# A hard integral identity on MATH.SE

$$\int_0^{1}\arctan\left(\frac{\mathrm{arctanh}\ x-\arctan{x}}{\pi+\mathrm{arctanh}\ x-\arctan{x}}\right)\frac{dx}{x}=\frac{\pi}{8}\log\frac{\pi^2}{8}$$

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?

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I'm waiting for Noam Elkies's solution :-) – Suvrit Jan 18 '14 at 19:20
A related problem can be found here. – Lucian Jan 21 '14 at 15:08
What happens if $\pi$ in the integral is replaced by some other real number? Can one give a closed-form formula for the integral in this case? – Zurab Silagadze Jan 23 '14 at 5:32
Where does this problem come from and why do you care about it? – Lev Borisov Jan 24 '14 at 3:46
@LevBorisov: This identity is a long-lasting problem(which was raised by another user) on Math.SE, and I think it is just the type of identity which may appear in Ramanujan's notebooks. Although I don't know what this identity may related to, it would be better to draw more attention to this elegent identity. – zy_ Jan 28 '14 at 12:15

## 1 Answer

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, you can get it at 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.

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That's great. I suggest that you at least copy the definitions of the key functions $f(z)$ and $G(z)$ into your answer. – Neil Strickland Feb 18 '14 at 8:57
@larry, I would like to know how this question was generated. I have seen so much structure in your integrand that I will be surprised if there is not something hidden. – juan Feb 18 '14 at 9:49
Nice observation! Thanks for your answer to this identity! – zy_ Feb 19 '14 at 14:42
Thank you for the fantastic contribution. – Ron Gordon Feb 19 '14 at 14:55
Awesome answer! --- it's more than just an observation, I'd say :-) – Suvrit Feb 19 '14 at 16:32