Closed- form of $\int_0^1 \frac{{\text{Li}}_3^2(-x)}{x^2}\,dx$

Solution 1:

$\def\Li{{\rm Li}}$ Repeatedly integrate by parts. \begin{align} &\int^1_0\frac{\Li_3^2(-x)}{x^2}{\rm d}x\\ =&-\frac{\Li_3^2(-x)}{x}\Bigg{|}^1_0+\int^1_0\frac{2\Li_2(-x)\Li_3(-x)}{x^2}{\rm d}x\\ =&-\frac{9}{16}\zeta^2(3)-\frac{2\Li_2(-x)\Li_3(-x)}{x}\Bigg{|}^1_0+\int^1_0\frac{2\Li_2^2(-x)}{x^2}{\rm d}x-\int^1_0\frac{2\Li_3(-x)\ln(1+x)}{x^2}{\rm d}x\\ =&-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{2\Li_2^2(-x)}{x}\Bigg{|}^1_0-\int^1_0\frac{4\Li_2(-x)\ln(1+x)}{x^2}{\rm d}x\\ &+\frac{2\Li_3(-x)\ln(1+x)}{x}\Bigg{|}^1_0-\int^1_0\frac{2\Li_3(-x)}{x(1+x)}{\rm d}x-\int^1_0\frac{2\Li_2(-x)\ln(1+x)}{x^2}{\rm d}x\\ =&-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{2}\zeta^2(2)+\frac{7}{4}\zeta(4)-\frac{3}{2}\zeta(3)\ln{2}+\int^1_0\frac{2\Li_3(-x)}{1+x}{\rm d}x\\ &+\frac{6\Li_2(-x)\ln(1+x)}{x}\Bigg{|}^1_0-\int^1_0\frac{6\Li_2(-x)}{x(1+x)}{\rm d}x+\int^1_0\frac{6\ln^2(1+x)}{x^2}{\rm d}x\\ =&-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{2}\zeta^2(2)+\frac{7}{4}\zeta(4)-\frac{3}{2}\zeta(3)\ln{2}+\frac{9}{2}\zeta(3)-3\zeta(2)\ln{2}\\ &+\int^1_0\frac{6\Li_2(-x)}{1+x}{\rm d}x+\int^1_0\frac{2\Li_3(-x)}{1+x}{\rm d}x-\frac{6\ln^2(1+x)}{x}\Bigg{|}^1_0+\int^1_0\frac{12\ln(1+x)}{x(1+x)}{\rm d}x\\ =&-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{2}\zeta^2(2)+\frac{7}{4}\zeta(4)-\frac{3}{2}\zeta(3)\ln{2}+\frac{9}{2}\zeta(3)-3\zeta(2)\ln{2}\\ &+6\zeta(2)-12\ln^2{2}+\int^1_0\frac{6\Li_2(-x)}{1+x}{\rm d}x+\int^1_0\frac{2\Li_3(-x)}{1+x}{\rm d}x \end{align} Since \begin{align} \int^1_0\frac{6\Li_2(-x)}{1+x}{\rm d}x =&6\Li_2(-x)\ln(1+x)\Bigg{|}^1_0+\int^1_0\frac{6\ln^2(1+x)}{x}{\rm d}x\\ =&\frac{3}{2}\zeta(3)-3\zeta(2)\ln{2} \end{align} (see Log Integrals I)

and \begin{align} \int^1_0\frac{2\Li_3(-x)}{1+x}{\rm d}x =&2\Li_3(-x)\ln(1+x)\Bigg{|}^1_0-\int^1_0\underbrace{\frac{2\Li_2(-x)\ln(1+x)}{x}{\rm d}x}_{\displaystyle\small{2\Li_2(-x)\ {\rm d}\Li_2(-x)}}\\ =&\frac{1}{4}\zeta^2(2)-\frac{3}{2}\zeta(3)\ln{2} \end{align} The closed form is \begin{align} \int^1_0\frac{\Li_3^2(-x)}{x^2}{\rm d}x =&-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{4}\zeta^2(2)+\frac{7}{4}\zeta(4)-3\zeta(3)\ln{2}\\&+6\zeta(3)-6\zeta(2)\ln{2}+6\zeta(2)-12\ln^2{2}\\ \end{align}

Solution 2:

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?