frac{e^{-ikx}}{-ik} \, dx = -\frac{2i}{k}. \end{align*}

Indeed, this manipulation can be justified in the context of tempered distributions. (Just observe what happens when the Fourier transform and the differentiation are interchanged.)

The remaining part is now integrable in ordinary sense, and we have

\begin{align*} \int_{\Bbb{R}} e^{-ikx} \left( \frac{1-e^{-x}}{1+e^{-x}} - \mathrm{sgn}(x) \right) \, dx &= 4i \int_{0}^{\infty} \frac{\sin (kx) \, e^{-x}}{1+e^{-x}} \, dx \\ &= 4i \sum_{n=1}^{\infty}(-1)^{n-1} \int_{0}^{\infty} \sin (kx) \, e^{-nx} \, dx \\ &= 4i \sum_{n=1}^{\infty}(-1)^{n-1} \frac{k}{k^2 + n^2} \end{align*}

Combining,

\begin{align*} f(k) &= -2i \sum_{n=-\infty}^{\infty} \frac{k}{n^2 + k^2} = -\frac{2\pi i}{\sinh (\pi k)}. \end{align*}

The poles are at $x=(2n+1)\pi\mathrm i$, and the corresponding residues are $-2\mathrm e^{-(2n+1)\pi k}$. If we ignore issues of convergence, this suggests that the integral is

$$ -4\pi\mathrm i\sum_{n=0}^\infty\mathrm e^{-(2n+1)\pi k}=\frac{-4\pi\mathrm i\mathrm e^{-\pi k}}{1-\mathrm e^{-2\pi k}}=\frac{-2\pi\mathrm i}{\sinh\pi k}\;. $$

To make this rigorous in the context of tempered distributions, you can take the derivative of the function being Fourier-transformed, calculate the resulting convergent integral, and divide by $-\mathrm ik$ (see Tempered distributions and Fourier transform at Wikipedia). The residues are slightly more complicated to evaluate, but as one might expect the result is the same.