<s>[INST] Integral $\int_0^\infty \frac{\sin^2 ax}{x(1-e^x)}dx=\frac{1}{4}\log\left( \frac{2a\pi}{\sinh 2a\pi}\right)$ How can we prove this ${\it{interesting}}$ integral
$$
I:=\int_0^\infty \frac{\sin^2 ax}{x(1-e^x)}dx=\frac{1}{4}\log\left( \frac{2a\pi}{\sinh 2a\pi}\right)
$$
I to write
$$
I=\frac{1}{2}\int_0^\infty \frac{(1-\cos a x)}{x}dx \sum_{n=0}^\infty e^{nx}
$$
simplifying
$$
I=\frac{1}{2} \int_0^\infty \sum_{n=0}^\infty e^{nx}\frac{dx}{x} -\frac{1}{2}\int_0^\infty \frac{\cos a x}{x}dx \sum_{n=0}^\infty e^{nx}
$$
but didn't help much.  Thank you
 [/INST] Consider the integral
\begin{align}
I = \int_{0}^{\infty} \frac{\sin^{2}(ax)}{x(1-e^{x})} \ dx.
\end{align}
This can be seen as
\begin{align}
- I &= \int_{0}^{\infty} \frac{ e^{-x} \sin^{2}(ax) }{ x ( 1 - e^{-x}) } \ dx \\
&= \sum_{n=0}^{\infty} \ \int_{0}^{\infty} e^{-(n+1)x} \sin^{2}(ax) \ \frac{dx}{x} \\
&= \sum_{n=0}^{\infty} I_{n}
\end{align}
where $I_{n}$ is given by
\begin{align}
I_{n} = \int_{0}^{\infty} e^{-(n+1)x} \sin^{2}(ax) \ \frac{dx}{x}.
\end{align}
Now $I_{n}$ can be evaluated as follows.
\begin{align}
\partial_{n} I_{n} &= - \int_{0}^{\infty} e^{-(n+1)x} \sin^{2}(ax) \ dx \\
&= - \frac{1}{2} \ \int_{0}^{\infty} e^{-(n+1)x} (1 - \cos(2ax)) \ dx \\
&= - \frac{1}{2} \left[ \frac{1}{n+1} - \frac{(n+1)}{4a^{2} + (n+1)^{2}} \right] 
\end{align}
and upon integration yields
\begin{align}
I_{n} &= - \frac{1}{2} \ln(n+1) + \frac{1}{4} \ln (4 a^{2} + (n+1)^{2}) \\
&= \frac{1}{4} \ln\left( \frac{4 a^{2} + (n+1)^{2}}{(n+1)^{2}} \right). 
\end{align}
Now, returning to the summation, it is seen that
\begin{align}
- I &= \frac{1}{4} \sum_{n=0}^{\infty} \ln\left( \frac{4 a^{2} + (n+1)^{2}}{(n+1)^{2}} \right) \\
&= \frac{1}{4} \ln \left( \prod_{n=0}^{\infty} \left\{ \frac{4 a^{2} + (n+1)^{2}}{(n+1)^{2}} 
\right\} \right).
\end{align}
Using the product formula
\begin{align}
\frac{\sinh(x)}{x} = \prod_{k=1}^{\infty} \left( 1 + \frac{x^{2}}{k^{2} \pi^{2}} \right)
\end{align}
then the final result is
\begin{align}
I = \frac{1}{4} \ln \left( \frac{2 a \pi}{\sinh(2a \pi)} \right). 
\end{align}
Hence,
\begin{align}
\int_{0}^{\infty} \frac{\sin^{2}(ax)}{x(1-e^{x})} \ dx = \frac{1}{4} \ln \left( \frac{2 a \pi}
{\sinh(2a \pi)} \right).
\end{align}
</s>