Fourier transformation

Let $\xi \in \mathbb{R}$ and $\alpha \in (0, 1)$ . Show that

$\int_{- \infty}^{\infty} {e}^{-2 \pi i x \xi} \frac{ \sin \pi \alpha }{ \cosh \pi x + \cos \pi \alpha } \mathrm{d} x = \frac{2 \sinh 2 \pi \alpha \xi }{ \sinh 2 \pi \xi }$

Solution

We note that

$\begin{align*} \frac{\sin \pi \alpha}{\cosh \pi x + \cos \pi \alpha} &= \frac{1}{1+e^{-\pi(x+i\alpha)}}-\frac{1}{1+e^{-\pi(x - i\alpha)}}\\ &= 2\sum\limits_{n=1}^{\infty} (-1)^{n}e^{-n\pi x}\sin n\pi\alpha \end{align*}$

Thus,

$\begin{align*} \int_{-\infty}^{\infty} \frac{e^{-2\pi i \xi x}\sin \pi \alpha}{\cosh \pi x + \cos \pi \alpha}\,\mathrm{d}x &= \int_{0}^{\infty} \frac{2\sin \pi \alpha \cos 2\pi\xi x}{\cosh \pi x + \cos \pi \alpha}\,\mathrm{d}x\\ &= 4\int_0^{\infty}\sum\limits_{n=1}^{\infty}(-1)^n e^{-n\pi x}\cos 2\pi\xi x \sin n\pi \alpha\,\mathrm{d}x\\ &= \frac{4}{\pi}\sum\limits_{n=1}^{\infty}(-1)^n \frac{n\sin n\pi\alpha}{n^2+4\xi^2}\\ &= \frac{2}{\pi}\sum\limits_{n=1}^{\infty} (-1)^n\left(\frac{1}{n+2i\xi}+\frac{1}{n-2i\xi}\right)\sin n\pi \alpha\\ &= \frac{2}{\pi}\sum\limits_{n = -\infty}^{\infty} \frac{(-1)^n\sin n\pi\alpha}{n+2i\xi}\\ &= \frac{2 \sin 2\pi i \alpha\xi}{\sin 2\pi i\xi} \qquad \textrm{ (by residue theorem) }\\ &= \frac{2 \sinh 2\pi \alpha\xi}{\sinh 2\pi \xi} \end{align*}$