TITLE: Integral $\int_0^1\frac{\ln x}{x-1}\ln\left(1+\frac1{\ln^2x}\right)dx$
QUESTION [26 upvotes]: Is it possible to evaluate this integral in a closed form?
$$
I \equiv \int_{0}^{1}{\ln\left(x\right) \over x - 1}\,
\ln\left(1 + {1 \over \ln^{2}\left(x\right)}\right)\,{\rm d}x
$$
Numerically,
$$I\approx2.18083278090426462584033339029703713513\dots$$

REPLY [16 votes]: $$I=\left(\pi-\frac12\right)\ln(2\pi)-\frac34-4\pi^2\left(\ln\Gamma\left(\frac1{2\pi}\right)+\ln G\left(\frac1{2\pi}\right)\right),$$
where $G(z)$ denotes the Barnes G-function.