Alternative ways to evaluate $\int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$

Solution 1:

By series expansion

$$\displaystyle \int^1_0 \frac{\text{Li}_2(x)^2}{x}\,dx=\sum_{k,n\geq 1}\frac{1}{(nk)^2}\int^1_0x^{n+k-1}\,dx =\sum_{k,n\geq 1}\frac{1}{(nk)^2(n+k)}$$

By some manipulations

$$\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{k}{n^2(n+k)}= \sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{1}{n^2}-\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{1}{n(n+k)}$$

Now use that

$$\frac{H_k}{k} = \sum_{n\geq 1}\frac{1}{n(n+k)}$$

Hence we conclude that

$$\int^1_0 \frac{\mathrm{Li}^2_2(x)}{x}\,dx = \zeta(2)\zeta(3)-\sum_{k\geq 1}\frac{H_k}{k^4} $$

The euler some is known

$$\sum_{k\geq 1}\frac{H_k}{k^4} = 3\zeta(5)-\zeta(2)\zeta(3)$$

Finally we get

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

The other integral is very complicated to evaluate. I obtained formula using non-linear euler some here.

$$ \int^1_0\frac{\mathrm{Li}_{2}(x)^3}{x}\, dx = \zeta(3)\zeta(2)^2- \zeta(2) S_{3,2} +\sum_{k\geq 1} \frac{H_k^{(3)} H_k}{k^3}\\-\mathscr{H}(3,3)+\zeta(3) \zeta(4)-\zeta(3)\mathscr{H}(2,1)$$

where

$$ S_{p \, , \, q} = \sum_{n\geq 1} \frac{H^{(p)}}{n^q}$$

$$\begin{align}\mathscr{H}(p,q) = \int^1_0 \frac{\mathrm{Li}_p(x)\mathrm{Li}_q(x)}{x}\,dx \end{align}$$