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Hi I am trying to prove the following$$ I:=\int_1^\infty \log \log \left(x\right)\frac{dx}{1-x+x^2}=\frac{2\pi}{\sqrt 3}\left(\frac{5}{6}\log (2\pi)-\log \Gamma \big(\frac{1}{6}\big)\right). $$ I am not sure at all how to get started on this one. This looks quite intimidating. Something I realized was $$ \int_1^\infty \log \log \left(x\right)\frac{dx}{1-x+x^2}=\int_0^1 \log \log \left(\frac{1}{x}\right)\frac{dx}{1-x+x^2}. $$Thanks.

Note the Gamma function is given by $$ \Gamma(n)=(n-1)!,\quad \Gamma(z)=\int_0^\infty t^{z-1}e^{-t}\, dt. $$ EDIT: THE incorrect integral I first posted was because of a typo. The result of it is given by (notice the denominator sign mistake I made) $$ I_2:=\int_0^1 \log \log \left(\frac{1}{x}\right)\frac{dx}{1+x+x^2}=\frac{\pi}{\sqrt 3}\log\left(\frac{\sqrt[3]{2\pi}\Gamma(2/3)}{\Gamma(1/3)}\right). $$ as you can see the results are different, enjoy both. Obviously, I am ONLY interested in solving I thanks.

Jeff Faraci
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    $\displaystyle\large\ln\left(1 \over x\right) < 0$ when $\displaystyle\large x > 1$. – Felix Marin May 14 '14 at 04:46
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    @integrals : Can you check the question? Changing the limits to 0..1 also does not match with the answer numerically. – gar May 14 '14 at 05:20
  • -1. Why the upvotes? There are obvious problems with this question. (Not to mention the usual lack of context.) – mrf May 14 '14 at 06:37
  • I see Omran Kouba edited the question. Is the stated result correct now? – user111187 May 14 '14 at 07:52
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    This is explicitly calculated in a paper of Victor Adamchik. "A class of logarithmic integrals. Proceedings ISSAC, 1–8, 1997". It goes back to Bierens de Haan, "Nouvelles Tables d’Intégrales Définies" (Table 148 (5)-page 208) (1867). – Omran Kouba May 14 '14 at 07:56
  • @OmranKouba : Thanks for the paper. It's available here, and the result is actually $$\frac{2, \pi}{\sqrt{3}}, \left(\frac{2}{3}\log{(2, \pi)}-\frac{1}{4}\log{(3)}-\log{\Gamma{\left(\frac{1}{3}\right)}}\right)\approx -0.126321481706210$$ – gar May 14 '14 at 09:20
  • @gar You are right, the answer as is given corresponds to $\int_0^1\frac{\log \log(1/x)}{1-x+x^2}dx$. – Omran Kouba May 14 '14 at 09:48
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    @mrf it was merely a typo, i meant to write $1-x+x^2$ in denominator. What context are you looking for? "The usual lack of context", what does this refer to? Thanks for your rude comment:) – Jeff Faraci May 14 '14 at 16:42
  • @OmranKouba Yes small typo, sorry about that. ANd sorry to everyone, I meant to have 1-x in denominator. Sorry about that. – Jeff Faraci May 14 '14 at 16:42
  • @gar sorry I meant to have 1-x in denominator not 1+x. Thanks I just saw all the messages. – Jeff Faraci May 14 '14 at 16:42
  • @gar I added what that original integral I posted is equal to. Sorry for the typo thanks for catching it – Jeff Faraci May 14 '14 at 17:06
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    This integral is also computed in Adamchick's paper, page 8, formula (31). http://repository.cmu.edu/cgi/viewcontent.cgi?article=1095&context=compsci – Leucippus May 15 '14 at 04:20
  • @Leucippus Thanks a lot for this paper – Jeff Faraci May 15 '14 at 05:09
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    @Integrals : That's okay, typos happen! Thanks to OmranKouba for pointing out the paper, we got exposed to many more interesting integrals! – gar May 15 '14 at 09:06
  • @gar Yes many more interesting integrals in this paper. Thank you my friend – Jeff Faraci May 16 '14 at 03:56
  • @Integrals, please go to http://math.stackexchange.com/questions/874431/a-closed-form-of-int-01-frac-ln-ln-left1-x-rightx2-x1-mathrm-dx/874795#874795 to see my answer. – xpaul Sep 02 '14 at 14:18

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Here is an answer. Clear, letting $x\to 1/x$, we have \begin{eqnarray*} I=\int_0^1 \log(-\log x)\frac{1}{1-x+x^2}dx. \end{eqnarray*} Then the rest follows from A closed form of $\int_0^1\frac{\ln\ln\left({1}/{x}\right)}{x^2-x+1}\mathrm dx$.

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