Integral $\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx$ I would like to know how to evaluate the integral
$$\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx$$
I tried expanding the integrand  as a series but made little progress as I do not know how to evaluate the resulting sum.
\begin{align}
\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx
&=\int^1_0\sum_{n \ge 1}\frac{x^n}{n^2}\sum_{k \ge 0}x^k\ln{x}dx\\
&=\sum_{n \ge 1}\frac{1}{n^2}\sum_{k \ge 0}\int^1_0x^{n+k}\ln{x}dx\\
&=-\sum_{n \ge 1}\frac{1}{n^2}\sum_{k \ge 0}\frac{1}{(n+k+1)^2}
\end{align}
I am aware that a similar question has been answered here, however, I find that the answers are not detailed enough for someone who has a shallow understanding on Euler sums, such as myself, to fully comprehend. Hence, I would like to seek your help on the techniques that can be used to evaluate this integral. Thank you.
 A: \begin{align}
I&=\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1-x}\ dx\\
&=\sum_{n=1}^\infty\left(H_n^{(2)}-\frac1{n^2}\right)\int_0^1x^{n-1}\ln x\ dx\\
&=\sum_{n=1}^\infty\frac1{n^4}-\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^2}\\
&=\zeta(4)-\frac74\zeta(4)\\
&=-\frac34\zeta(4).
\end{align}

Proof of the last sum: Using $\displaystyle\sum_{m=1}^\infty \sum_{n=1}^m a_mb_n=\sum_{n=1}^\infty \sum_{m=n}^\infty a_mb_n$, we have
\begin{align}
\sum_{m=1}^\infty\frac{H_m^{(r)}}{m^s}&=\sum_{m=1}^\infty\sum_{n=1}^m \frac{1}{n^r m^s}\\
&=\sum_{n=1}^\infty\left(\sum_{m=n}^\infty\frac{1}{m^s}\right)\frac1{n^r}\\
&=\sum_{n=1}^\infty\left(\sum_{m=1}^\infty\frac{1}{m^s}-\sum_{m=1}^n\frac1{m^s}+\frac1{n^s}\right)\frac1{n^r}\\
&=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac1{n^rm^s}-\sum_{n=1}^\infty\frac{H_n^{(s)}}{n^r}+\sum_{n=1}^\infty\frac{1}{n^{r+s}}\\
&=\zeta(r)\zeta(s)-\sum_{n=1}^\infty\frac{H_n^{(s)}}{n^r}+\zeta(r+s),
\end{align}
or
$$\sum_{n=1}^\infty\frac{H_n^{(s)}}{n^r}+\sum_{n=1}^\infty\frac{H_n^{(r)}}{n^s}=\zeta(r)\zeta(s)+\zeta(r+s).$$
Setting $r=s=2$ gives $\displaystyle \sum_{n=1}^\infty\frac{H_n^{(2)}}{n^2}=\frac12\zeta^2(2)+\frac12\zeta(4)=\frac74\zeta(4).$
A: \begin{align}\text{J}&=\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx\\
&\overset{\text{IBP}}=\underbrace{\Big[-\ln(1-x)\ln x\mathrm{Li}_2(x)\Big]_0^1}_{=0}+\underbrace{\int_0^1\frac{\ln(1-x)\text{Li}_2(x)}{x}dx}_{=-\frac{1}{2}\text{Li}^2_2(1)}-\underbrace{\int_0^1 \frac{\ln^2(1-x)\ln x}{x}dx}_{\text{IBP}}\\
