I have two forms of solution for this integral:
First : $$\displaystyle \begin{aligned}\int_{0}^{1}{\ln\left(1+\dfrac{\ln^2x}{4π^2}\right)\dfrac{\ln(1-x)}{x}\ \mathrm{d}x}&=2π\int_{0}^{\infty}{\ln(1+x^2)\ln(1-e^{-2πx})\ \mathrm{d}x}\\\\&=-2\int_{0}^{\infty}{\dfrac{x\ \mathrm{Li}_{2}(e^{-2πx})}{1+x^2}}\ \mathrm{d}x\\\\&=-2\sum_{k=1}^{\infty}{\dfrac{1}{k^2}\int_{0}^{\infty}{\dfrac{xe^{-2πkx}}{1+x^2}}\ \mathrm{d}x}\\\\&=2\sum_{k=1}^{\infty}{\dfrac{\mathrm{Ci}(2πk)}{k^2}}\end{aligned}$$
Second: $$\begin{aligned}\int_{0}^{1}{\ln\left(1+\dfrac{\ln^2x}{4π^2}\right)\dfrac{\ln(1-x)}{x}\ \mathrm{d}x}&=2π\int_{0}^{\infty}{\ln(1+x^2)\ln(1-e^{-2πx})\ \mathrm{d}x}\\\\&=-4π^2\int_{1}^{\infty}\int_{0}^{\infty}{\dfrac{x\ln(1+x^2)}{e^{2πxt}-1}\ \mathrm{d}x}\ \mathrm{d}t\end{aligned}$$
It seems that Abel-Plana formula is the possible direction, but I can't find the right function to match the form.
Which is the better solution? Or are there any other methods to solve? Thanks!!!
