Integral of $\ln(\tanh(x))$ I'd like a hint toward how I could evaluate this definite integral. I'm aware it's likely to be non elementary and I haven't found a way to evaluate it yet:$$\int_0^\infty \ln(\tanh(x))\,\,\mathrm{d}x$$
If you're curious where this came from, I was looking at an integral involving $\ln(\sin(x))$ and I thought of this one.
Thanks.
 A: Through the substitution $x=\text{arctanh}(t)$ we have $I=\int_{0}^{+\infty}\log\tanh x\,dx = \int_{0}^{1}\frac{\log t}{1-t^2}\,dt$. 
Since $\int_{0}^{1}t^{2n}\log(t)\,dt = -\frac{1}{(2n+1)^2}$ we have
$$ I = -\sum_{n\geq 0}\frac{1}{(2n+1)^2}=-\left[\zeta(2)-\frac{1}{4}\zeta(2)\right]=-\frac{3}{4}\cdot\frac{\pi^2}{6}=\color{red}{-\frac{\pi^2}{8}}. $$
A: Here is a slight variation on a theme.
Making use of the result $\tanh^2 x = 1 - \mbox{sech}^2 x$, we can write the integral as
$$I = \frac{1}{2} \int_0^\infty \ln (1 - \text{sech}^2 x) \, dx.$$
Setting $\text{sech}^2 x \mapsto x$ gives
$$I =  \frac{1}{4} \int_0^1 \frac{\ln (1 - x)}{x \sqrt{1 - x}} \, dx.$$
There are many ways to evaluate this integral. One way is by enforcing a substitution of $x \mapsto 1 - x^2$. Doing so we arrive at
$$I = \int_0^1 \frac{\ln x}{1 - x^2} \, dx,$$
which is exactly the same point @Jack D'Aurizio arrived at in his solution.
Departing from Jack, we now employ a self-similar substitution of $u = \dfrac{1 - x}{1 + x}$.
Thus
$$I = \frac{1}{2} \int_0^1 \frac{1}{u} \ln \left (\frac{1 - u}{1 + u} \right ) \, du = \frac{1}{2} \int_0^1 \frac{\ln (1 - u)}{u} \, du - \frac{1}{2} \int_0^1 \frac{\ln (1 + u)}{u} \, du.$$
In the second of these integrals let $u \mapsto -u$
\begin{align*}
I &= \frac{1}{2} \int_0^1 \frac{\ln (1 - u)}{u} \, du - \frac{1}{2} \int_0^{-1} \frac{\ln (1 - u)}{u} \, du\\
&= -\frac{1}{2} \text{Li}_2 (1) + \frac{1}{2} \text{Li}_2 (-1)\\
&= -\frac{1}{2} \cdot \frac{\pi^2}{6} + \frac{1}{2} \cdot -\frac{\pi^2}{12}\\
&= -\frac{\pi^2}{8},
\end{align*}
where use of the dilogarithm function has been made.
A: Another option: $$\ln\tanh x=-2\operatorname{artanh}\exp -2x=-2\sum_{k\ge 0}\frac{\exp -(4k+2)x}{2k+1},$$so$$\int_0^\infty\ln\tanh x dx=-\sum_{k\ge 0}\frac{1}{(2k+1)^2}=-\frac{\pi^2}{8}.$$
A: By applying the definition of the hyperbolic tangent function in terms of the exponential we obtain
$$\begin{align}
\int_0^{\infty}\log(\tanh(x))dx&=\int_0^{\infty}\log\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right)dx \\
&=\int_0^{\infty}\log\left(\frac{1-e^{-2x}}{1+e^{-2x}}\right)dx\\
&=\int_0^{\infty}\log\left(1-e^{-2x}\right)-\log\left(1+e^{-2x}\right)dx
\end{align}$$
Now by expanding the logarithm as a series $($!$)$ we further get
$$\begin{align}
\int_0^{\infty}\log\left(1-e^{-2x}\right)-\log\left(1+e^{-2x}\right)dx&=\int_0^{\infty}-\sum_{n=1}^{\infty}\frac1{n}e^{-2nx}+\sum_{n=1}^{\infty}\frac{(-1)^n}{n}e^{-2nx}dx\\
&=-\sum_{n=1}^{\infty}\left[-\frac{e^{-2nx}}{2n^2}\right]_0^{\infty}+\sum_{n=1}^{\infty}\left[(-1)^{n+1}\frac{e^{-2nx}}{2n^2}\right]_0^{\infty}\\
&=-\sum_{n=1}^{\infty}\frac1{2n^2}-\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n^2}\\
&=-\frac12\zeta(2)-\frac12\eta(2)\\
&=-\frac12\frac{\pi^2}6-\frac12\frac{\pi^2}{12}\\
&=-\frac{\pi^2}8
\end{align}$$
Which is the desired result. $\zeta(s)$ denotes the Riemann Zeta Function and $\eta(s)$ the Dirichlet Eta Function respectively for which the values are known. 
Anyway this solution is kind of unsteady since I cannot really justify the validity of the power series of the logarithmic functions $($which is normally restricted to $|x|<1$ $)$ nor the termwise integration. Nevertheless it leads to the right solution.

EDIT:
As illustrated by ComplexYetTrivial within the comments the validity of the series expansion of the logarithm is guaranteed due the the fact that $e^{-2x}<1$ for all $x>0$ from where the series converges. Whereas the termwise integration is justified by the monotone dominate convergence theorem. Thus my proposed solution seems to be entirely fine.
