Is this closed-form of $\int_0^1 \operatorname{Li}_3^2(x)\,dx$ correct?

Question

According to Freitas' paper at page $11$.

$$\int_0^1 \operatorname{Li}_3^2(x)\,dx = 20-8\zeta(2)-10\zeta(3)-\frac{15}{2}\zeta(4)-2\zeta(2)\zeta(3)+\zeta^2(3).$$

I evaluated the LHS and it is $0.427714784290824$ to me, but the RHS is $-15.8071337213762487846272$. Where am I wrong? Is this closed-form correct? If not, what is the correct closed-form?

Edit. The correct closed-form is $$\int_0^1 \operatorname{Li}_3^2(x)\,dx = 20-8\zeta(2)-10\zeta(3)+\frac{15}{2}\zeta(4)-2\zeta(2)\zeta(3)+\zeta^2(3).$$

Answer 1

The trilogarithm has the antiderivative,

$$\int\operatorname{Li}_{3}{\left(x\right)}\,\mathrm{d}x=x\operatorname{Li}_{3}{\left(x\right)}-x\operatorname{Li}_{2}{\left(x\right)}+x-x\ln{\left(1-x\right)}+\ln{\left(1-x\right)}+\color{grey}{constant}.$$

So, integrating by parts we find after integrating all the terms with simple antiderivatives:

$$\begin{align} \int_{0}^{1}\operatorname{Li}_{3}^2{\left(x\right)}\,\mathrm{d}x &=\left[\left(x\operatorname{Li}_{3}{\left(x\right)}-x\operatorname{Li}_{2}{\left(x\right)}+x-x\ln{\left(1-x\right)}+\ln{\left(1-x\right)}\right)\operatorname{Li}_{3}{\left(x\right)}\right]_{0}^{1}\\ &~~~~~ -\int_{0}^{1}\left(x\operatorname{Li}_{3}{\left(x\right)}-x\operatorname{Li}_{2}{\left(x\right)}+x-x\ln{\left(1-x\right)}+\ln{\left(1-x\right)}\right)\frac{\operatorname{Li}_{2}{\left(x\right)}}{x}\,\mathrm{d}x\\ &=\zeta{(3)}-\zeta{(2)}\zeta{(3)}+\zeta{(3)}^2-\int_{0}^{1}\operatorname{Li}_{3}{\left(x\right)}\operatorname{Li}_{2}{\left(x\right)}\,\mathrm{d}x+\int_{0}^{1}\operatorname{Li}_{2}^2{\left(x\right)}\,\mathrm{d}x\\ &~~~~ -\int_{0}^{1}\operatorname{Li}_{2}{\left(x\right)}\,\mathrm{d}x+\int_{0}^{1}\ln{\left(1-x\right)}\operatorname{Li}_{2}{\left(x\right)}\,\mathrm{d}x-\int_{0}^{1}\frac{\ln{\left(1-x\right)}\operatorname{Li}_{2}{\left(x\right)}}{x}\,\mathrm{d}x\\ &=4-2\zeta{(2)}-\zeta{(3)}+\frac54\zeta{(4)}-\zeta{(2)}\zeta{(3)}+\zeta{(3)}^2\\ &~~~~ +\int_{0}^{1}\operatorname{Li}_{2}^2{\left(x\right)}\,\mathrm{d}x-\int_{0}^{1}\operatorname{Li}_{3}{\left(x\right)}\operatorname{Li}_{2}{\left(x\right)}\,\mathrm{d}x.\\ \end{align}$$

For the remaining two integrals, you may substitute the values reported in the paper you cited.

