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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.

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  • $\begingroup$ What does WFA give for the antiderivative? The integrand is unbounded – are you sure the integral converges? $\endgroup$ Commented May 4, 2023 at 12:53
  • $\begingroup$ It converges since I get a numeric value from WFA $\endgroup$
    – Anik Patel
    Commented May 4, 2023 at 13:39
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    $\begingroup$ @AnikPatel WFA is not perfect, and I've definitely had it give numerical values for divergent integrals before. $\endgroup$ Commented May 4, 2023 at 14:33

2 Answers 2

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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}

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Using the series expansion of the dilogarithm:

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

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