TITLE: How to evaluate: $\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x) \, \mathrm dx$
QUESTION [6 upvotes]: $$\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x)\,\mathrm dx=1.03693\ldots$$
This number looks like $\zeta(5)$ value. 
We expand the terms
$$\int_0^1\frac{\frac{\pi^2}{6}}{1-x}\cdot \ln^2(x) \, \mathrm dx-\int_0^1 \frac{\operatorname{Li}_2(1-x)}{1-x} \cdot \ln^2(x) \, \mathrm dx$$
$$2\zeta(2)\zeta(3)-\int_0^1 \frac{\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x) \, \mathrm dx$$
this last integral it is very complicate to compute...
Can any user please help to show that whether this value is equal to $\zeta(5)$ or not?

REPLY [2 votes]: A solution that is not using harmonic series.
\begin{align}J&=\int_0^1 \frac{\Big(\operatorname{Li}_2(1)-\operatorname{Li}_2(1-x)\Big)\ln^2(x)}{1-x} \, \mathrm dx\\
&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)\Big(\operatorname{Li}_2(1)-\operatorname{Li}_2(1-x)\Big)\right]_0^1+\int_0^1 \left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)\frac{\ln x}{1-x}dx\\
&=2\zeta(3)\zeta(2)+\int_0^1 \left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)\frac{\ln x}{1-x}dx\\
&\overset{\text{IBP}}=-\int_0^1 \left(\int_0^x \frac{\ln t}{1-t}dt\right)\frac{\ln^2 x}{1-x}dx\\
&=-\int_0^1 \int_0^1 \frac{x\ln(tx)\ln^2 x}{(1-tx)(1-x)}\,dtdx\\
&=\int_0^1 \int_0^1\left(\frac{\ln(tx)\ln^2 x}{(1-t)(1-tx)}-\frac{\ln(tx)\ln^2 x}{(1-t)(1-x)}\right)dt dx\\
&=2\zeta(2)\zeta(3)+\int_0^1 \int_0^1\left(\frac{\ln^3(tx)-2\ln t\ln^2(tx)+\ln^2 t\ln(tx)}{(1-t)(1-tx)}-\frac{\ln^3 x}{(1-t)(1-x)}\right)dt dx\\
&=2\zeta(2)\zeta(3)-\int_0^1 \frac{1}{1-t}\left(\int_t^1 \frac{\ln^3 u}{1-u}du\right)dt+\int_0^1\frac{1}{t}\left(\int_0^t \frac{\ln^3 u}{1-u}du\right)dt\\&-\int_0^1 \frac{2\ln t}{t}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)dt -2\underbrace{\int_0^1 \frac{\ln t}{1-t}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)dt}_{J-2\zeta(2)\zeta(3)}+\\&\int_0^1\frac{\ln^2 t}{t}\left(\int_0^t \frac{\ln u}{1-u}du\right)dt+\underbrace{\int_0^1\frac{\ln^2 t}{1-t}\left(\int_0^t \frac{\ln u}{1-u}du\right)dt}_{-J}\\
&=2\zeta(2)\zeta(3)-\int_0^1 \frac{1}{1-t}\left(\int_t^1 \frac{\ln^3 u}{1-u}du\right)dt+\int_0^1\frac{1}{t}\left(\int_0^t \frac{\ln^3 u}{1-u}du\right)dt\\&-\int_0^1 \frac{2\ln t}{t}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)dt -2\Big(J-2\zeta(2)\zeta(3)\Big)+\int_0^1\frac{\ln^2 t}{t}\left(\int_0^t \frac{\ln u}{1-u}du\right)dt-J\\
&\overset{\text{IBP}}=6\zeta(2)\zeta(3)-3J+\int_0^1 \frac{\ln(1-t)\ln^3 t}{1-t}dt-\frac{1}{3}\int_0^1 \frac{\ln^4 t}{1-t}dt\\
&=6\zeta(2)\zeta(3)-3J+\int_0^1 \frac{\ln(1-t)\ln^3 t}{1-t}dt-8\zeta(5)
\end{align}
On the other hand,
\begin{align}K&=\int_0^1 \frac{\ln(1-t)\ln^3 t}{1-t}dt,C=\int_0^1 \frac{\ln^3 t}{1-t}dt\\
&\overset{\text{IBP}}=\left[\left(\int_0^t\frac{\ln^3 u}{1-u}du-C\right)\ln(1-t)\right]_0^1+\int_0^1 \frac{1}{1-t}\left(\int_0^t\frac{\ln^3 u}{1-u}du-C\right)dt\\
&=\int_0^1\int_0^1\left(\frac{t\ln^3 (tu)}{(1-tu)(1-t)}-\frac{C}{1-t}\right)dt du\\
&=\int_0^1\int_0^1\left(\frac{\ln^3 (tu)}{(1-t)(1-u)}-\frac{\ln^3 (tu)}{(1-u)(1-tu)}-\frac{C}{1-t}\right)dt du\\
&=-12\zeta(2)\zeta(3)+\int_0^1 \frac{1}{1-u}\left(\int_u^1\frac{\ln^3 t}{1-t}dt\right)du-\int_0^1 \frac{1}{u}\left(\int_0^u\frac{\ln^3 t}{1-t}dt\right)du+\\&\int_0^1 \int_0^1 \left(\frac{\ln^3 u}{(1-t)(1-u)}-\frac{C}{1-t}\right)dt du\\
&\overset{\text{IBP}}=-12\zeta(2)\zeta(3)-J+\int_0^1\frac{\ln^4 u}{1-u}du+\underbrace{\int_0^1 \int_0^1 \left(\frac{\ln^3 u}{(1-t)(1-u)}-\frac{C}{1-t}\right)dt du}_{L}\\
&0\leq A<1\\
L(A)&=\int_0^A \int_0^A \left(\frac{\ln^3 u}{(1-t)(1-u)}-\frac{C}{1-t}\right)dt du\\
&=\ln(1-A)\int_0^A \left(C-\frac{\ln^3 u}{1-u}\right)du\\
&=\ln(1-A)\int_A^1\frac{\ln x}{1-x}dx-(1-A)\ln(1-A)C\\
L&=\lim_{A\rightarrow 1}L(A)\\&=0\\
K&=\boxed{-6\zeta(2)\zeta(3)+12\zeta(5)}\\
\end{align}
Therefore,
$\boxed{\displaystyle J=\zeta(5)}$
I assume that,
\begin{align}\int_0^1 \frac{\ln t}{1-t}dt&=-\zeta(2)=-\frac{1}{6}\pi^2\\
\int_0^1 \frac{\ln^2 t}{1-t}dt&=2\zeta(3)\\
\int_0^1 \frac{\ln^4 t}{1-t}dt&=24\zeta(5)\\
\end{align}