TITLE: Showing that $2\int_0^\infty {\cosh(x)-1\over x(e^{ax}-1)}\,\mathrm dx=\ln\left({\pi \over a\sin\left({\pi\over a }\right)}\right)$
QUESTION [5 upvotes]: $$2\int_{0}^{\infty}{\cosh(x)-1\over x(e^{ax}-1)}\,\mathrm{d}x=\ln\left({\pi \over a\sin\left({\pi\over a }\right)}\right)$$
$$\cosh(x)=1+{x^2\over 2!}+{x^4\over 4!}+{x^6\over 6!}+\cdots$$
$${\cosh(x)-1\over x}={x\over 2!}+{x^3\over 4!}+{x^5\over 6!}+\cdots=\sum_{n=1}^{\infty}{x^{2n-1}\over (2n)!}$$
$$I=\sum_{n=1}^{\infty}{2\over (2n)!}\int_{0}^{\infty}{x^{2n-1}\over e^{ax-1}}\,\mathrm{d}x$$
Recall the Zeta function
$${a^k\over(k-1)!}\int_{0}^{\infty}{x^{k-1}\over e^{ax}-1}\,\mathrm{d}x=\zeta(k)$$
Now setting $k=2n$
$${a^{2k}\over(2k-1)!}\int_{0}^{\infty}{x^{2k-1}\over e^{ax}-1}\text{d}x=\zeta(2k)$$
Rearrange 
$$I=\sum_{n=1}^{\infty}{2\over (2n)!}\cdot{(2n-1)!\zeta(2n)\over a^{2n}}=\sum_{n=1}^{\infty}{\zeta(2n)\over a^{2n}\cdot{n}}$$
So finally we have
$$\sum_{n=1}^{\infty}{\zeta(2n)\over a^{2n}\cdot{n}}=\ln\left({\pi \over a\sin\left({\pi\over a }\right)}\right)$$
This does not answer my question, it is only showing the transformation from integral into an infinite sum.
Anyone can help here to show that the infinite sum produced this closed form?

REPLY [2 votes]: Jack D'Aurizio comments on the answer given by Olivier Oloa and I expand his remark into an answer. Thanks to him for this simple and beautiful observation.

We know that $$\sin \pi x = \pi x\prod_{n = 1}^{\infty}\left(1 - \frac{x^{2}}{n^{2}}\right)\tag{1}$$ and hence
\begin{align}
\log\left(\frac{\pi x}{\sin \pi x}\right) &= -\sum_{n = 1}^{\infty}\log\left(1 - \frac{x^{2}}{n^{2}}\right)\notag\\
&= \sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty}\frac{x^{2k}}{kn^{2k}}\notag\\
&= \sum_{k = 1}^{\infty}\frac{x^{2k}}{k}\sum_{n = 1}^{\infty}\frac{1}{n^{2k}}\notag\\
&= \sum_{n = 1}^{\infty}\frac{\zeta(2n)}{n}x^{2n}\tag{2}
\end{align}
and this is what you need in your last equation (with $1/a$ in place of $x$).