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.


  • 2
    $\begingroup$ Related: math.stackexchange.com/questions/2182256/integral-of-ln-cosh-x Did you find a solution for $\int_0^{\infty}\log(\sinh(x))$? Mathematica gives $\int_0^{\infty}\log(\tanh(x))=-\frac{\pi^2}{8}$ so there might be a nice way to solve this one $\endgroup$ – Václav Mordvinov Dec 5 '18 at 20:13
  • $\begingroup$ I believe that both $\int_0^\infty{\ln\left(\sinh\left(x\right)\right)}\,\,\mathrm{d}x$ and $\int_0^\infty{\ln\left(\cosh\left(x\right)\right)}\,\,\mathrm{d}x$ diverge but that result for tanh seems promising for a really satisfying solution. $\endgroup$ – Sheepe Dec 5 '18 at 20:19
  • $\begingroup$ In the light of some of the solutions below, why not now consider a generalization of your original integral: $\int_0^\infty \left(\ln(\tanh(x)) \right)^n\, dx$ where n is a positive integer. $\endgroup$ – James Arathoon Dec 8 '18 at 17:32

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.


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.

  • $\begingroup$ Absolutely amazing, thank you very much. $\endgroup$ – Sheepe Dec 5 '18 at 20:40
  • $\begingroup$ @Sheepe Thank you. Howsoever I am still unsure about the convergence radius of the logarithmic series expansion which is dismissed withtin this solution ^^ $\endgroup$ – mrtaurho Dec 5 '18 at 20:41
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    $\begingroup$ Since $\mathrm{e}^{-2 x} < 1$ for $x>0$, the series expansion is valid. Termwise integration is justified by the monotone/dominated convergence theorem, so the solution is perfectly fine! $\endgroup$ – ComplexYetTrivial Dec 5 '18 at 20:42
  • $\begingroup$ I did get up until the series substitution in my own attempts, but out of my own concerns of the same idea I couldn't see how to continue but it seems to work out nevertheless. $\endgroup$ – Sheepe Dec 5 '18 at 20:43
  • $\begingroup$ @ComplexYetTrivial This completes my evaluation and clears my doubts! Thank you! Firstly I was sceptical about my own attempt but out of pure curiosity I finished the evaluation and indeed it turned out to work :) $\endgroup$ – mrtaurho Dec 5 '18 at 20:50

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

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


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: $$\ln\tanh x=-2\operatorname{artanh}\exp -2x=-2\sum_{k\ge 0}\frac{\exp -(4k+2)x}{2k+1},$$so$$\int_0^\infty\ln\tanh x dx=-\sum_{k\ge 0}\frac{1}{(2k+1)^2}=-\frac{\pi^2}{8}.$$


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