TITLE: Closed-form of $\int_{0}^{\infty} \frac{{\text{Li}}_2^3(-x)}{x^3}\,dx$
QUESTION [14 upvotes]: Is there a possibility to find a closed-form for $$\int_{0}^{\infty} \frac{{\text{Li}}_2^3(-x)}{x^3}\,dx$$
 We have
$$I=\int_0^1\frac{Li_2^3(-x)+x^4Li_2^3(-\frac{1}{x})}{x^3}\,dx$$
 After repeatedly integrate by parts, I found
 $$I=\frac{51}{8}\zeta(4)+\frac{15}{32}\zeta(4)\ln(2)+\frac{3}{8}\zeta(2)ln(2)+\frac{39}{8}\zeta(2)\zeta(3)+\frac{3}{2}\zeta(5)-\frac{15}{4}\zeta(3)-\frac{33}{4}\zeta(2)$$
 Could anyome here help me to ascertain this result?Thanks

REPLY [13 votes]: As tired indicates, the proposed closed form isn't correct. The actual value is actually a little more compact and can be found fairly directly without resorting to series methods. I find:
$$\begin{align}
\mathcal{I}
&=\int_{0}^{\infty}\frac{\operatorname{Li}_{2}{\left(-x\right)}^{3}}{x^3}\,\mathrm{d}x\\
&=\int_{1}^{\infty}\frac{\operatorname{Li}_{2}{\left(1-y\right)}^{3}}{\left(y-1\right)^3}\,\mathrm{d}y;~~~\small{\left[1+x=y\right]}\\
&=\int_{0}^{1}\frac{\operatorname{Li}_{2}{\left(1-\frac{1}{z}\right)}^{3}}{z^2\left(\frac{1}{z}-1\right)^3}\,\mathrm{d}z;~~~\small{\left[\frac{1}{y}=z\right]}\\
&=\int_{0}^{1}\frac{z\operatorname{Li}_{2}{\left(-\frac{1-z}{z}\right)}^{3}}{\left(1-z\right)^3}\,\mathrm{d}z\\
&=\int_{0}^{1}\frac{\left(1-w\right)\operatorname{Li}_{2}{\left(\frac{w}{w-1}\right)}^{3}}{w^3}\,\mathrm{d}w;~~~\small{\left[1-z=w\right]}\\
&=\frac32\int_{0}^{1}\frac{\left(1-w\right)\ln{\left(1-w\right)}\operatorname{Li}_{2}{\left(\frac{w}{w-1}\right)}^{2}}{w^3}\,\mathrm{d}w;~~~I.B.P.\\
&=\small{\frac32\int_{0}^{1}\frac{\left[\left(\frac{1-w}{w}\right)^2\ln{\left(1-w\right)}-\ln{\left(w\right)}-\left(\frac{1-w}{w}\right)\right]\ln{\left(1-w\right)}\operatorname{Li}_{2}{\left(\frac{w}{w-1}\right)}}{w(1-w)}\,\mathrm{d}w};~I.B.P.\\
&=\frac32\int_{0}^{1}\frac{\left(1-w\right)\ln^{2}{\left(1-w\right)}\operatorname{Li}_{2}{\left(\frac{w}{w-1}\right)}}{w^3}\,\mathrm{d}w\\
&~~~~~-\frac32\int_{0}^{1}\frac{\left[\ln{\left(w\right)}+\left(\frac{1-w}{w}\right)\right]\ln{\left(1-w\right)}\operatorname{Li}_{2}{\left(\frac{w}{w-1}\right)}}{w(1-w)}\,\mathrm{d}w\\
&=-\frac32\int_{0}^{1}\frac{\left(1-w\right)\ln^{2}{\left(1-w\right)}\left[\operatorname{Li}_{2}{\left(w\right)}+\frac12\ln^{2}{\left(1-w\right)}\right]}{w^3}\,\mathrm{d}w\\
&~~~~~\small{+\frac32\int_{0}^{1}\frac{\left[\ln{\left(w\right)}+\left(\frac{1-w}{w}\right)\right]\ln{\left(1-w\right)}\left[\operatorname{Li}_{2}{\left(w\right)}+\frac12\ln^{2}{\left(1-w\right)}\right]}{w(1-w)}\,\mathrm{d}w}\\
