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?
 A: Let $a>0$ and $p>0$ and let $I(a,p)$ be given by
$$\begin{align}
I(a,p)&=\int_{-\infty}^\infty \frac{\sin(px)}{\sinh(ax)}\,dx\\\\
&=\text{Im}\left(\text{PV}\int_{-\infty}^\infty \frac{e^{ipx}}{\sinh(ax)}\,dx\right)\tag1
\end{align}$$
Next, we move to the complex plane.  Let $J(a,p)$ be given by the contour integral
$$\begin{align}
J(a,p)&=\oint_C \frac{e^{ipx}}{\sinh(ax)}\,dx\\\\
&=\int_{\varepsilon \le |x|\le R}\frac{e^{ipx}}{\sinh(ax)}\,dx\\\\
&+\int_0^\pi \frac{e^{ipRe^{i\phi}}}{\sinh(aRe^{i\phi})}\,iRe^{i\phi}\,d\phi\\\\
&-\int_0^\pi \frac{e^{ip\varepsilon e^{i\phi}}}{\sinh(a \varepsilon e^{i\phi})}\,i\varepsilon e^{i\phi}\,d\phi\tag2
\end{align}$$
As $R\to \infty$ and $\varepsilon\to 0^+$, it is easy to show that the first integral on the right-hand side of $(2)$ goes to $iI(a,p)$, the second integral goes to $0$, and the third integral on the right-hand side of $(2)$ goes to $-i\pi/a$.
Noting that $\frac1{\sinh(az)}$ has poles at values of $z=in\pi$, the residue theorem guarantees that
$$\begin{align}
\lim_{N\to\infty}J(a,p)&=2\pi i \left(\frac1a \sum_{n=1}^\infty (-1)^n e^{-np\pi/a}\right)\\\\
&=-2\pi i \frac{1}{a(1+e^{p\pi/a})}
\end{align}$$
Putting it all together yields
$$\begin{align}
I(a,p)&=\frac{\pi}{a}-\frac{2\pi}{a}\frac1{1+e^{p\pi/a}}\\\\
&=\frac{\pi \text{tanh}(p\pi/2a)}{a}
\end{align}$$
Therefore, we find that
$$\int_{-\infty}^\infty \frac{\sin(px)}{e^{ax}-e^{-ax}}\,dx=\frac{\pi \text{tanh}(p\pi/2a)}{2a}$$
which agrees with the result reported by @GEdgar!
A: 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
$$
\frac{\pi}{2 a}\,\tanh\frac{p\pi}{2 a}
$$
A: Let $ a,p\in\mathbb{R} $, let's start by using $ \frac{1}{1-x} $ series expansion, we'll then switch the sum and the integral signs, we can easily prove that we're allowed to do so.
\begin{aligned}\int_{-\infty}^{+\infty}{\frac{\sin{\left(px\right)}}{\operatorname{e}^{ax}-\operatorname{e}^{-ax}}\,\mathrm{d}x}&=2\int_{0}^{+\infty}{\frac{\sin{\left(px\right)}}{\operatorname{e}^{ax}-\operatorname{e}^{-ax}}\,\mathrm{d}x}\\ &=2\int_{0}^{+\infty}{\sum_{n=0}^{+\infty}{\operatorname{e}^{-\left(2n+1\right)ax}\sin{\left(px\right)}}\,\mathrm{d}x}\\ &=2\sum_{n=0}^{+\infty}{\int_{0}^{+\infty}{\operatorname{e}^{-\left(2n+1\right)ax}\sin{\left(px\right)}\,\mathrm{d}x}}\\ &=2\sum_{n=0}^{+\infty}{\frac{p}{a^{2}\left(2n+1\right)^{2}+p^{2}}}\end{aligned}
