The [following identity on MathSE](https://math.stackexchange.com/questions/464769/how-to-prove-int-01-tan-1-left-frac-tanh-1x-tan-1x-pi-tanh-1)


$$\int\_0^{1}\arctan\left(\frac{\mathrm{arctanh}\ x-\arctan{x}}{\pi+\mathrm{arctanh}\ x-\arctan{x}}\right)\frac{dx}{x}=\frac{\pi}{8}\log\frac{\pi^2}{8}$$


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?



A hard integral identity on MathSE
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](http://arxiv.org/abs/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.


