# Evaluating $\int_0^1 \frac{\ln x\text{Li}_2(1-x)}{1+x}dx$

i came across the integral

$$\displaystyle J= \int_0^1 \frac{\ln x\text{Li}_2(1-x)}{1+x}dx$$

Probably

$$\displaystyle J=4\text{Li}_4\left(\frac{1}{2}\right)-\frac{33}{16}\zeta(4)-\zeta(2)\ln^2 2+\frac{1}{6}\ln^4 2$$

Is it possible to compute it using (generalised) harmonic series?

NB: Probably i can compute it using only integration techniques.

By using $$\text{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)-\text{Li}_2(x)$$ we have

$$\int_0^1\frac{\ln x\text{Li}_2(1-x)}{1+x}dx=\zeta(2)\underbrace{\int_0^1\frac{\ln x}{1+x}dx}_{I_1}-\underbrace{\int_0^1\frac{\ln^2x\ln(1-x)}{1+x}dx}_{I_2}-\underbrace{\int_0^1\frac{\ln x\text{Li}_2(x)}{1+x}dx}_{I_3}$$

where

$$I_1=\int_0^1\frac{\ln x}{1+x}dx=-\eta(2)=-\frac12\zeta(2)$$

$$I_2=\int_0^1\frac{\ln^2x\ln(1-x)}{1+x}dx=\sum_{n=1}^\infty (-1)^{n-1}\int_0^1 x^{n-1}\ln^2x \ln(1-x)dx$$

$$=\sum_{n=1}^\infty (-1)^{n-1}\left(2\frac{\zeta(3)-H_n^{(3)}}{n}+2\frac{\zeta(2)-H_n^{(2)}}{n^2}-\frac{2H_n}{n^3}\right)$$

$$I_3=\int_0^1\frac{\ln x\text{Li}_2(x)}{1+x}dx=\sum_{n=1}^\infty (-1)^{n-1}\int_0^1 x^{n-1}\ln x\text{Li}_2(x)dx$$

$$=\sum_{n=1}^\infty (-1)^{n-1}\left(\frac{H_n^{(2)}}{n^2}+\frac{2H_n}{n^3}-\frac{2\zeta(2)}{n^2}\right)$$

Collecting all integrals we have

$$\int_0^1\frac{\ln x\text{Li}_2(1-x)}{1+x}dx=-\frac54\zeta(4)-2\zeta(3)\ln2-2\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n}-\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^2}$$

and I am sure you are well familiar with these two sums.