Question. Prove the following polylogarithmic integral has the stated value:
$$I:=\int_{0}^{1}\frac{\operatorname{Li}_2{(1-x)}\log^2{(1-x)}}{x}\mathrm{d}x=-11\zeta{(5)}+6\zeta{(3)}\zeta{(2)}.$$
I was able to arrive at the proposed value for the integral through a combination of educated guessing and after-the-fact numerical checking (the value of the integral was approximately $I\approx0.457621...$), but I'm at a loss for how to derive it rigorously. Thoughts?
Since no answers have been posted, I'll expand on my comment above.
There is a general formula that states $$\sum_{n=1}^{\infty}\frac{H_{n}^{(r)}}{n^{q}}=\zeta(r)\zeta(q)-\frac{(-1)^{r-1}}{(r-1)!}\int_{0}^{1}\frac{\text{Li}_{q}(x) \log^{r-1}(x) }{1-x}dx $$ where $$ H_{n}^{(r)} = \sum_{k=1}^{n} \frac{1}{k^{r}} .$$
A proof can be found here.
Making the substitution $u = 1-x$, $$ \sum_{n=1}^{\infty}\frac{H_{n}^{(r)}}{n^{q}}=\zeta(r)\zeta(q)-\frac{(-1)^{r-1}}{(r-1)!}\int_{0}^{1}\frac{\text{Li}_{q}(1-u) \log^{r-1}(1-u)}{u}dx .$$
Therefore, $$ \int_{0}^{1} \frac{\text{Li}_{2}(1-x)\log^{2}(1-x)}{x} \ dx = 2 \zeta (3) \zeta (2) - 2 \sum_{n=1}^{\infty} \frac{H_{n}^{(3)}}{n^{2}} .$$
To simplify the evaluation of that Euler sum slightly, I'm first going to evaluate $ \displaystyle \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}}$ and then use the identity $$\sum_{n=1}^{\infty} \frac{H_{n}^{(r)}}{n^{q}} + \sum_{n=1}^{\infty} \frac{H_{n}^{(q)}}{n^{r}} = \zeta(r) \zeta(q) + \zeta(r+q) . \tag{1} $$
Consider $$f(z) = \frac{ \pi \cot (\pi z) \ \psi_{1}(-z)}{z^{3}} $$ where $\psi_{1}(z)$ is the trigamma function.
The function $f(z)$ has poles of order $3$ at the positive integers, a pole of order 6 at the origin, and simple poles at the negative integers.
On the sides of a square with vertices at $\pm \left( N+ \frac{1}{2} \right) \pm i \left( N+ \frac{1}{2} \right)$, $\cot (\pi z)$ is uniformly bounded.
And when $z$ is large in magnitude and not on the positive real axis, $\psi_{1}(-z)$ is approximately $ \displaystyle - \frac{1}{z}$.
So as $N \to \infty$ through the integers, $ \displaystyle \int f(z) \ dz$ will vanish on all four sides of the square.