&=-\frac{\pi^4}{72}-\int_0^1 \frac{\ln(1-x)\ln^2 x}{1-x}dx\\
&\overset{\text{IBP}}=-\frac{\pi^4}{72}-\left[\left(\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)-\int_0^1\frac{\ln^2 t}{1-t}dt\right)\ln(1-x)\right]_0^1-\\&\int_0^1 \frac{1}{1-x}\left(\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)-\int_0^1 \frac{\ln^2 t}{1-t}dt\right)dx\\
&=-\frac{\pi^4}{72}-\int_0^1 \frac{1}{1-x}\left(\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)-\int_0^1 \frac{\ln^2 t}{1-t}dt\right)dx\\
&=-\frac{\pi^4}{72}-\int_0^1 \int_0^1 \left(\frac{x\ln^2(tx)}{(1-tx)(1-x)}-\frac{\ln^2 t}{(1-x)(1-t)}\right)dtdx\\
&=-\frac{\pi^4}{72}+\int_0^1\int_0^1\left(\frac{\ln^2(tx)}{(1-t)(1-tx)}+\frac{\ln^2 t}{(1-t)(1-x)}-\frac{\ln^2(tx)}{(1-t)(1-x)}\right)dtdx\\
&=-\frac{\pi^4}{72}+\int_0^1\int_0^1\left(\frac{\ln^2(tx)}{(1-t)(1-tx)}-\frac{\ln^2 x}{(1-t)(1-x)}\right)dtdx-\\&2\underbrace{\left(\int_0^1 \frac{\ln t}{1-t}dt\right)}_{=-\frac{\pi^2}{6}}\left(\int_0^1 \frac{\ln x}{1-x}dx\right)\\
&=-\frac{5\pi^4}{72}+\int_0^1 \left(\frac{1}{t(1-t)}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)-\frac{1}{1-t}\left(\int_0^1\frac{\ln^2 x}{1-x}dx\right)\right)dt\\
&=-\frac{5\pi^4}{72}+\int_0^1 \frac{1}{t}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)dt-\int_0^1 \frac{1}{1-t}\left(\int_t^1 \frac{\ln^2 u}{1-u}du\right)dt\\
&\overset{\text{IBP}}=-\frac{5\pi^4}{72}-\underbrace{\int_0^1 \frac{\ln^3 t}{1-t}dt}_{=-6\zeta(4)=-\frac{\pi^4}{15}}+\underbrace{\int_0^1 \frac{\ln(1-t)\ln^2 t}{1-t}dt}_{-\frac{\pi^4}{72}-\text{J}}\\
&=-\frac{\pi^4}{60}-\text{J}\\
&=\boxed{-\dfrac{\pi^4}{120}}
\end{align}
A: $$2\sum_{n\geq 1}\frac{1}{n^2}\sum_{m>n}\frac{1}{m^2}=\left(\sum_{n\geq 1}\frac{1}{n^2}\right)^2 -\sum_{n\geq 1}\frac{1}{n^4}=\zeta(2)^2-\zeta(4)=\frac{\pi^4}{60},$$
hence the value of your integral is just $-\frac{\pi^4}{120}$. Pretty nice.
A: Evaluate the integral $I$ below in two ways
\begin{align}
I=&\int_0^1 \frac{\ln^2x \ln (1-x)}{1-x}dx\\
= &\int_0^1 \frac{\ln^2x}{1-x}\left(\int_0^1 \frac {-x}{1-x y}dy\right) dx
=\int_0^1 \frac{1}{1-y}\left(\int_0^1 \frac {\ln^2x}{1-yx}dx - 2Li_3(1)\right) dy \\
=& \>2 \int_0^1\frac{Li_3(y)}{y}dy
 + 2 \int_0^1\frac{Li_3(y)-Li_3(1)}{1-y}\>\overset{ibp}{dy}
= 2Li_4(1) - Li_2^2(1)\\\\
I =&\int_0^1 \frac{\ln x \ln^2(1-x)}{x}dx
\overset{ibp}=- \int_0^1 \frac{\ln x Li_2(x)}{1-x}dx - \frac12 Li_2^2(1)
\end{align}
which leads to
$$\int_0^1 \frac{\ln x Li_2(x)}{1-x}dx=-2Li_4(1)+\frac12Li_2^2(1)=-\frac{\pi^4}{120}
$$
