Finding the closed form of $\int_0^1 \frac{(1-x+x\log x)\operatorname{Li}_3(x)}{x(x-1) \log x} \ dx$

Solution 1:

I believe the closed form proposed in Chris's sis' answer is correct. Here's the derivation I used:

$$\begin{align} \mathcal{I} &=\int_{0}^{1}\frac{\left(1-x+x\ln{(x)}\right)\operatorname{Li}_{3}{\left(x\right)}}{(x-1)x\ln{(x)}}\,\mathrm{d}x\\ &=\lim_{z\to1^-}\int_{0}^{z}\left(\frac{\operatorname{Li}_{3}{\left(x\right)}}{x-1}-\frac{\operatorname{Li}_{3}{\left(x\right)}}{x\ln{(x)}}\right)\,\mathrm{d}x\\ &=\lim_{z\to1^-}\left[\int_{0}^{z}\frac{\operatorname{Li}_{3}{\left(x\right)}}{x-1}\,\mathrm{d}x-\int_{0}^{z}\frac{\operatorname{Li}_{3}{\left(x\right)}}{x\ln{(x)}}\,\mathrm{d}x\right]\\ &=\lim_{z\to1^-}\left[\frac12\operatorname{Li}_{2}^2{\left(z\right)}+\operatorname{Li}_{3}{\left(z\right)}\ln{\left(1-z\right)}-\int_{0}^{z}\frac{\operatorname{Li}_{3}{\left(x\right)}}{x\ln{(x)}}\,\mathrm{d}x\right]\\ &=\lim_{z\to1^-}\left[\frac12\operatorname{Li}_{2}^2{\left(z\right)}+\operatorname{Li}_{3}{\left(z\right)}\ln{\left(1-z\right)}-\ln{\left(-\ln{\left(z\right)}\right)}\operatorname{Li}_{3}{\left(z\right)}+\int_{0}^{z}\frac{\ln{\left(-\ln{\left(x\right)}\right)}\operatorname{Li}_{2}{\left(x\right)}}{x}\,\mathrm{d}x\right]\\ &=\frac{\pi^4}{72}+\int_{0}^{1}\frac{\ln{\left(-\ln{\left(x\right)}\right)}\operatorname{Li}_{2}{\left(x\right)}}{x}\,\mathrm{d}x\\ &=\frac{\pi^4}{72}-\int_{0}^{1}\frac{\ln{\left(x\right)}\ln{\left(1-x\right)}}{x}\,\mathrm{d}x+\int_{0}^{1}\frac{\ln{\left(x\right)}\ln{\left(1-x\right)}\ln{\left(-\ln{\left(x\right)}\right)}}{x}\,\mathrm{d}x\\ &=\frac{\pi^4}{72}-\zeta{(3)}+\int_{0}^{1}\frac{\ln{\left(x\right)}\ln{\left(1-x\right)}\ln{\left(-\ln{\left(x\right)}\right)}}{x}\,\mathrm{d}x\\ &=\frac{\pi^4}{72}-\zeta{(3)}-\frac14\int_{0}^{1}\frac{\ln^2{\left(x\right)}}{1-x}\,\mathrm{d}x+\frac12\int_{0}^{1}\frac{\ln^2{\left(x\right)}\ln{\left(-\ln{\left(x\right)}\right)}}{1-x}\,\mathrm{d}x\\ &=\frac{\pi^4}{72}-\frac32\zeta{(3)}+\frac12\int_{0}^{1}\frac{\ln^2{\left(x\right)}\ln{\left(\ln{\left(\frac{1}{x}\right)}\right)}}{1-x}\,\mathrm{d}x\\ &=\frac{\pi^4}{72}-\frac32\zeta{(3)}+\frac12\left[\int_{0}^{1}\frac{x^{p-1}\ln^2{\left(x\right)}\ln{\left(\ln{\left(\frac{1}{x}\right)}\right)}}{1-x}\,\mathrm{d}x\right]_{p=1}\\ &=\frac{\pi^4}{72}-\frac32\zeta{(3)}+\frac12\left[\int_{0}^{1}\sum_{k=0}^{\infty}x^{k+p-1}\ln^2{\left(x\right)}\ln{\left(\ln{\left(\frac{1}{x}\right)}\right)}\,\mathrm{d}x\right]_{p=1}\\ &=\frac{\pi^4}{72}-\frac32\zeta{(3)}+\frac12\left[\sum_{k=0}^{\infty}\int_{0}^{1}x^{k+p-1}\ln^2{\left(x\right)}\ln{\left(\ln{\left(\frac{1}{x}\right)}\right)}\,\mathrm{d}x\right]_{p=1}\\ &=\frac{\pi^4}{72}-\frac32\zeta{(3)}+\frac12\left[\sum_{k=0}^{\infty}\frac{\partial^2}{\partial p^2}\int_{0}^{1}x^{k+p-1}\ln{\left(\ln{\left(\frac{1}{x}\right)}\right)}\,\mathrm{d}x\right]_{p=1}\\ &=\frac{\pi^4}{72}-\frac32\zeta{(3)}+\frac12\left[\sum_{k=0}^{\infty}\frac{\partial^2}{\partial p^2}\left(-\frac{\gamma+\ln{\left(k+p\right)}}{k+p}\right)\right]_{p=1}\\ &=\frac{\pi^4}{72}-\frac32\zeta{(3)}+\frac12\left[\sum_{k=0}^{\infty}\left(\frac{3-2\gamma}{(k+p)^3}-\frac{2\ln{\left(k+p\right)}}{(k+p)^3}\right)\right]_{p=1}\\ &=\frac{\pi^4}{72}-\frac32\zeta{(3)}+\frac12\sum_{k=0}^{\infty}\left(\frac{3-2\gamma}{(k+1)^3}-\frac{2\ln{\left(k+1\right)}}{(k+1)^3}\right)\\ &=\frac{\pi^4}{72}-\frac32\zeta{(3)}+\frac12\sum_{k=1}^{\infty}\left(\frac{3-2\gamma}{k^3}-\frac{2\ln{\left(k\right)}}{k^3}\right)\\ &=\frac{\pi^4}{72}-\gamma\zeta{(3)}-\sum_{k=1}^{\infty}\left(\frac{\ln{\left(k\right)}}{k^3}\right)\\ &=\frac{\pi^4}{72}-\gamma\zeta{(3)}+\zeta^{\prime}{(3)}.\\ \end{align}$$

Solution 2:

$\def\Li{{\rm{Li}}}$Thanks to @David H's comment I got that

$$\int_0^1 \frac{(1-x+x\log x)\Li_3(x)}{x(x-1) \log x} \ dx=\frac{5}{4}\zeta(4)-\gamma \zeta(3)+\zeta'(3)$$ that is proving to be numerically correct. Here I have a question I just received, and still trying to find a proper starting point

An alternative way

$$\int_{0}^{1}\sum_{n=1}^{\infty}\left(\frac{x^n}{n^3(x-1)}-\frac{x^{n-1}}{n^3\ln{(x)}}\right) \ dx $$ $$=-\sum_{n=1}^{\infty}\frac{1}{n^3}\int_{0}^{1}\left(\frac{x^{n-1}}{\ln{(x)}}+\frac{x^n}{1-x}\right) \ dx \tag{*}$$ $$=-\sum_{n=1}^{\infty}\frac{\log(n)-\psi(n+1)}{n^3}$$ $$=\frac{5}{4}\zeta(4)-\gamma \zeta(3)+\zeta'(3)$$ where in $(*)$ I use the Proposition $2.16.$ from this paper.

Q.E.D.