MHB Integration using Beta and Gamma Functions

[alyafey22]

Here is the evaluation of the last integral. There might be a mistake in a constant

$$\int^1_0 \frac{\operatorname{Li}^2_2(x)\log^2(x)}{x}\,dx = \sum_{n,k\geq 1}\frac{1}{(nk)^2}\int^1_0x^{n+k-1}\log^2(x)\,dx= 2\sum_{n,k\geq 1}\frac{1}{(nk)^2(n+k)^3}$$

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

First sum

$$\begin{align}\sum_{n,k\geq 1}\frac{1}{n^3k^2(n+k)^2} &= \sum_{n,k\geq 1}\frac{1}{n^4k^2(n+k)}-\sum_{n,k\geq 1}\frac{1}{n^4k(n+k)^2}\\
&=\sum_{n,k\geq 1}\frac{1}{n^5k^2}-\sum_{n,k\geq 1}\frac{1}{n^5k(n+k)}-\sum_{n,k\geq 1}\frac{1}{n^5k(n+k)}+\sum_{n,k\geq 1}\frac{1}{n^5(n+k)^2}\\
&=\zeta(5)\zeta(2)-2\sum_{n \geq 1}\frac{H_n}{n^5}+\sum_{n \geq 1}\frac{H^{(2)}_n}{n^5}
\end{align}$$

$$\begin{align}\sum_{n,k\geq 1}\frac{1}{n^3k(n+k)^3} &= \sum_{n,k\geq 1}\frac{1}{n^4k(n+k)^2}-\sum_{n \geq 1}\frac{H_n^{(3)}}{n^4}\\
&= \sum_{n,k\geq 1}\frac{H^{(2)}_n}{n^5}-\sum_{n,k\geq 1}\frac{H_n}{n^5}-\sum_{n \geq 1}\frac{H_n^{(3)}}{n^4}
\end{align}$$