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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.

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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.

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