Ytukyg

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.

By applying the definition of the hyperbolic tangent function in terms of the exponential we obtain

$$begin{align}
int_0^{infty}log(tanh(x))dx&=int_0^{infty}logleft(frac{e^x-e^{-x}}{e^x+e^{-x}}right)dx \
&=int_0^{infty}logleft(frac{1-e^{-2x}}{1+e^{-2x}}right)dx\
&=int_0^{infty}logleft(1-e^{-2x}right)-logleft(1+e^{-2x}right)dx
end{align}$$

Now by expanding the logarithm as a series $($!$)$ we further get

$$begin{align}
int_0^{infty}logleft(1-e^{-2x}right)-logleft(1+e^{-2x}right)dx&=int_0^{infty}-sum_{n=1}^{infty}frac1{n}e^{-2nx}+sum_{n=1}^{infty}frac{(-1)^n}{n}e^{-2nx}dx\
&=-sum_{n=1}^{infty}left[-frac{e^{-2nx}}{2n^2}right]_0^{infty}+sum_{n=1}^{infty}left[(-1)^{n+1}frac{e^{-2nx}}{2n^2}right]_0^{infty}\
&=-sum_{n=1}^{infty}frac1{2n^2}-sum_{n=1}^{infty}frac{(-1)^{n+1}}{2n^2}\
&=-frac12zeta(2)-frac12eta(2)\
&=-frac12frac{pi^2}6-frac12frac{pi^2}{12}\
&=-frac{pi^2}8
end{align}$$

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.

Through the substitution $x=text{arctanh}(t)$ we have $I=int_{0}^{+infty}logtanh x,dx = int_{0}^{1}frac{log t}{1-t^2},dt$.

Since $int_{0}^{1}t^{2n}log(t),dt = -frac{1}{(2n+1)^2}$ we have

$$ I = -sum_{ngeq 0}frac{1}{(2n+1)^2}=-left[zeta(2)-frac{1}{4}zeta(2)right]=-frac{3}{4}cdotfrac{pi^2}{6}=color{red}{-frac{pi^2}{8}}. $$

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.

Another option: $$lntanh x=-2operatorname{artanh}exp -2x=-2sum_{kge 0}frac{exp -(4k+2)x}{2k+1},$$so$$int_0^inftylntanh x dx=-sum_{kge 0}frac{1}{(2k+1)^2}=-frac{pi^2}{8}.$$