Therefore, $$ \sum_{n=-\infty}^{\infty} \text{Res} [f(z), n] = 0.$$
The Laurent expansion of $\psi_{1}(-z)$ at the positive integers (including $0$) is $$ \psi_{1}(-z) = \frac{1}{(z-n)^{2}} + \sum_{m=0}^{\infty} (m+1) \left( (-1)^{m} H_{n}^{(m+2)} + \zeta(m+2) \right) (z-n)^{m} . $$
And the Laurent expansion of $\pi \cot \pi z$ at the integers is
$$ \pi \cot (\pi z) = \frac{1}{z-n} - 2 \sum_{m=1}^{\infty} \zeta(2m) (z-n)^{2m-1} .$$
So at the positive integers, $$f(z) = \frac{1}{z^{3}} \left(\frac{1}{(z-n)^{3}} + \frac{H_{n}^{(2)} - \zeta(2)}{(z-n)} + \mathcal{O}(1) \right) $$
which implies
$$ \begin{align} \text{Res} [f(z), n] &= \text{Res} \left[\frac{1}{z^{3}(z-n)^{3}}, n \right] + \text{Res} \left[\frac{H_{n}^{(2)}-\zeta(2)}{z^{3}(z-n)}, n \right] \\ &= \frac{6}{n^{3}} + \frac{H_{n}^{(2)}}{n^{3}} -\frac{\zeta(2)}{n^{3}} . \end{align} $$
At the negative integers,
$$\text{Res} [f(z), -n] = - \frac{\psi_{1}(n)}{n^{3}} = \frac{H_{n-1}^{(2)} - \zeta(2)}{n^{3}} = \frac{H_{n}^{(2)}}{n^{3}} - \frac{1}{n^{5}} - \frac{\zeta(2)}{n^{3}} .$$
And at the origin,
$$ f(z) = \frac{1}{z^{6}} - \frac{\zeta(2)}{z^{4}} + \frac{2 \zeta(3)}{z^{3}} + \left(\zeta(4) - 2 \zeta^{2}(2) \right) \frac{1}{z^{2}} + \Big(4 \zeta(5) - 4 \zeta(3) \zeta(2) \Big) \frac{1}{z} + \mathcal{O}(1)$$
while implies
$$ \text{Res}[f(z),0] = 4 \zeta(5) - 4 \zeta(3) \zeta(2) .$$
Summing up all the residues,
$$6 \zeta(5) + \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}} - \zeta(3) \zeta(2) + \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}} - \zeta(5) - \zeta(3) \zeta(2) + 4 \zeta(5) - 4 \zeta(3) \zeta(2) = 0$$
which implies
$$ \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}} = 3 \zeta(3) \zeta(2) - \frac{9}{2} \zeta(5) .$$
Then using $(1)$,
$$\sum_{n=1}^{\infty} \frac{H_{n}^{(3)}}{n^{2}} = \zeta(3) \zeta(2) + \zeta(5) - 3 \zeta(3) \zeta(2) + \frac{9}{2} \zeta(5) = - 2 \zeta(3) \zeta(2) + \frac{11}{2} \zeta(5). $$
So finally we have
$$ \begin{align} \int_{0}^{1} \frac{\text{Li}_{2}(1-x)\log^{2}(1-x)}{x} \ dx &= 2 \zeta (3) \zeta (2) - 2 \Big(- 2 \zeta(3) \zeta(2) + \frac{11}{2} \zeta(5) \Big) \\ &= 6 \zeta(3) \zeta(2) - 11 \zeta(5) . \end{align}$$
\begin{align} I&=\int_0^1\frac{\operatorname{Li}_2(1-x)\ln^2(1-x)}{x}\ dx=\int_0^1\frac{\operatorname{Li}_2(x)\ln^2x}{1-x}\ dx\\ &=\sum_{n=1}^\infty H_n^{(2)}\int_0^1x^n\ln^2x\ dx=2\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)^3}\\ &=2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}-2\zeta(5)\\ &=2\left(3\zeta(2)\zeta(5)-\frac92\zeta(5)\right)-2\zeta(5)\\ &=6\zeta(2)\zeta(5)-11\zeta(5) \end{align}
Proof:
