Computation of integral $\int_{0}^{1}\ln(p)\ln(1-p)p^{2}\,dp$

Question

I want to compute this integral:

\begin{equation*} J=\int_{0}^{1}\ln(p)\ln(1-p)p^{2}dp \end{equation*} It will be great if you can detail the proof.

I tried to do change of variable it does not work, and also integration by part. Thanks.

Answer 1

Change variables: $p=\mathrm{e}^{-t}$ $$ J = \int_0^1 \ln(p) \ln(1-p) p^2 \mathrm{d}p = \int_0^\infty t \mathrm{e}^{-3t} \ln\left(1-\mathrm{e}^{-t}\right) \mathrm{d}t = \int_0^\infty t \mathrm{e}^{-3t} \sum_{k=1}^\infty \left(- \frac{1}{k} \mathrm{e}^{-k t}\right) \mathrm{d}t $$ Using $$ \int_0^\infty t \mathrm{e}^{-(k+3) t} \mathrm{d}t = \left. - \frac{1+(k+3)t}{k+3}\mathrm{e}^{-(k+3)t} \right|_0^{\infty} = \frac{1}{(k+3)^2} $$ Hence: $$\begin{eqnarray} J &=& - \sum_{k=1}^\infty \frac{1}{k} \frac{1}{(k+3)^2} = - \sum_{k=1}^\infty \left( \frac{1}{3} \frac{1}{(k+3)^2} - \frac{1}{9k} + \frac{1}{9(k+3)}\right) \\ &=& - \frac{1}{3} \sum_{k=1}^\infty \frac{1}{(k+3)^2} +\frac{1}{9} \sum_{k=1}^\infty \left( \frac{1}{k} - \frac{1}{k+3}\right) \\ &=& - \frac{1}{3} \left( \underbrace{\sum_{k=1}^\infty \frac{1}{k^2}}_{\zeta(2) = \pi^2/6} - \frac{1}{1^2} - \frac{1}{2^2} - \frac{1}{3^2} \right) +\frac{1}{9} \underbrace{\sum_{k=1}^\infty \left( \frac{1}{k} - \frac{1}{k+3}\right)}_{\frac{1}{1} + \frac{1}{2} + \frac{1}{3}} = \frac{71}{108} - \frac{\pi^2}{18} \end{eqnarray} $$