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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.

Community
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math110
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    OK you do not agree. Nice to know. But why? And what do you suggest instead? – Did Jul 17 '14 at 07:29
  • because this two $ln(ln())$ – math110 Jul 17 '14 at 07:30
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    why closed? Thank you – math110 Jul 17 '14 at 07:37
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    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. – Did Jul 17 '14 at 07:44
  • Please try the change of variable proposes in the chatroom and show us that it gives nothing. – user37238 Jul 17 '14 at 07:54
  • @user37238,hello,you can go in the room see it – math110 Jul 17 '14 at 07:56
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    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.) – Did Jul 17 '14 at 08:02
  • @Did,Because I don't want to say which person.and second I want introduce this room. – math110 Jul 17 '14 at 08:07
  • .and I hope @Did can in this room to discuss – math110 Jul 17 '14 at 08:08
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    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". – Did Jul 17 '14 at 09:16
  • I think \dfrac should not be used in titles. See: [Guidelines for good use of $\LaTeX$ in question titles}(http://meta.math.stackexchange.com/questions/9687/guidelines-for-good-use-of-latex-in-question-titles) – Martin Sleziak Jul 17 '14 at 09:50
  • @Martin I think \dfrac should not be used, other than in exceptional circumstances! – user1729 Jul 17 '14 at 12:22
  • (Also, I should say that \dfrac and \frac render the same in this post, because the fractions are taking new lines.) – user1729 Jul 17 '14 at 12:42

1 Answers1

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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.

Aecturus
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