TITLE: How to evaluate $\int_0^1\frac{\ln x\ln^2(1-x)}{1+x}dx$ in an elegant way?
QUESTION [8 upvotes]: How to prove, in an elegant way that

$$I=\int_0^1\frac{\ln x\ln^2(1-x)}{1+x}dx=\frac{11}{4}\zeta(4)-\frac14\ln^42-6\operatorname{Li}_4\left(\frac12\right)\ ?$$


First, let me show you how I did it
\begin{align}
I&=\int_0^1\frac{\ln x\ln^2(1-x)}{1+x}\ dx\overset{1-x\ \mapsto x}{=}\int_0^1\frac{\ln(1-x)\ln^2x}{2-x}\ dx\\
&=\sum_{n=1}^\infty\frac1{2^n}\int_0^1x^{n-1}\ln^2x\ln(1-x)\ dx\\
&=\sum_{n=1}^\infty\frac1{2^n}\frac{\partial^2}{\partial n^2}\int_0^1x^{n-1}\ln(1-x)\ dx\\
&=\sum_{n=1}^\infty\frac1{2^n}\frac{\partial^2}{\partial n^2}\left(-\frac{H_n}{n}\right)\\
&=\sum_{n=1}^\infty\frac1{2^n}\left(\frac{2\zeta(2)}{n^2}+\frac{2\zeta(3)}{n}-\frac{2H_n}{n^32^n}-\frac{2H_n^{(2)}}{n^22^n}-\frac{2H_n^{(3)}}{n2^n}\right)\\
&=2\zeta(2)\operatorname{Li}_2\left(\frac12\right)+2\ln2\zeta(3)-2\sum_{n=1}^\infty\frac{H_n}{n^32^n}-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}-2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}
\end{align}
By substituting 
$$S_1=\sum_{n=1}^\infty \frac{H_n}{n^32^n}=\operatorname{Li}_4\left(\frac12\right)+\frac18\zeta(4)-\frac18\ln2\zeta(3)+\frac1{24}\ln^42$$
$$
S_2=\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n^22^n}}=\operatorname{Li_4}\left(\frac12\right)+\frac1{16}\zeta(4)+\frac14\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac1{24}\ln^42$$
$$S_3=\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}=\operatorname{Li_4}\left(\frac12\right)-\frac{5}{16}\zeta(4)+\frac78\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac{1}{24}\ln^42$$
along with $\operatorname{Li}_2(1/2)=\frac12\zeta(2)-\frac12\ln^22$ we get the closed form on $I$. Note that $S_1$, $S_2$ and $S_3$ can be found here, here and here respectively.

Now we can see how boring and tedious our calculations are as we used results of three harmonic series with powers of 2 in the denominator. A friend ( who proposed this problem ) suggested that the integral can be done without using harmonic series, so any idea how to do it that way?
Thanks

REPLY [2 votes]: Integrate as follows
\begin{align}
I=&\int_0^1 \frac{\ln x\ln^2(1-x)}{1+x}dx\\
= &\int_0^1 \frac{\ln^2 x\ln(1-x)}{2-x}dx
= \int_0^1 \ln(1-x)\> d\left( \int_1^x \frac{\ln^2 t}{2-t}dt\right) \\
=& \int_0^1 \frac1{1-x}\left( \int_0^x \frac{\ln^2 t}{2-t}\overset{t=xy}{dt}- \int_0^1 \frac{\ln^2 t}{2-t}dt \right) dx\\
=& \int_0^1 \left( \int_0^1 \frac{\ln^2 (xy)}{2-y} 
\left( \frac1{1-x}-\frac2{2-xy}\right){dy}- \frac1{1-x}\int_0^1 \frac{\ln^2 t}{2-t}dt \right) dx\\
 =& \int_0^1 \int_0^1 \frac{\ln^2 x+2\ln x\ln y}{(1-x)(2-y)} dydx - \int_0^1 \int_0^1 \frac{2\ln^2(xy)}{(2-y)(2-xy)} \overset{t=xy}{dx}dy\\
=& \> 2\ln2 Li_3(1)+2Li_2(1)Li_2(\frac12)
 + \int_0^1 \overset{ibp}d\left(\ln \frac y{2-y}\right)\int_0^y\frac{\ln^2t}{2-t}dt\\
=& \>2\ln2 Li_3(1)+2Li_2(1)Li_2(\frac12)-6Li_4(\frac12)-K\tag1
\end{align}
where
\begin{align}
K=&\int_0^1 \frac{\ln^2x \ln (2-x)}{2-x}dx\\
= &\>\ln2 \int_0^1 \frac{\ln^2x}{2-x}dx
 + \int_0^1 \frac{\ln^2x \ln\frac{2-x}2}{2-x} \overset{x\to 2x}{dx}\\
=&\>2\ln2Li_3(\frac12)+\ln2 \int_0^{\frac12} \frac{\ln(2x^2)\ln(1-x)}{1-x}\overset{ibp}{dx}
 + \int_0^{\frac12} \frac{\ln^2x\ln(1-x)}{1-x}dx\\
=&\>2\ln2Li_3(\frac12)+\ln2 \left( \frac12\ln^32 +\int_0^{\frac12} \frac{\ln^2(1-x)}{x}dx\right)
-\frac14\ln^42+ \frac12J\\
=& \>2\ln2Li_3(1)-2\ln^22Li_2(\frac12)-\frac34\ln^42+\frac12J\tag2\\
\\
J=&\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)
\end{align}
Plug the result for $J$ into (2) and then $K$ into (1) to obtain
\begin{align}
I= &-6Li_4(\frac12) -Li_4(1)+2(Li_2(1)+\ln^22)Li_2(\frac12)+\frac12Li_2^2(1)+\frac34\ln^42\\
=& -6Li_4(\frac12)+\frac{11\pi^4}{360}-\frac14\ln^42
\end{align}