Answer 2

There is a wrong sign, the correct closed form is

$$20-8\,\zeta \left( 2 \right) -10\,\zeta \left( 3 \right) +{\frac {15}{2}}\,\zeta \left( 4 \right) -2\,\zeta \left( 2 \right) \zeta \left( 3 \right) + \left( \zeta \left( 3 \right) \right) ^{2} $$

Answer 3

$$I=\int_{0}^{1}(Li_{3}(x))^{2}dx=x(Li_{3}(x))^2\ _{0}^{1}\ -\int_{0}^{1}x.2Li_{3}(x).\frac{Li_{2}(x)}{x}dx\ \ \ \ \ \ (1)\\ \\ \\ \therefore I=(Li_{3}(1))^2-2\int_{0}^{1}Li_{2}(x).Li_{3}(x)dx\ \ \ \ \ \ \ \ \ \ \ \\ \\ =(Li_{3}(1))^2-2(Li_{3}(x)(1-x)-(1-x)ln(1-x)+xLi_{2}(x)_{0}^{1})\\ \\ -\int_{0}^{1}[(1-x)-(1-x)ln(1-x)+xLi_{2}(x).\frac{Li_{2}(x)}{x}]dx\\ \\ \\$$

thus we have

$$\therefore I=(Li_{3}(1))^2-2Li_{2}(1).Li_{3}(1)+2\int_{0}^{1}Li_{2}(x)(Li_{2}(x)+\frac{1}{x}-1\frac{ln(1-x)}{x}+ln(1-x))dx\\ \\ =(Li_{3}(1))^2-2Li_{2}(1).Li_{3}(1)+2\int_{0}^{1}[Li_{2}(x)]^2dx+2\int_{0}^{1}\frac{Li_{2}(x)}{x}dx\\ \\ -2\int_{0}^{1}Li_{2}(x)dx-2\int_{0}^{1}\frac{ln(1-x)}{x}.Li_{2}(x)dx+2\int_{0}^{1}ln(1-x).Li_{2}(x)\\ \\ \\ \therefore I=\zeta ^2(3)-2\zeta (2)\zeta (3)+2[6-2\zeta (2)+\zeta ^2(2)-4\zeta (3)]+[Li_{2}(x)]^2\\ \\ +2\zeta (3)-2[xLi_{2}(x)_{1}^{0}+\int_{0}^{1}\frac{xln(1-x)}{x}dx]+2[Li_{2}(x)((1-x))-(1-x)ln(1-x)]_{0}^{1}\\ \\ +\int_{0}^{1}[(1-x)-(1-x)ln(1-x).\frac{ln(1-x)}{x}]dx$$

thus

$$\therefore I=\zeta ^2(3)-2\zeta (2)\zeta (3)+12-4\zeta (2)+2\zeta ^2(2)-8\zeta (3)+\zeta ^2(2)+2\zeta (3)-2\zeta (2)\\ \\ -2\int_{0}^{1}ln(1-x)dx+2\int_{0}^{1}\frac{ln(1-x)}{x}dx-2\int_{0}^{1}ln(1-x)dx-2\int_{0}^{1}\frac{ln^2(1-x)}{x}dx+2\int_{0}^{1}ln^2(1-x)dx\\ \\ but\ \ \int_{0}^{1}ln(1-x)dx=-1\ \ \ \ \ , and\ \int_{0}^{1}ln^2(1-x)dx=2\ \ \ ,\\ \\\ \ \ \ \ , \ \ \ \ \ \ and\ \ \ \ \ \ \int_{0}^{1}\frac{ln(1-x)}{x}dx=-\zeta (2)\ \ \ \ ,and\ \int_{0}^{1}\frac{ln^2(1-x)}{x}dx=2\zeta (3)\\ \\$$

so we have

$$\therefore I=\zeta ^2(3)-2\zeta (2)\zeta (3)+12-4\zeta (2)+2\zeta ^2(2)-8\zeta (3)+\zeta ^2(2)+2\zeta (3)-2\zeta (2)+2+2-2\zeta (2)+4-4\zeta (3)\\ \\ \\ \\ \therefore I=20+\frac{\pi ^{4}}{12}+\zeta (3)[\zeta (3)-10]-\frac{\pi ^{2}}{3}[4+\zeta (3)]$$