\begin{align} S&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=\sum_{n=1}^\infty\frac1{n^3}\left(\zeta(2)-\sum_{k=1}^\infty\frac1{(n+k)^2}\right)=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{n^3(n+k)^2}\\ &=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\sum_{n=1}^\infty\left(\frac{3}{k^4}\left(\frac1{n}-\frac1{n+k}\right)-\frac2{k^3n^2}-\frac1{k^3(n+k)^2}+\frac1{k^2n^3}\right)\\ &=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\left(\frac{3H_k}{k^4}-\frac{2\zeta(2)}{k^3}-\frac{\zeta(2)-H_k^{(2)}}{k^3}+\frac{\zeta(3)}{k^2}\right)\\ &=\zeta(2)\zeta(3)-3\sum_{k=1}^\infty\frac{H_k}{k^4}+2\zeta(2)\zeta(3)+\zeta(2)\zeta(3)-S-\zeta(2)\zeta(3)\\ 2S&=3\zeta(2)\zeta(3)-3\sum_{k=1}^\infty\frac{H_k}{k^4}\\ &=3\zeta(2)\zeta(3)-3\left(3\zeta(5)-\zeta(2)\zeta(3)\right)\\ &=6\zeta(2)\zeta(3)-9\zeta(5)\\ S&=3\zeta(2)\zeta(3)-\frac92\zeta(5) \end{align}
Addendum
Different proof:
By Cauchy product we have
$$\operatorname{Li}_2(x)\operatorname{Li}_3(x)=\sum_{n=1}^\infty\left(\frac{6H_n}{n^4}+\frac{3H_n^{(2)}}{n^3}+\frac{H_n^{(3)}}{n^2}-\frac{10}{n^5}\right)x^n$$
set $x=1$ to get
$$\zeta(2)\zeta(3)=6\sum_{n=1}^\infty\frac{H_n}{n^4}+3\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}-10\zeta(5)$$
Now lets use the well-known identity
$$\sum_{n=1}^\infty\frac{H_n^{(p)}}{n^q}+\sum_{n=1}^\infty\frac{H_n^{(q)}}{n^p}=\zeta(p)\zeta(q)+\zeta(p+q)$$
set $p=2$ and $q=3$
$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\zeta(2)\zeta(3)+\zeta(5)-\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}\tag{*}$$
So
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=\frac92\zeta(5)-3\sum_{n=1}^\infty\frac{H_n}{n^4}$$
substitute $\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ we get
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(5)-\frac92\zeta(2)\zeta(3)$$
and as a bonus, substitute the last result in (*) we get
$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\frac{11}2\zeta(5)-2\zeta(2)\zeta(3)$$
An alternative computation, 100% Euler sums free.
\begin{align}J&=\int_{0}^{1}\frac{\operatorname{Li}_2{(1-x)}\log^2{(1-x)}}{x}\mathrm{d}x\\ &\overset{u=1-x}=\int_0^1 \frac{\text{Li}_2(u)\ln^2 u }{1-u}du\\ &=-\int_0^1 \frac{\ln^2 u }{1-u}\left(\underbrace{\int_0^u \frac{\ln(1-t)}{t}dt}_{\text{IBP}}\right)du\\ &=-\underbrace{\int_0^1 \frac{\ln(1-u)\ln^3 u}{1-u}du}_{=A}-\underbrace{\int_0^1 \frac{\ln^2 u}{1-u}\left(\int_0^u \frac{\ln t}{1-t}dt\right)du}_{=B}\\ \end{align} Computation of $B$.
