How to evaluate $\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$ I want to evaluate $$\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$$
But I've not been successful in doing so, what I tried is
$$\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx=\int _0^1\left(x\right)'\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$$
$$\overset{\operatorname{IBP}}=\ln \left(\zeta (2)\right)+\int _0^1\frac{\ln \left(1-x\right)}{\operatorname{Li}_2\left(x\right)}\:dx$$
I also tried using identities for the dilogarithm but they dont do much.
I'm not sure what to do now, any help will be very well regarded, thank you.
 A: I do not know if there is a closed form.
Being myself quite lazy, I should compose Taylor series  starting from
$$\text{Li}_2(x)=\sum_{n=1}^\infty \frac {x^n}{n^2}$$ and being very patient (this is cheating ! I used a CAS); write
$$\log \left(\operatorname{Li}_2\left(x\right)\right)=\log(x)+\sum_{n=1}^\infty a_n x^n$$ where, unfortunately, all $a_n$'s are positive. The coefficients make the sequence
$$\left\{\frac{1}{4},\frac{23}{288},\frac{23}{576},\frac{50119}{2073600},\frac{33769
   }{2073600},\frac{257986487}{21946982400},\frac{130265243}{14631321600},\cdots\right\}$$ which decrease quite fast; halas, nothing being recognized by $OEIS$.
This gives
$$\int_0^1\log \left(\operatorname{Li}_2\left(x\right)\right)\,dx=-1+\sum_{n=1}^\infty \frac{a_n}{n+1}$$ Unfortunately, the convergence is not very fast
$$\left(
\begin{array}{cc}
p & 1+\sum_{n=1}^p \frac{a_n}{n+1} \\
 10 & -0.826294 \\
 20 & -0.824783 \\
 30 & -0.824456 \\
 40 & -0.824334 \\
 50 & -0.824276 \\
 60 & -0.824243 \\
 70 & -0.824223 \\
 80 & -0.824210 \\
 90 & -0.824201 \\
 100 & -0.824194 \\
\dots & \dots \\
\infty &-0.824166
\end{array}
\right)$$
However, it is much faster than the same work done using your last expression.
