Integral $ \int_{0}^{\infty} \frac{2axdx}{(x^{2}+a^{2})(e^{2\pi x}-1)} $ how could i evaluate the following integral ?=
$$ \int_{0}^{\infty} \frac{2axdx}{(x^{2}+a^{2})(e^{2\pi x}-1)} $$
for positive 'a' ??
i have tried the expansion of the integran $ exp(2\pi x) -1 $ into a power series of exponentials  $ \sum_{n=0}^{\infty}e^{-2n\pi x} $ but i would like to know if there is an easier method exists or if the integral is computed in tables.. ?
 A: It seems there is an easier method using contours as Ron Gordon has showed us with this integral being treated identical, except he does the triple pole version.  I do not think these integrals are in a table, I am familiar with most tables and they are not in the ones I have. 
You can write 
$$
\int_0^\infty \frac{2ax\, dx}{(x^{2}+a^{2})(e^{2\pi x}-1)}=2a \int_0^\infty \frac{x\, dx}{(x^{2}+a^{2})(e^{2\pi x}-1)}
$$
and define u=$2\pi x$ to obtain
$$
2a\int_0^\infty \frac{u\, du}{4\pi^2} \frac{1}{e^u-1}\frac{1}{\frac{u^2}{4\pi^2}+a^2}=\frac{a}{2\pi^2}\int_0^\infty \frac{u\, du}{(e^u-1)(a^2+\beta^2 u^2)},\quad
\beta^2\equiv (4\pi^2)^{-1}
$$
Now we can write
$$
I=\frac{a}{2\pi^2} \int_0^\infty \frac{u \, du}{(e^u-1)\big(\beta^2(u^2+\frac{a^2}{\beta^2})\big)}=\frac{a}{2\pi^2\beta^2}\int_0^\infty \frac{u\, du}{(e^u-1)(u^2+c^2)},\quad c\equiv \frac{a^2}{\beta^2}.
$$
Thus you want to solve
$$
I=\frac{c}{2\pi^2a}\int_0^\infty \frac{u \, du}{(e^u-1)(u^2+c^2)}.
$$
This is now the same problem as here http://residuetheorem.com/2014/01/15/integral-of-function-with-deceptive-triple-pole/, except yours is simpler.  That is the user $\bf{@Ron \ Gordon}$.  You can follow his solution till he evaluates the "ugly" triple pole.  Yours is clean because you have simple pole at $u=\pm i c$ and $u=i 2 n \pi, n\in\mathbb{Z}$.  Of course we need to consider $c=2n\pi$ which will at most give us a double pole, which is still simpler than what Ron had done for us.  Thanks to Bennett Gardiner for making this clear
Also see this integral, I posted it, still no solution but they are very similar.  Yours is my case for $n=1$ and I have different denominator.  Integral $\int_0^\infty \frac{x^n}{(x^2+\alpha^2)^2(e^x-1)^2}dx$
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
&\color{#00f}{\large%
\int_{0}^{\infty}{2ax\,\dd x \over \pars{x^{2} + a^{2}}\pars{\expo{2\pi x} - 1}}}
=2a\bracks{\half\,\ln\pars{a} - {1 \over 4a} - \half\,\Psi\pars{a}}
\\[3mm]&=\color{#00f}{\large a\bracks{\ln\pars{a} - \Psi\pars{a}} - \half}
\end{align}

See $\ds{\bf\mbox{6.3.21}}$ in A&S Table.

