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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?

YCor
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Y. Zhao
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1 Answers1

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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.

David Roberts
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juan
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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 08:57
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    @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 09:49
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    Nice observation! Thanks for your answer to this identity! – Y. Zhao Feb 19 '14 at 14:42
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    Awesome answer! --- it's more than just an observation, I'd say :-) – Suvrit Feb 19 '14 at 16:32