\begin{align}B&=\int_0^1\int_0^1\frac{u\ln^2 u\ln(tu)}{(1-u)(1-tu)}du\\ &=\int_0^1\int_0^1\left(\frac{\ln^2 u\ln(tu)}{(1-t)(1-u)}-\frac{\ln^2 u\ln(tu)}{(1-t)(1-tu)}\right)dtdu\\ &=\left(\int_0^1\frac{\ln t}{1-t}dt\right)\left(\int_0^1\frac{\ln^2 u}{1-u}du\right)+\\& \int_0^1 \int_0^1 \left(\frac{\ln^3 u}{(1-t)(1-u)}-\frac{\left(\ln(tu)-\ln t\right)^2\ln(tu)}{(1-t)(1-tu)}\right)dtdu\\ &=-2\zeta(2)\zeta(3)+2\int_0^1\frac{\ln t}{t(1-t)}\left(\int_0^t\frac{\ln^2 u}{1-u}du\right)dt+\\&\int_0^1\left(\frac{1}{1-t}\left(\int_0^1\frac{\ln^3 u}{1-u}du\right)-\frac{1}{t(1-t)}\left(\int_0^t\frac{\ln^3 u}{1-u}du\right)\right)dt-\\&\int_0^1\frac{\ln^2 t}{t(1-t)}\left(\int_0^t\frac{\ln u}{1-u}du\right)dt\\ &\overset{\text{IBP}}=-2\zeta(2)\zeta(3)+2\int_0^1\frac{\ln t}{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^3 u}{1-u}du\right)dt+\\&\frac{1}{3}\int_0^1 \frac{\ln^4 t}{1-t}dt-B\\ &\overset{\text{IBP}}=-2\zeta(2)\zeta(3)+2\left(\int_0^1 \frac{\ln t}{1-t}dt\right)\left(\int_0^1 \frac{\ln^2 u}{1-u}du\right)-2B+8\zeta(5)-A-B\\ &=-6\zeta(2)\zeta(3)+8\zeta(5)-A-3B\\ &=\boxed{2\zeta(5)-\frac{3}{2}\zeta(2)\zeta(3)-\frac{A}{4}}\\ \end{align} Computation of $A$ \begin{align}A&\overset{\text{IBP}}=\underbrace{\left[\left(\int_0^u\frac{\ln^3 t}{1-t}dt-\int_0^1\frac{\ln^3 t}{1-t}dt\right)\ln(1-u)\right]_0^1}_{=0}+\\&\int_0^1 \frac{1}{1-u}\left(\int_0^u\frac{\ln^3 t}{1-t}dt-\int_0^1\frac{\ln^3 t}{1-t}dt\right)du\\ &=\int_0^1\int_0^1\left(\frac{u\ln^3(tu)}{(1-u)(1-tu)}-\frac{\ln^3 t}{(1-t)(1-u)}\right)dtdu\\ &=\int_0^1\int_0^1\left(\frac{\ln^3(tu)}{(1-t)(1-u)}-\frac{\ln^3(tu)}{(1-t)(1-tu)}-\frac{\ln^3 t}{(1-t)(1-u)}\right)dtdu\\ &=\int_0^1\int_0^1\left(\frac{\ln^3 u+3\ln t\ln^2 u+3\ln^2 t\ln u}{(1-t)(1-u)}-\frac{\ln^3(tu)}{(1-t)(1-tu)}\right)dtdu\\ &\overset{\text{Fubini}}=\int_0^1 \left(\frac{1}{1-t}\left(\int_0^1 \frac{\ln^3 u}{1-u}du\right)-\frac{1}{t(1-t)}\left(\int_0^t \frac{\ln^3 u}{1-u}du\right)\right)dt-12\zeta(2)\zeta(3)\\ &\overset{\text{IBP}}=\underbrace{\int_0^1 \frac{1}{1-t}\left(\int_t^1 \frac{\ln^3 u}{1-u}du\right)dt}_{\text{IBP}}+\int_0^1\frac{\ln^4 t}{1-t}dt-12\zeta(2)\zeta(3)\\ &=24\zeta(5)-12\zeta(2)\zeta(3)-A\\ &=\boxed{12\zeta(5)-6\zeta(2)\zeta(3)} \end{align} Therefore, \begin{align}J&=-A-B\\&=-A-2\zeta(5)+\frac{3}{2}\zeta(2)\zeta(3)+\frac{A}{4}\\ &=\frac{3}{2}\zeta(2)\zeta(3)-\frac{3A}{4}-2\zeta(5)\\ &=\frac{3}{2}\zeta(2)\zeta(3)-\frac{3}{4}\Big(12\zeta(5)-6\zeta(2)\zeta(3)\Big)-2\zeta(5)\\ &=\boxed{6\zeta(2)\zeta(3)-11\zeta(5)} \end{align}
\begin{align*} I &= \sum_{n=1}^\infty \frac{1}{n^2}\int_0^1 x^{n-1}\log^2(1-x) \,dx \\ &= \sum_{n=1}^\infty \frac{1}{n^2}\left( \frac{(H_n)^2 + H_n^{(2)}}{n} \right) \\ &= \sum_{n=1}^\infty \frac{(H_n)^2}{n^3} + \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3} \end{align*}
We can prove using residues that
\begin{align*} \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3} &= 3\zeta(2)\zeta(3) - \frac{9}{2}\zeta(5) \\ \sum_{n=1}^\infty \frac{(H_n)^2}{n^3} &= -\zeta(2)\zeta(3) + \frac{7}{2}\zeta(5) \end{align*}
Adding them gives
\begin{equation*} I = 2\zeta(2)\zeta(3) - \zeta(5) \end{equation*}
