# Integral$\int_1^\infty \log \log \left(x\right)\frac{dx}{1-x+x^2}=\frac{2\pi}{\sqrt 3}\left(\frac{5}{6}\log (2\pi)-\log \Gamma \frac{1}{6}\right)$

UPDATED

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) $$I_2:=\int_0^1 \log \log \left(\frac{1}{x}\right)\frac{dx}{1+x+x^2}=\frac{\pi}{\sqrt 3}\log\left(\frac{\sqrt{2\pi}\Gamma(2/3)}{\Gamma(1/3)}\right).$$ as you can see the results are different, enjoy both. Obviously, I am ONLY interested in solving I thanks.

• $\displaystyle\large\ln\left(1 \over x\right) < 0$ when $\displaystyle\large x > 1$. – Felix Marin May 14 '14 at 4:46
• @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 5:20
• 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 7:56
• @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
• This integral is also computed in Adamchick's paper, page 8, formula (31). repository.cmu.edu/cgi/… – Leucippus May 15 '14 at 4:20

## 1 Answer

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 $\int_0^1\frac{\ln\ln\left({1}/{x}\right)}{x^2-x+1}\mathrm dx$.

• You're cheating Mr. xpaul! +1 ≧◠◡◠≦✌ – Anastasiya-Romanova 秀 Sep 2 '14 at 15:23
• @V-Moy, what do you mean? – xpaul Sep 3 '14 at 22:18