TITLE: Integral$\int_1^\infty \log \log \left(x\right)\frac{dx}{1-x+x^2}=\frac{2\pi}{\sqrt 3}\left(\frac{5}{6}\log (2\pi)-\log \Gamma \frac{1}{6}\right)$
QUESTION [15 upvotes]: 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)
$$
I_2:=\int_0^1 \log \log \left(\frac{1}{x}\right)\frac{dx}{1+x+x^2}=\frac{\pi}{\sqrt 3}\log\left(\frac{\sqrt[3]{2\pi}\Gamma(2/3)}{\Gamma(1/3)}\right).
$$
as you can see the results are different, enjoy both.  Obviously, I am ONLY interested in solving I thanks.

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