&=-\frac32\int_{0}^{1}\frac{\left(1-w\right)\ln^{2}{\left(1-w\right)}\operatorname{Li}_{2}{\left(w\right)}}{w^3}\,\mathrm{d}w\\
&~~~~~-\frac34\int_{0}^{1}\frac{\left(1-w\right)\ln^{4}{\left(1-w\right)}}{w^3}\,\mathrm{d}w\\
&~~~~~+\frac32\int_{0}^{1}\frac{\left[\ln{\left(w\right)}+\left(\frac{1-w}{w}\right)\right]\ln{\left(1-w\right)}\operatorname{Li}_{2}{\left(w\right)}}{w(1-w)}\,\mathrm{d}w\\
&~~~~~+\frac34\int_{0}^{1}\frac{\left[\ln{\left(w\right)}+\left(\frac{1-w}{w}\right)\right]\ln^{3}{\left(1-w\right)}}{w(1-w)}\,\mathrm{d}w\\
&=\small{\frac32\int_{0}^{1}\frac{\left(1-w\right)\ln{\left(1-w\right)}\operatorname{Li}_{2}{\left(w\right)}}{w^{2}}\,\mathrm{d}w+\frac32\int_{0}^{1}\frac{\left(\frac{1-w}{w}\right)^{2}\ln^{3}{\left(1-w\right)}}{2w}\,\mathrm{d}w}\\
&~~~~~+\frac32\int_{0}^{1}\frac{\left(1-w\right)\ln^{3}{\left(1-w\right)}}{w^{2}}\,\mathrm{d}w\\
&~~~~~\small{+\frac32\int_{0}^{1}\frac{\ln{\left(w\right)}\ln{\left(1-w\right)}\operatorname{Li}_{2}{\left(w\right)}}{w(1-w)}\,\mathrm{d}w+\frac32\int_{0}^{1}\frac{\ln{\left(1-w\right)}\operatorname{Li}_{2}{\left(w\right)}}{w^2}\,\mathrm{d}w}\\
&~~~~~+\frac34\int_{0}^{1}\frac{\ln{\left(w\right)}\ln^{3}{\left(1-w\right)}}{w(1-w)}\,\mathrm{d}w+\frac34\int_{0}^{1}\frac{\ln^{3}{\left(1-w\right)}}{w^2}\,\mathrm{d}w\\
&=\small{\frac32\int_{0}^{1}\frac{\left(2-w\right)\ln{\left(1-w\right)}\operatorname{Li}_{2}{\left(w\right)}}{w^{2}}\,\mathrm{d}w+\frac34\int_{0}^{1}\frac{\left(1+w-w^2\right)\ln^{3}{\left(1-w\right)}}{w^{3}}\,\mathrm{d}w}\\
&~~~~~\small{+\frac32\int_{0}^{1}\frac{\ln{\left(w\right)}\ln{\left(1-w\right)}\operatorname{Li}_{2}{\left(w\right)}}{w}\,\mathrm{d}w+\frac32\int_{0}^{1}\frac{\ln{\left(w\right)}\ln{\left(1-w\right)}\operatorname{Li}_{2}{\left(w\right)}}{1-w}\,\mathrm{d}w}\\
&~~~~~\small{+\frac34\int_{0}^{1}\frac{\ln{\left(w\right)}\ln^{3}{\left(1-w\right)}}{w}\,\mathrm{d}w+\frac34\int_{0}^{1}\frac{\ln{\left(w\right)}\ln^{3}{\left(1-w\right)}}{1-w}\,\mathrm{d}w}\\
&=\small{\frac32\int_{0}^{1}\frac{\left(2-w\right)\ln{\left(1-w\right)}\operatorname{Li}_{2}{\left(w\right)}}{w^{2}}\,\mathrm{d}w+\frac34\int_{0}^{1}\frac{\left(1+w-w^2\right)\ln^{3}{\left(1-w\right)}}{w^{3}}\,\mathrm{d}w}\\
&~~~~~\small{+\frac32\int_{0}^{1}\frac{\ln{\left(w\right)}\ln{\left(1-w\right)}\operatorname{Li}_{2}{\left(w\right)}}{w}\,\mathrm{d}w+\frac32\int_{0}^{1}\frac{\ln{\left(v\right)}\ln{\left(1-v\right)}\operatorname{Li}_{2}{\left(1-v\right)}}{v}\,\mathrm{d}v}\\
