Fourier sine transform of $ \frac{1}{e^{ax } - e^{-ax}} $ Consider the function $ \frac{1}{e^{ax} - e^{-ax}}.$ We are try to find the Fourier sine transform. After putting the values we get $$\int_{-\infty}^{\infty}\frac{1}{e^{ax}-e^{-ax}}\sin px\, dx$$
Now I try to solve it by different techniques but every time got stuck, for example I convert $(e^{ax}-e^{-ax})$ in $\sinh x$, but when I take it in nominator, it become $cosec (hx)$. But again I am stuck in finding the integral of $(\sin(px))(cosec(hx))$.
Similarly I tried integration by parts, but I could not solve.  Can I get some help?

We can find it in
Gradshteyn, I. S.; Ryzhik, I. M.; Zwillinger, Daniel (ed.); Moll, Victor (ed.), Table of integrals, series, and products. Translated from the Russian. Translation edited and with a preface by Victor Moll and Daniel Zwillinger, Amsterdam: Elsevier/Academic Press (ISBN 978-0-12-384933-5/hbk; 978-0-12-384934-2/ebook). xlv, 1133 p. (2015). ZBL1300.65001.
3.981.1 is
$$
\int_0^\infty\frac{\sin \alpha x}{\sinh \beta x}\, dx 
= \frac{\pi}{2\beta}\,\tanh\frac{\alpha\pi}{2\beta},
\qquad \operatorname{Re} \beta > 0, \alpha > 0.
$$
Going from $\int_0^\infty$ to $\int_{-\infty}^\infty$ doubles the result, but going from $e^{ax}-e^{-ax}$ to $\sinh(ax)$ halves the result.  So your answer is
$$
\frac{\pi}{2 a}\,\tanh\frac{p\pi}{2 a}
$$
