Closed- form of $\int_0^1 \frac{{\text{Li}}_3^2(-x)}{x^2}\,dx$ Is there a possibility to find a closed-form for
 $$\int_0^1 \frac{{\text{Li}}_3^2(-x)}{x^2}\,dx$$
 A: Focus on the relation
$$\frac{d}{dx}{\rm Li}_k(x)=\frac{1}{x}{\rm Li}_{k-1}(x)$$
Let's look at the derivative of ${\rm Li}_m {\rm Li}_n$ in general:
$$\frac{d}{dx}({\rm Li}_m(-x){\rm Li}_n(-x))=\frac{1}{x}({\rm Li}_{m-1}(-x){\rm Li}_n(-x)+{\rm Li}_m(-x){\rm Li}_{n-1}(-x))$$
Let's define
$$f(m,n)=\int_0^1 \frac{{\rm Li}_m(-x){\rm Li}_n(-x)}{x^2}dx$$
Integration by parts:
$$f(m,n)=-\frac{1}{x}{\rm Li}_m(-x){\rm Li}_n(-x)|_0^1 +\int_0^1 \frac{{\rm Li}_{m-1}(-x){\rm Li}_n(-x)+{\rm Li}_{m}(-x){\rm Li}_{n-1}(-x)}{x^2}dx$$
$$f(m,n)=-\frac{1}{x}{\rm Li}_m(-x){\rm Li}_n(-x)|_0^1 +f(m-1,n)+f(m,n-1)$$
This is a recursive relation that can express $f(3,3)$ with lower terms that are easier to express analytically. Note that the nonintegral part above has to be taken as a limit at $x=0$. You can imagine having $\epsilon$ for the lower integral bound and taking $\epsilon\to 0$ at the end.
Anyway, you can see from the power series definition that
$$\lim_{\epsilon\to 0}\frac{1}{\epsilon}{\rm Li}_m(\epsilon){\rm Li}_n(\epsilon)=0$$
Additionally, ${\rm Li}_n(-1)=-\eta(n)$ where $\eta$ is the Dirichlet eta function. Simplification:
$$f(m,n)=-\eta(m)\eta(n) +f(m-1,n)+f(m,n-1)$$
First of all, $f$ is symmetric in the arguments. Secondly, ${\rm Li}_0(x)=\frac{x}{1-x}$ so recursion can end at
$$f(n,0)=-\int_0^1 \frac{{\rm Li}_n(-x)}{x(1+x)}dx=-\int_0^1 \frac{{\rm Li}_n(-x)}{x}dx+\int_0^1 \frac{{\rm Li}_n(-x)}{1+x}dx=$$
$$=-{\rm Li}_{n+1}(-1)+\int_0^1 \frac{{\rm Li}_n(-x)}{1+x}dx$$
The first one is again the eta function. But the second one I don't know what to do with. Maybe this wasn't such a good idea. Any suggestions?
