How to evaluate the integral $ \int_0^\infty \frac{\sin(\omega\tau)\sin(\omega y)\sinh\,(\omega x)}{\sinh(\omega a)} \,d\omega$

Question

How to evaluate the following integral

$$\displaystyle \int_0^\infty \frac{\sin{(\omega\tau)}\sin{(\omega y)}\sinh\,(\omega x)}{\sinh{(\omega a)}} \,\text{d}\omega$$

where $a > 0$, $x \in (0,\, a)$ , $y \in (0,\,\infty)$ and $\tau \in (0,\,\infty)$? The solution should be a function of $x\,,y\,,\tau\,,a$.

Any clues? I heard that it has a closed form and can be expressed by elementary functions. Any idea will help! Thanks.

Is it equivalent to $\frac{\sin{(\dfrac{\pi}{a} x)}\sinh\,(\dfrac{\pi}{a}y)\sinh(\dfrac{\pi}{a} \tau)}{\sin^2(\dfrac{\pi}{a} x)\sinh^2(\dfrac{\pi}{a} y) \,+\, [\cos\,(\dfrac{\pi}{a} \tau)\,+\,\cos\,(\dfrac{\pi}{a} x)\cosh\,(\dfrac{\pi}{a} y)]^2}$ ?

Answer 1

The integral $$f(z)=\int_0^\infty\frac{\sinh zx}{\sinh x}\,dx\qquad(z\in\mathbb{C},|\Re z|<1)$$ can be evaluated using the residue theorem (like this), or like this: \begin{align} f(z)&=\int_0^\infty(e^{zx}-e^{-zx})\frac{e^{-x}}{1-e^{-2x}}\,dx=\sum_{n=0}^{\infty}\int_0^\infty(e^{zx}-e^{-zx})e^{-(2n+1)x}\,dx\\&=\sum_{n=0}^{\infty}\left(\frac{1}{2n+1-z}-\frac{1}{2n+1+z}\right)=\sum_{n=0}^{\infty}\frac{2z}{(2n+1)^2-z^2}\\&=-\frac{d}{dz}\sum_{n=0}^{\infty}\ln\left(1-\frac{z^2}{(2n+1)^2}\right)=-\frac{d}{dz}\ln\cos\frac{\pi z}{2}=\color{blue}{\frac{\pi}{2}\tan\frac{\pi z}{2}}. \end{align} By $\sinh(iz)=i\sin z$ and product-to-sum formulae for $\sinh$/$\cosh$, your integral is equal to $$-\frac{1}{4a}\left(f\Big(\frac{x+i(y+\tau)}{a}\Big)+f\Big(\frac{x-i(y+\tau)}{a}\Big)-f\Big(\frac{x+i(y-\tau)}{a}\Big)-f\Big(\frac{x-i(y-\tau)}{a}\Big)\right).$$ It remains to simplify this expression...