TITLE: Showing that $\int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi x} \, \cos bx \, dx = \frac{\sin a}{\cos a + \cosh b}$
QUESTION [15 upvotes]: I want to show that $$ \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi x} \, \cos bx \, dx = \frac{\sin a}{\cos a + \cosh b} \, , \quad -\pi < a < \pi . $$
I tried integrating $f \displaystyle (z) = e^{ibz} \, \frac{\sinh az}{\sinh \pi z} $ around an indented rectangular contour with vertices at $\pm R, \pm R+ i$.
I  ended up with  the equation $$ (1+ e^{-b} \cos a) \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi x} \, \cos bx \, dx  - e^{-b}\sin a \int_{-\infty}^{\infty} \frac{\cosh ax}{\sinh \pi x} \, \sin bx \ dx = e^{-b} \sin a . $$ 
But I don't see how to extract the value of the integral from this equation.
Did I integrate the wrong function?

REPLY [9 votes]: I have a simple way to calculate your old question. Note
$$  \frac{\sinh(ax)}{\sinh(\pi x)} \cos(bx) = \frac{\sinh(ax)}{\sinh(\pi x)}\cosh(ibx)=\frac{\sinh((a+bi)x)+\sinh((a-bi)x)}{\sinh(\pi x)} $$
and 
$$ \int_0^\infty\frac{\sinh(ax)}{\sinh(bx)}dx=\frac{\pi}{2b}\tan\frac{a\pi}{2b}. $$
So
\begin{eqnarray}
I&=&\int_0^\infty\frac{\sinh(ax)}{\sinh(\pi x)} \cos(bx)dx\\
&=&\int_0^\infty\frac{\sinh((a+bi)x)+\sin((a-bi)x)}{\sinh(\pi x)}dx\\
&=&\frac{1}{2}\tan\frac{a+bi}{2}+\frac{1}{2}\tan\frac{a-bi}{2}\\
&=&\frac{\sin a}{\cos a + \cosh b}.
\end{eqnarray}