TITLE: A hard integral identity on MathSE
QUESTION [82 upvotes]: The following identity on MathSE
$$\int_0^{1}\arctan\left(\frac{\mathrm{arctanh}\ x-\arctan{x}}{\pi+\mathrm{arctanh}\ x-\arctan{x}}\right)\frac{dx}{x}=\frac{\pi}{8}\log\frac{\pi^2}{8}$$
seems to be very difficult to prove.
Question: I worked on this identity for several days without any success. Is there any clue how to prove this integral identity?

REPLY [88 votes]: I have proved this equality by means of Cauchy’s Theorem
applied to an adequate function. Since my solution is too long to post it
here, I posted it in arXiv:

Juan Arias de Reyna, Computation of a Definite Integral, arXiv:1402.3830.

The function
$$G(z)=\frac{\log(1+(1+i)\,f(z)\,)}z$$
where
$$f(x)=\frac{\operatorname{arctanh}(x)-\arctan(x)}{\pi}$$
extended analytically.