# How to show that $\int_{0}^{\infty}\ln^2\left(\frac{e^x+1}{e^x-1}\right)dx=\frac{7\zeta(3)}{2}$?

Inspired by the integral $$\int_{0}^{\infty}\ln\left(\frac{e^x+1}{e^x-1}\right)dx=\frac{\pi^2}{4}$$ I further found out that

$$I=\int_{0}^{\infty}\ln^2\left(\frac{e^x+1}{e^x-1}\right)dx=\frac{7\zeta(3)}{2}\tag{1}$$

I am trying to prove $$(1)$$

Let $$u=e^{-x}$$, $$u(0)=1,u(\infty)=0$$ $$du=-e^{-x}dx,du=-udx$$ $$dx=-\frac{1}{u}du$$ \begin{align} I&=\int_{0}^{1}\ln^2\left(\frac{1/u+1}{1/u-1}\right)\left(\frac{1}{u}\right)du\\\\ &=\int_{0}^{1}\frac{[\ln(1+u)-\ln(1-u)]^2}{u}du\\\\ &=\int_{0}^{1}\frac{\ln^2(1+u)}{u}du-\int_{0}^{1}\frac{2\ln(1+u)\ln(1-u)}{u}du+\int_{0}^{1}\frac{\ln^2(1-u)}{u}du \end{align} I know that $$-\ln(1-u)=\sum_{n=0}^{\infty}\frac{u^{n+1}}{n+1}$$ $$\ln(1+u)=\sum_{n=0}^{\infty}\frac{(-1)^nu^{n+1}}{n+1}$$ What should I do from this point on? Also, I think that trig substitution is another possible way to approach this problem.

First, rewrite the integrand as$$\mathfrak{I}=\int\limits_0^{\infty}\mathrm dx\,\log^2\left(\frac {1+e^{-x}}{1-e^{-x}}\right)$$Now make the substitution $$z=e^{-x}$$ so that $$x=-\log z$$. Thus\begin{align*}\mathfrak{I} & =\int\limits_0^1\mathrm dz\,\frac {\log^2\left(\frac {1+z}{1-z}\right)}z\end{align*}From here, make the transformation $$z\mapsto\tfrac {1-z}{1+z}$$ and the integral transforms into\begin{align*}\mathfrak{I} & =\int\limits_0^1\frac {2\,\mathrm dz}{(1+z)^2}\frac {1+z}{1-z}\log^2z\\ & =2\int\limits_0^1\mathrm dz\,\frac {\log^2 z}{1-z^2}\end{align*}Expand the denominator as an infinite sequence and integrate by parts twice to see that\begin{align*}\mathfrak{I} & =2\sum\limits_{n\geq0}\int\limits_0^1\mathrm dz\, z^{2n}\log^2 z\\ & =4\sum\limits_{n\geq0}\frac 1{(2n+1)^3}\end{align*}It can be shown, through some manipulation, that the infinite sum is equal to$$\tfrac 78\zeta(3)$$. Thus$$\int\limits_0^{\infty}\mathrm dx\,\log^2\left(\frac {e^x+1}{e^x-1}\right)\color{blue}{=\frac 72\zeta(3)}$$

• \begin{align}\frac{2}{1-x^2}=\frac{2}{1-x}-\frac{2x}{1-x^2}\end{align} therefore the integral is expressible in term of $\int_0^1 \frac{\ln^2 x}{1-x}\,dx$ (the change of variable $y=x^2$ is required) – FDP Feb 20 at 17:05

You could use the substitution $$y=\log \left( \frac{e^x+1}{e^x-1}\right)$$

$$x=\log \left( \frac{e^y+1}{e^y-1}\right)$$
with $$\frac{dx}{dy}=-\frac{2}{e^y-e^{-y}}$$
$$\frac{1}{2(n-1)!} \int_0^\infty \log \left( \frac{e^x+1}{e^x-1}\right)^{n-1} \,dx=\frac{1}{(n-1)!} \int_0^\infty \frac{x^{n-1}}{e^y-e^{-y}} \,dy=\lambda(n)$$
where $$\lambda(n)=\sum_{k=1}^\infty \frac{1}{(2k-1)^n}$$
$$\frac{1}{(n-1)!} \int_0^\infty \frac{x^{n-1}}{e^y-e^{-y}} \,dy$$ being the standard and easily determined integral for $$\lambda(n)$$