A logarithmic integral $\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx$

How to prove the following

$$\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx=\frac{\pi^2}{2}$$

I thought of separating the two integrals and use the beta or hypergeometric functions but I thought these are not best ideas to approach the problem.

Any other ideas ?

Solution 1:

After the change of variables $x=\tanh u$ (suggested by the square root) this integral reduces to $$\mathcal{I}=\int_0^{\infty}\frac{2u\,du}{\sinh u}.$$ Expanding $\displaystyle\frac{1}{\sinh u}=2\sum_{k=0}^{\infty}e^{-(2k+1)u}$ and exchanging summation and integration, we find that $$\mathcal{I}=4\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}.$$ Standard manipulations express the last sum in terms of $\zeta(2)=\frac{\pi^2}{6}$: $$\zeta(2)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}+\frac{\zeta(2)}{4}\quad \Longrightarrow \quad \displaystyle\mathcal{I}=3\zeta(2).$$

Solution 2:

Let $-1\le a \le 1$ and: \begin{align*} I(a) &= \int_{0}^{1} \, \log\left(\frac{1+a\,x}{1-a\, x}\right)\frac{1}{x\sqrt{1-x^2}}\, dx \tag 1\\ \frac{\partial}{\partial a}I(a) &= \int_{0}^{1} \, \frac{1}{(1+a\, x)\sqrt{1-x^2}} + \frac{1}{(1-a\, x)\sqrt{1-x^2}} \, dx\\ &= \frac{1}{\sqrt{1-a^2}}\, \left(\arcsin\left(\frac{x+a}{1+a\, x}\right)+\arcsin\left(\frac{x-a}{1-a\, x}\right) \right) \Big|_0^1\\ &= \frac{\pi}{\sqrt{1-a^2}}\\ \therefore I(a) &= \pi\, \arcsin{a} + C \tag 2\\ \end{align*} Putting $a=0$, in $(1)$ and $(2)$, we see that $C=0$

Hence, \begin{align*} I(a) &= \int_{0}^{1} \, \log\left(\frac{1+a\,x}{1-a\, x}\right)\frac{1}{x\sqrt{1-x^2}}\, dx = \pi\, \arcsin{a} \end{align*}

and for this problem $$I(1)=\frac{\pi^2}{2}$$