Let me define $J=\int_{0}^{1} \frac{\ln(t)\ln^{2}(1-t)}{t}dt$ and the function $g$ defined on $[0,1]$ : $g(x) = \int_{0}^{x} \frac{\ln(1-t)}{t}dt$ where $\ln^{2}(1-t)$ means $\ln(1-t) \times \ln(1-t)$. In the exercice I was trying to do, we first prove that for all $x \in ]0,1]$, $g(x) = \sum_{n=1}^{+\infty} \frac{x^{n}}{n^{2}}$. Then we prove that $\sum_{1 \leq m < n} \frac{1}{m^{2}n^{2}}$ can be expressed using $\zeta(2)$ and $\zeta(4)$. I found that $\sum_{1 \leq m < n} \frac{1}{m^{2}n^{2}} = \frac{\zeta(2)^{2}-\zeta(4)}{2}$.
In the last question of the exercice, I have to compute $J$. (To do so, I assumed there was some relation between $J$ and $g$... but I couldn't find it). So, my question is : how can I use $g$ to compute $J$ ?
Thanks !
