TITLE: Definite Dilogarithm integral $\int^1_0 \frac{\operatorname{Li}_2^2(x)}{x}\, dx $
QUESTION [13 upvotes]: Prove the following 
$$\int^1_0 \frac{\operatorname{Li}_2^2(x)}{x}\, dx = -3\zeta(5)+\pi^2 \frac{\zeta(3)}{3}$$
where 
$$\operatorname{Li}^2_2(x) =\left(\int^x_0 \frac{\log(1-t)}{t}\,dt \right)^2$$

REPLY [2 votes]: $$I=\int_0^1\frac{\operatorname{Li}_2^2(x)}{x}\ dx=\sum_{n=1}^\infty\frac1{n^2}\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\overset{IBP}{=}\sum_{n=1}^\infty\frac1{n^2}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)$$
$$=\zeta(2)\zeta(3)-(3\zeta(5)-\zeta(2)\zeta(3))=\boxed{2\zeta(2)\zeta(3)-3\zeta(5)}$$