Computing $\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx$

Question

Any idea how ot approach

$$I=\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx\ ?$$

I came across this integral while I was trying to find a different solution for $\Re\ \text{Li}_4(1+i)$ posted here.

here is how I came across it;

using the identity

$$\int_0^1\frac{\ln(x)\text{Li}_2(x)}{1-ax}dx=\frac{\text{Li}_2^2(a)}{2a}+3\frac{\text{Li}_4(a)}{a}-2\zeta(2)\frac{\text{Li}_2(a)}{a}$$

multiply both sides by $\frac{a}{3}$ then replace $a$ by $1+i$ and consider the the real parts of both sides we have

$$\Re\ \text{Li}_4(1+i)=-\frac16\Re\ \text{Li}_2^2(1+i)+\frac23\zeta(2)\Re\ \text{Li}_2(1+i)+\frac13\Re \int_0^1\frac{(1+i)}{1-(1+i)x}\ln(x)\text{Li}_2(x)dx$$

For the integral, use $\Re\frac{1+i}{1-(1+i)x}=\frac{1-2x}{2x^2-2x+1}$ which gives $I$.

What I tried is subbing $1-2x=y$ which gives

$$I=\int_{-1}^1\frac{-y}{1+y^2}\ln\left(\frac{1+y}{2}\right)\text{Li}_2\left(\frac{1+y}{2}\right)dy=\int_{-1}^1 f(y)dy=\underbrace{\int_{-1}^0 f(y)dy}_{y\to\ -y}+\int_{0}^1 f(y)dy$$

$$=\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1-y}{2}\right)\text{Li}_2\left(\frac{1-y}{2}\right)dy-\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1+y}{2}\right)\text{Li}_2\left(\frac{1+y}{2}\right)dy$$

I think I made it more complicated. Any help would be appriciated.

Answer 1

$$I=\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1-y^2}{4}\right)\text{Li}_2\left(\frac{1-y}{2}\right)dy+\frac{7}{32} \zeta (3) \log (2)+\frac{ \text{Li}_4\left(\frac{1}{2}\right)}{8}-\frac{157 \pi ^4}{46080}-\frac{ 11\log ^4(2)}{48}+\frac{19}{384} \pi ^2 \log ^2(2)$$ Put $$y=\frac{1-x}{1+x}$$ $$\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1-y^2}{4}\right)\text{Li}_2\left(\frac{1-y}{2}\right)dy=\int_0^1\frac{\ln(x)}{1+x}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx-2\int_0^1\frac{\ln(1+x)}{1+x}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx-\int_0^1\frac{x\ln(x)}{1+x^2}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx+2\int_0^1\frac{x\ln(1+x)}{1+x^2}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx$$ be continued $$\int_0^1\frac{\ln(1+x)}{1+x}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=3\operatorname{Li}_4\left(\frac12\right)-\frac{\pi^4}{30}+\frac{21}8\ln2\zeta(3)-\frac{\pi^2}{12}\ln^22$$ $$\int_0^1\frac{x\ln(x)}{1+x^2}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=\frac{C^2}{2}+\frac{15 \text{Li}_4\left(\frac{1}{2}\right)}{16}-\frac{701 \pi ^4}{46080}+\frac{7}8\ln2\zeta(3)+\frac{5 \log ^4(2)}{128}-\frac{3}{64} \pi ^2 \log ^2(2)$$

Answer 2

$$\int_0^1\frac{\ln(1+x)}{1+x}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=3\operatorname{Li}_4\left(\frac12\right)-\frac{\pi^4}{30}+\frac{21}8\ln2\zeta(3)-\frac{\pi^2}{12}\ln^22$$ $$\int_0^1\frac{x\ln(x)}{1+x^2}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=\frac{C^2}{2}+\frac{15 \text{Li}_4\left(\frac{1}{2}\right)}{16}-\frac{701 \pi ^4}{46080}+\frac{7}8\ln2\zeta(3)+\frac{5 \log ^4(2)}{128}-\frac{3}{64} \pi ^2 \log ^2(2)$$

Answer 3

$$\int_0^1\frac{\ln(x)}{1+x}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=3\text{Li}_4(2)+\text{Li}_2(2)\log^22-3\text{Li}_3(2)\log2+6\operatorname{Li}_4\left(\frac12\right)+\frac{21}4\ln2\zeta(3)-\frac{\pi^2}{8}\log^22+\frac{1}{4}\log^42-\frac{29\pi^4}{288}$$ I being known,I deduce $$\int_0^1\frac{x\ln(1+x)}{1+x^2}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=-\frac{1}{16}\operatorname{Li}_4\left(\frac12\right)+\frac{21}{64}\ln2\zeta(3)-\frac{41}{768} \pi ^2 \log ^2(2)-\frac{1}{96}\log^42+\frac{1609\pi^4}{92160}-\frac{3}{2}\text{Li}_4(2)-\frac{1}{2}\text{Li}_2(2)\log^22+\frac{3}{2}\text{Li}_3(2)\log2$$ Sorry,i couldn't deduct this integral. $$3\text{Li}_4(2)+\text{Li}_2(2)\log^22-3\text{Li}_3(2)\log2=-3\operatorname{Li}_4\left(\frac12\right)-\frac{21}8\ln2\zeta(3)-\frac{1}{8}\log^42+\frac{\pi^4}{15}$$

Answer 4

$$=\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1-y}{2}\right)\text{Li}_2\left(\frac{1-y}{2}\right)dy=2\operatorname{Li}_4\left(\frac12\right)+\frac{133}{64}\ln2\zeta(3)-\frac{37}{768} \pi ^2 \log ^2(2)+\frac{77}{384}\log^42-\frac{3197\pi^4}{92160}-\frac{C^2}{2}-\frac{1}{8} \pi C \log (2)+\frac{3}{2}\text{Li}_4(2)+\frac{1}{2}\text{Li}_2(2)\log^22-\frac{3}{2}\text{Li}_3(2)\log2$$ $$\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1+y}{2}\right)\text{Li}_2\left(\frac{1+y}{2}\right)dy=\frac{47}{16}\operatorname{Li}_4\left(\frac12\right)+\frac{133}{64}\ln2\zeta(3)-\frac{61}{768} \pi ^2 \log ^2(2)+\frac{23}{96}\log^42-\frac{4367\pi^4}{92160}+\frac{1}{8} \pi C \log (2)+\frac{3}{2}\text{Li}_4(2)+\frac{1}{2}\text{Li}_2(2)\log^22-\frac{3}{2}\text{Li}_3(2)\log2$$