Now, we can prove that $ \left(\forall k\in\mathbb{N}^{*}\right)$ : $$\frac{p}{a^{2}\left(2k+1\right)^{2}+p^{2}}=\frac{1}{a\left(1+\operatorname{e}^{-\frac{p\pi}{a}}\right)}\int_{0}^{\pi}{\operatorname{e}^{-\frac{px}{a}}\cos{\left(\left(2k+1\right)x\right)}\,\mathrm{d}x}$$
Thus, if $ n\in\mathbb{N}^{*} $, we have : \begin{aligned}2\sum_{k=0}^{n-1}{\frac{p}{a^{2}\left(2k+1\right)^{2}+p^{2}}}&=\frac{2}{a\left(1+\operatorname{e}^{-\frac{p\pi}{a}}\right)}\int_{0}^{\pi}{\operatorname{e}^{-\frac{px}{a}}\sum_{k=0}^{n-1}{\cos{\left(\left(2k+1\right)x\right)}}\,\mathrm{d}x}\\ &=\frac{1}{a\left(1+\operatorname{e}^{-\frac{p\pi}{a}}\right)}\int_{0}^{\pi}{\operatorname{e}^{-\frac{px}{a}}\frac{\sin{\left(2nx\right)}}{\sin{x}}\,\mathrm{d}x}\\ &=\frac{1}{a\left(1+\operatorname{e}^{-\frac{p\pi}{a}}\right)}\int_{0}^{\frac{\pi}{2}}{\left(\operatorname{e}^{-\frac{px}{a}}-\operatorname{e}^{-\frac{p\left(\pi-x\right)}{a}}\right)\frac{\sin{\left(2nx\right)}}{\sin{x}}\,\mathrm{d}x}\\ &=\frac{1}{a\cosh{\left(\frac{p\pi}{2a}\right)}}\int_{0}^{\frac{\pi}{2}}{\sinh{\left(\frac{p\left(\frac{\pi}{2}-x\right)}{a}\right)}\frac{\sin{\left(2nx\right)}}{\sin{x}}\,\mathrm{d}x}\\ &\small=\frac{\tanh{\left(\frac{p\pi}{2a}\right)}}{a}\int_{0}^{\frac{\pi}{2}}{\frac{\sin{\left(2nx\right)}}{\sin{x}}\,\mathrm{d}x}-\frac{1}{a\cosh{\left(\frac{p\pi}{2a}\right)}}\int_{0}^{\frac{\pi}{2}}{\frac{\sinh{\left(\frac{p\pi}{2a}\right)}-\sinh{\left(\frac{p\left(\frac{\pi}{2}-x\right)}{a}\right)}}{\sin{x}}\sin{\left(2nx\right)}\,\mathrm{d}x}\\ 2\sum_{k=0}^{n-1}{\frac{p}{a^{2}\left(2k+1\right)^{2}+p^{2}}}&=\frac{2\tanh{\left(\frac{p\pi}{2a}\right)}}{a}\sum_{k=0}^{n-1}{\frac{\left(-1\right)^{k}}{2k+1}}-\int_{0}^{\frac{\pi}{2}}{\varphi_{p,a}\left(x\right)\sin{\left(2nx\right)}\,\mathrm{d}x}\end{aligned}
Where $ \varphi_{p,a} : x\mapsto\frac{\sinh{\left(\frac{p\pi}{2a}\right)}-\sinh{\left(\frac{p\left(\frac{\pi}{2}-x\right)}{a}\right)}}{a\cosh{\left(\frac{p\pi}{2a}\right)}\sin{x}} $, and can be extended by continuity to a $ \mathcal{C}^{1} $ function on $ \left[0,\frac{\pi}{2}\right] $, which means that : $$ \int_{0}^{\frac{\pi}{2}}{\varphi_{p,a}\left(x\right)\sin{\left(2nx\right)}\,\mathrm{d}x}\underset{n\to +\infty}{\longrightarrow}0 $$
Hence, taking $ n\to +\infty $, we get that $ 2\sum\limits_{n=0}^{+\infty}{\frac{p}{a^{2}\left(2n+1\right)^{2}+p^{2}}}=\frac{\pi\tanh{\left(\frac{p\pi}{2a}\right)}}{2a} $, thus : $$ \fbox{$\begin{array}{rcl}\displaystyle\int_{-\infty}^{+\infty}{\frac{\sin{\left(px\right)}}{\operatorname{e}^{ax}-\operatorname{e}^{-ax}}\,\mathrm{d}x}=\frac{\pi\tanh{\left(\frac{p\pi}{2a}\right)}}{2a}\end{array}$} $$
