TITLE: Evaluate $\int^1_0 \log^2(1-x) \log^2(x) \, dx$
QUESTION [15 upvotes]: I have no idea where to even start. WolframAlpha cant compute it either.
$$\int^1_0 \log^2(1-x) \log^2(x) \, dx$$
I think it can be done with series, but I am not sure, can someone help a little so I can get a start??
Thanks!

REPLY [6 votes]: First, letting $x=e^{-t}$ gives
$$
\begin{align}
\int_0^1x^k\log(x)^2\,\mathrm{d}x
&=\int_0^\infty e^{-(k+1)t}\,t^2\,\mathrm{d}t\\
&=\frac2{(k+1)^3}\tag{1}
\end{align}
$$
Also, squaring the power series yields
$$
\log(1-x)^2=\sum_{k=1}^\infty\frac{2H_{k-1}}{k}x^k\tag{2}
$$
Therefore,
$$
\begin{align}
&\int_0^1\log(1-x)^2\log(x)^2\,\mathrm{d}x\\
&=\int_0^1\sum_{k=1}^\infty\frac{2H_{k-1}}{k}x^k\log(x)^2\,\mathrm{d}x\tag{3a}\\
&=\sum_{k=1}^\infty\frac{4H_{k-1}}{k(k+1)^3}\tag{3b}\\
&=\sum_{k=1}^\infty\sum_{j=1}^{k-1}\frac4{jk(k+1)^3}\tag{3c}\\
&=\sum_{k=1}^\infty\sum_{j=1}^k\frac4{jk(k+1)^3}-\sum_{k=1}^\infty\frac4{k^2(k+1)^3}\tag{3d}\\
&=\sum_{j=1}^\infty\frac4j\sum_{k=j}^\infty\left(\frac1k-\frac1{k+1}-\frac1{(k+1)^2}-\frac1{(k+1)^3}\right)\tag{3e}\\
&-4\sum_{k=1}^\infty\left(-\frac3k+\frac3{k+1}+\frac1{k^2}+\frac2{(k+1)^2}+\frac1{(k+1)^3}\right)\tag{3f}\\
&=4\zeta(2)-\sum_{j=1}^\infty\frac4j\left[\sum_{k=j}^\infty\left(\frac1{k^2}+\frac1{k^3}\right)-\frac1{j^2}-\frac1{j^3}\right]\tag{3g}\\[6pt]
&-4(-6+3\zeta(2)+\zeta(3))\tag{3h}\\[15pt]
&=24-8\zeta(2)-4\zeta(3)\tag{3i}\\[4pt]
&-\sum_{k=1}^\infty\sum_{j=1}^k\frac4j\left(\frac1{k^2}+\frac1{k^3}\right)+4\zeta(3)+4\zeta(4)\tag{3j}\\[2pt]
&=24-8\zeta(2)+4\zeta(4)\tag{3k}\\[6pt]
&-\sum_{k=1}^\infty\frac{4H_k}{k^2}-\sum_{k=1}^\infty\frac{4H_k}{k^3}\tag{3l}\\
&=24-8\zeta(2)-6\zeta(4)-8\zeta(3)+2\zeta(2)^2\tag{3m}\\[6pt]
&=24-\frac{4\pi^2}3-\frac{\pi^4}{90}-8\zeta(3)\tag{3n}
\end{align}
$$
Explanation:
$\text{(3a)}$: apply $(2)$
$\text{(3b)}$: apply $(1)$
$\text{(3c)}$: expand $H_{k-1}$
$\text{(3d)}$: add and subtract the $j=k$ term
$\text{(3e)}$: partial fractions: $\frac1{k(k+1)^3}=\frac1k-\frac1{k+1}-\frac1{(k+1)^2}-\frac1{(k+1)^3}$
$\text{(3f)}$: partial fractions: $\frac1{k^2(k+1)^3}=-\frac3k+\frac3{k+1}+\frac1{k^2}+\frac2{(k+1)^2}+\frac1{(k+1)^3}$
$\text{(3g)}$: apply $\sum\limits_{j=1}^\infty\frac4j\sum\limits_{k=j}^\infty\left(\frac1k-\frac1{k+1}\right)=\sum\limits_{j=1}^\infty\frac4{j^2}=4\zeta(2)$ and
$\hphantom{\text{(3g):}}$ $\sum\limits_{k=j}^\infty\left(\frac1{(k+1)^2}+\frac1{(k+1)^3}\right)=\sum\limits_{k=j}^\infty\left(\frac1{k^2}+\frac1{k^3}\right)-\frac1{j^2}-\frac1{j^3}$ to $\text{(3e)}$
$\text{(3h)}$: evaluate $\text{(3f)}$
$\text{(3i)}$: collect the evaluated terms in $\text{(3g)}$ and $\text{(3h)}$
$\text{(3j)}$: change order of summation in $\text{(3g)}$ and evaluate sums
$\text{(3k)}$: collect the evaluated terms in $\text{(3i)}$ and $\text{(3j)}$
$\text{(3l)}$: collect $H_k$ in $\text{(3j)}$
$\text{(3m)}$: use this answer to evaluate the sums in $\text{(3l)}$
$\text{(3n)}$: use $\zeta(2)=\frac{\pi^2}6$ and $\zeta(4)=\frac{\pi^4}{90}$