Evaluate a Dilogarithmic integral $\int_0^1 \ln(x)\operatorname{Li}_2(x)dx$

Question

I believe a closed form for this integral exists since WFA is able to compute the antiderivative. It does not come out with a closed form when I give it bounds on the integral for some reason though. This integral is tricky. I tried Feynman's Trick by paramterising with $$I(a)=\int_0^1 \ln(ax)\operatorname{Li}_2(x)dx$$

however, $a=0$ does not give a value so this cannot be used. It would be very useful though since after differentiating we get $\frac{\operatorname{Li}_2(x)}{ax}$ as the integrand which is just $\operatorname{Li}_3(x) a^{-1}$ but of course this is not possible because of the $a=0$ problem. Are there any other ways to approach this? I think integration by parts may work but I'm not sure how to deal with the integrals that result from it.

Answer 1

Integrate by parts

\begin{align} \int_0^1 \ln x\operatorname{Li}_2(x)\ dx &= \int_0^1 \operatorname{Li}_2(x)\ d[x(\ln x-1)]\\ &=-\text{Li}_2(1) +\int_0^1 (\ln x-1)\ln(1-x)dx\\ &=-\frac{\pi^2}6 +\left(3-\frac{\pi^2}6 \right)=3-\frac{\pi^2}3 \end{align}

Answer 2

Using the series expansion of the dilogarithm:

$$\begin{align*} & \int_0^1 \ln(x) \operatorname{Li}_2(x) \, dx \\ &= \sum_{n=1}^\infty \frac1{n^2} \int_0^1 x^n \ln(x) \, dx \\ &= -\sum_{n=1}^\infty \frac1{n^2(n+1)^2} \\ &= 2 \sum_{n=1}^\infty \left(\frac1n-\frac1{n+1}\right) - \sum_{n=1}^\infty \frac1{n^2} - \sum_{n=1}^\infty \frac1{(n+1)^2} \\ &= 2 - \frac{\pi^2}6 - \left(\frac{\pi^2}6-1\right) = \boxed{3-\frac{\pi^2}3} \end{align*}$$