Ideas on approaching the integral $\int_0^1 \frac{\operatorname{artanh} x}{x} \mathrm d x$

Question

I stumbled across this integral while trying to prove, $\displaystyle \sum_{n=0}^\infty \frac 1 {(2n + 1)^2} = \frac{\pi^2} 8 $, a brief sketch of my method being,

$\displaystyle \sum_{n=0}^\infty \frac 1 {(2n + 1)^2} = \sum_{n=0}^\infty \int_0^1 x^{2n} \mathrm dx \int_0^1 y^{2n} \mathrm dy = \int_0^1 \int_0^1\sum_{n=0}^\infty (x^2 y^2)^n \mathrm d x \mathrm d y = \int_0^1 \int_0^1 \frac 1 {1 - x^2y^2}\mathrm d x \mathrm d y = \frac 1 2 \int_0^1 \frac 1 {x} \log\left(\frac{1+x}{1-x}\right) \mathrm dx = \int_0^1 \frac{\operatorname{artanh}(x)} x \mathrm d x$

Obviously, there's no elementary antiderivative, and I don't think using Wolfram to calculate dilogs would constitute a nice proof. I'm absolutely lost, so I would appreciate a nudge in the right direction. (maybe residues?)

Edit: looking for a method that doesn't assume or compute the result $\zeta(2) = \frac{\pi^2}{6}$

Answer 1

By integration by parts, one has \begin{eqnarray} \int_0^1 \frac{\operatorname{artanh} x}{x} \mathrm d x&=&\int_0^1\operatorname{artanh} x \mathrm d \ln x\\ &=&\operatorname{artanh} x\ln x|_0^1-\int_0^1\ln x\mathrm d\operatorname{artanh} x\\ &=&-\int_0^1\frac{\ln x}{1-x^2}dx. \end{eqnarray} The calculation of this integral follows this post.

Answer 2

Consider a substitution in your double integral: $$ u=\arccos\sqrt{\frac{1-x^2}{1-x^2y^2}}\quad v=\arccos\sqrt{\frac{1-y^2}{1-x^2y^2}}, $$ or $$ x=\frac{\sin u}{\cos v}\quad y=\frac{\sin v}{\cos u}. $$ The rest should be fairly easy (because the Jacobi determinant will be something nice).

Answer 3

There are other ways to evaluate,

$\displaystyle \int_0^1 \int_0^1 \dfrac{1}{1-x^2y^2}dxdy$

Define,

$\begin{align}P&:=\int_0^1\int_0^1\frac{1}{1-xy}\, dxdy\\ Q&:=\int_0^1\int_0^1\frac{1}{1+xy}\, dxdy \end{align}$

$\begin{align} Q+P&=\int_0^1\int_0^1 \frac{1}{1+xy}\, dxdy+\int_0^1\int_0^1 \frac{1}{1-xy}\, dxdy\\ \end{align}$

In the second integral perform the change of variable $u=-x$,

$\begin{align} Q+P&=\int_0^1\int_0^1 \frac{1}{1+xy}\, dxdy+\int_0^1\left(\int_{-1}^0 \frac{1}{1+xy}\,dx\right)\, dy\\ &=\int_0^1\left(\int_{-1}^1 \frac{1}{1+xy}\,dx\right)\, dy\\ \end{align}$

For $y\in[0;1]$, let,

$\displaystyle K(y)=\int_{-1}^1 \frac{1}{1+xy}\,dx$

If $y\in [0;1]$, define the function $\varphi$ on $[-1;1]$,

$\displaystyle \varphi(x)=x+\frac{1}{2}y(x^2-1)$

$\displaystyle \varphi^{-1}(u)=\frac{\sqrt{y^2+2uy+1}-1}{y}$

and,

$\displaystyle \frac{\partial{\varphi^{-1}}}{\partial{u}}(u)=\frac{1}{\sqrt{y^2+2uy+1}}$

Perform the change of variable $u=\varphi(x)$,

$\begin{align} K(y)&=\int_{-1}^1\frac{1}{y^2+2uy+1}\,du \end{align}$

Therefore,

$\begin{align} Q+P&=\int_0^1\left(\int_{-1}^1\frac{1}{y^2+2uy+1}\,du\right)\,dy \end{align}$

Perform the change of variable $u=\cos\theta$,

$\begin{align} Q+P&=\int_0^1\left(\int_{0}^\pi\frac{\sin \theta}{y^2+2y\cos\theta+1}\,d\theta\right)\,dy\\ &=\int_{0}^\pi \sin\theta\left(\int_0^1 \frac{1}{y^2+2y\cos\theta+1}\,dy\right)\,d\theta\\ &=\int_{0}^\pi \sin\theta\left[\frac{\arctan\left(\frac{\cos \theta+y}{\sqrt{1-\cos^2 \theta}}\right)}{\sqrt{1-\cos^2 \theta}}\right]_{y=0}^{y=1}\,d\theta\\ \end{align}$

Since for $y\in \left[0,\pi\right[$,

$\begin{align} \tan\left(\frac{\pi}{2}-\frac{\theta}{2}\right)&=\frac{\cos\left(\frac{\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\\ &=\frac{2\cos^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}\\ &=\frac{\cos^2\left(\frac{\theta}{2}\right)+\left(1-\sin^2\left(\frac{\theta}{2}\right)\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}\\ &=\frac{1+\cos^2\left(\frac{\theta}{2}\right)-\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}\\ &=\frac{1+\cos\left(2\times \frac{y}{2}\right)}{\sin\left(2\times \frac{\theta}{2}\right)}\\ &=\frac{1+\cos\theta}{\sin \theta} \end{align}$

then,

$\begin{align} Q+P&=\int_{0}^\pi \left(\left(\frac{\pi}{2}-\frac{\theta}{2}\right)-\left(\frac{\pi}{2}-\theta\right)\right)\,d\theta\\ &=\int_{0}^\pi\frac{\theta}{2}\,d\theta\\ &=\frac{\pi^2}{4} \end{align}$

Moreover,

$\begin{align} Q+P&=\int_0^1\int_0^1 \frac{1}{1+xy}\, dxdy+\int_0^1\int_0^1 \frac{1}{1-xy}\, dxdy\\ &=\int_0^1\int_0^1\frac{2}{1-x^2y^2}\, dxdy\\ \end{align}$

Therefore,

$\boxed{\displaystyle \int_0^1\int_0^1\frac{1}{1-x^2y^2}\, dxdy=\frac{\pi^2}{8}}$

(from Archiv der Mathematik und Physik,1913, p323-324, F. Goldscheider)

Another way,

https://algean2016.wordpress.com/2013/10/21/the-basel-problem-double-integral-method-i/

(from, A proof that Euler missed: evaluating $\zeta(2)$ the easy way. Tom M. Apostol, The mathematical intelligencer, vol. 5, number 3,1983. This article computes $P$ but it's easily proved that $P=2Q$)