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How do I compute this integral ?

$$I=\int_{0}^{1}\ln{\left(\ln{\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)}\right)}dx$$

In the math chatroom someone suggests setting $x=\operatorname{sech}(t)$ and that the result immediately follows.

I don't agree with it

because $$\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}=\frac{e^t+e^{-t}}{2}+\sqrt{\cosh^2{t}-1}=e^t$$ and $$dx=\frac{e^t}{(e^{2t}+1)^2}dt$$

so $$I=\int_{0}^{\infty}\ln(t)\frac{e^{t}}{(e^{2t}+1)^2}dt$$ Thanks for your help.

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marked as duplicate by Tunk-Fey, Najib Idrissi, AlexR, Gerry Myerson, Namaste integration Jul 17 '14 at 12:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ OK you do not agree. Nice to know. But why? And what do you suggest instead? $\endgroup$ – Did Jul 17 '14 at 7:29
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    $\begingroup$ why closed? Thank you $\endgroup$ – china math Jul 17 '14 at 7:37
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    $\begingroup$ Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts. $\endgroup$ – Did Jul 17 '14 at 7:44
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    $\begingroup$ What if I do not want to go and visit the chatroom? Please make your question self-contained. (Actually I did go to the chatroom, first return after a long period of absence, and well... one cannot recommand the experience.) $\endgroup$ – Did Jul 17 '14 at 8:02
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    $\begingroup$ Voted to reopen. After the additions you made, I think this should be put on hold as duplicate, not anymore as "unclear what you're asking". $\endgroup$ – Did Jul 17 '14 at 9:16
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Your derivative is not correct. Actually you should obtain $$\mathrm{d}x=-\mathrm{sech}(t)\tanh(t)\mathrm{d}t\,.$$ Your integral then becomes $$I=-\int_\infty^0\log(t)\mathrm{sech}(t)\tanh(t)\mathrm{d}t=\int_0^\infty\log(t)\mathrm{sech}(t)\tanh(t)\mathrm{d}t\,.$$

This is still not a trivial integral, but Mathematica tells me that it is $$I=-\gamma +\log \left(\Gamma \left(\frac{1}{4}\right)\right)-2 \log \left(\Gamma \left(\frac{3}{4}\right)\right)+\log \left(\Gamma \left(\frac{5}{4}\right)\right)\approx 0.205973\,. $$ Here, $\gamma$ is the Euler-Mascheroni constant and $\Gamma$ is the usual Gamma function.

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  • $\begingroup$ You can use the Hurwitz zeta function to evaluate the integral. See here. $\endgroup$ – Mhenni Benghorbal Jul 17 '14 at 9:46

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