&~~~~~\small{+\frac34\int_{0}^{1}\frac{\ln{\left(w\right)}\ln^{3}{\left(1-w\right)}}{w}\,\mathrm{d}w+\frac34\int_{0}^{1}\frac{\ln^{3}{\left(v\right)}\ln{\left(1-v\right)}}{v}\,\mathrm{d}v};~~~\small{\left[1-w=v\right]}\\
\end{align}$$
Continuing,
$$\begin{align}
\mathcal{I}
&=\small{3\int_{0}^{1}\frac{\ln{\left(1-w\right)}\operatorname{Li}_{2}{\left(w\right)}}{w^{2}}\,\mathrm{d}w+\frac34\int_{0}^{1}\frac{\left(1+w\right)\ln^{3}{\left(1-w\right)}}{w^{3}}\,\mathrm{d}w}\\
&~~~~~\small{-\frac32\int_{0}^{1}\frac{\ln{\left(1-w\right)}\operatorname{Li}_{2}{\left(w\right)}}{w}\,\mathrm{d}w-\frac34\int_{0}^{1}\frac{\ln^{3}{\left(1-w\right)}}{w}\,\mathrm{d}w}\\
&~~~~~+\frac32\int_{0}^{1}\frac{\ln{\left(w\right)}\ln{\left(1-w\right)}\left[\operatorname{Li}_{2}{\left(w\right)}+\operatorname{Li}_{2}{\left(1-w\right)}\right]}{w}\,\mathrm{d}w\\
&~~~~~+\frac92\,S_{2,3}{\left(1\right)}+\frac92\,S_{4,1}{\left(1\right)}\\
&=-\int_{0}^{1}\left[-\frac{1-w}{w}\ln{\left(1-w\right)}-\ln{\left(w\right)}\right]\frac{(-3)\ln{\left(1-w\right)}}{w}\,\mathrm{d}w\\
&~~~~~-\frac34\int_{0}^{1}\left(\frac32-\frac{2w+1}{2w^2}\right)\frac{(-3)\ln^{2}{\left(1-w\right)}}{1-w}\,\mathrm{d}w\\
&~~~~~+\frac34\,\operatorname{Li}_{2}{\left(1\right)}^{2}+\frac92\,S_{1,3}{\left(1\right)}\\
&~~~~~+\frac32\int_{0}^{1}\frac{\ln{\left(w\right)}\ln{\left(1-w\right)}\left[\zeta{(2)}-\ln{\left(w\right)}\ln{\left(1-w\right)}\right]}{w}\,\mathrm{d}w\\
&~~~~~+\frac92\,S_{2,3}{\left(1\right)}+\frac92\,S_{4,1}{\left(1\right)}\\
&=\small{-3\int_{0}^{1}\frac{\left(1-w\right)\ln^{2}{\left(1-w\right)}}{w^{2}}\,\mathrm{d}w-\frac94\int_{0}^{1}\frac{\left(1+3w\right)\ln^{2}{\left(1-w\right)}}{2w^2}\,\mathrm{d}w}\\
&~~~~~\small{+\frac34\,\zeta{(2)}^{2}+\frac92\,S_{1,3}{\left(1\right)}+\left(\frac32\zeta{(2)}-3\right)\int_{0}^{1}\frac{\ln{\left(w\right)}\ln{\left(1-w\right)}}{w}\,\mathrm{d}w}\\
&~~~~~\small{-\frac32\int_{0}^{1}\frac{\ln^{2}{\left(w\right)}\ln^{2}{\left(1-w\right)}}{w}\,\mathrm{d}w+\frac92\,S_{2,3}{\left(1\right)}+\frac92\,S_{4,1}{\left(1\right)}}\\
&=-\frac{33}{4}\int_{0}^{1}\frac{\ln^{2}{\left(1-w\right)}}{2w^{2}}\,\mathrm{d}w-\frac34\,S_{1,2}{\left(1\right)}\\
&~~~~~+\frac34\,\zeta{(2)}^{2}+\frac92\,S_{1,3}{\left(1\right)}+\left(\frac32\zeta{(2)}-3\right)S_{2,1}{\left(1\right)}\\
&~~~~~-6\,S_{3,2}{\left(1\right)}+\frac92\,S_{2,3}{\left(1\right)}+\frac92\,S_{4,1}{\left(1\right)}\\
&=\frac32\,\zeta{(5)}+\left(\frac{\pi^2}{2}-\frac{15}{4}\right)\zeta{(3)}+\frac{17\pi^{4}}{240}-\frac{11\pi^2}{8}.\\
\end{align}$$