Closed-form of $\int_{0}^{\infty} \frac{{\text{Li}}_2^3(-x)}{x^3}\,dx$

Solution 1:

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}$$