Integral$\int_1^\infty \log \log \left(x\right)\frac{dx}{1-x+x^2}=\frac{2\pi}{\sqrt 3}\left(\frac{5}{6}\log (2\pi)-\log \Gamma \frac{1}{6}\right)$
Solution 1:
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 $\int_0^1\frac{\ln\ln\left({1}/{x}\right)}{x^2-x+1}\mathrm dx$.