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Prove the following

$$\int^1_0 \frac{\operatorname{Li}_2^2(x)}{x}\, dx = -3\zeta(5)+\pi^2 \frac{\zeta(3)}{3}$$

where

$$\operatorname{Li}^2_2(x) =\left(\int^x_0 \frac{\log(1-t)}{t}\,dt \right)^2$$

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  • $\begingroup$ Note that, $\mathrm{Li}_2(x)=\mathrm{dilog(x)}$ which is the standard convention not the computer algebra systems convention. $\endgroup$ – Mhenni Benghorbal Aug 25 '13 at 5:27
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Integrating once by parts, we find \begin{align} I=\int_0^{1}\frac{\mathrm{Li}_2^2(x)\,dx}{x}&=\int_0^{1}\mathrm{Li}_2(x)\,d\left(\mathrm{Li}_3(x)\right)=\\ &=\mathrm{Li}_2(1)\mathrm{Li}_3(1)+\int_0^{1}\frac{\mathrm{Li}_3(x)\,\ln(1-x)\,dx}{x}=\\ &=\zeta(2)\zeta(3)-\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^3k(n+k)},\tag{1} \end{align} where the last line is obtained by expanding $\mathrm{Li}_3(x)$ and $\ln(1-x)$ into series and integrating. Now if we denote $H_n=\sum_{k=1}^n\frac{1}{k}$ the $n$th harmonic number, the sum with respect to $k$ in (1) can be written as $$\sum_{k=1}^{\infty}\frac{1}{k(n+k)}=\frac{1}{n}\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{n+k}\right)=\frac{H_n}{n},$$ so that $$I=\zeta(2)\zeta(3)-\sum_{n=1}^{\infty}\frac{H_n}{n^4}.\tag{2}$$ Now using that $\zeta(2)=\frac{\pi^2}{6}$ and the formula (20) here (it is a particular case of a more general Euler sum (24)), namely, $$ \sum_{n=1}^{\infty}\frac{H_n}{n^4} = 3\zeta(5)-\frac{\pi^2}{6}\zeta(3),$$ we finally obtain the desired result for $I$. $\blacksquare$

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  • 3
    $\begingroup$ Interestingly I proved that $$\int^1_0 \frac{\operatorname{Li}_2^2(x)}{x}\,dx= \frac{2}{3}\,\sum_{n\geq 1}\frac{H_n^{(2)}}{n^3}$$ $\endgroup$ – Zaid Alyafeai Aug 9 '13 at 0:03
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\begin{align} I&=\int_0^1\frac{\operatorname{Li}_2^2(x)}{x}\ dx=\sum_{n=1}^\infty\frac1{n^2}\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx=\sum_{n=1}^\infty\frac1{n^2}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)\\ &=\zeta(2)\zeta(3)-(3\zeta(5)-\zeta(2)\zeta(3))=2\zeta(2)\zeta(3)-3\zeta(5) \end{align}

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