Definite integral involving polylogarithm I was trying to compute the value of
$$\sum_{k=1}^\infty \frac{H_k}{k^4}$$
and I was able to reduce it down to
$$-\zeta(2)\zeta(3)+\int_0^1 \frac{\text{Li}_2^2(x)}{x}dx$$
However, I can't figure out how to compute the value of the integral
$$\int_0^1 \frac{\text{Li}_2^2(x)}{x}dx$$
How can I find its value?
Don't try integration by parts. This integral is what I ended up with after integration by parts.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
\sum_{k = 1}^{\infty}{H_{k} \over k^{4}} & =
\sum_{k = 1}^{\infty}H_{k}\
\overbrace{\bracks{-\,{1 \over 6}\int_{0}^{1}\ln^{3}\pars{x}\,x^{k - 1}\,\dd x}}
^{\ds{1 \over k^{4}}}
\\[5mm] & =
-\,{1 \over 6}\int_{0}^{1}\ln^{3}\pars{x}\
\overbrace{\sum_{k = 1}^{\infty}H_{k}\,x^{k}}
^{\ds{-\,{\ln\pars{1 - x} \over 1 - x}}}\ \,{\dd x \over x}
\\[5mm] & =
{1 \over 6}\int_{0}^{1}{\ln^{3}\pars{x}\ln\pars{1 - x} \over x\pars{1 - x}}
\,\dd x
\\[5mm] & =
{1 \over 6}\int_{0}^{1}{\ln^{3}\pars{x}\ln\pars{1 - x} \over x}\,\dd x +
{1 \over 6}\int_{0}^{1}{\ln^{3}\pars{x}\ln\pars{1 - x} \over 1 - x}\,\dd x
\\[5mm] & =
{1 \over 6}\
\underbrace{\int_{0}^{1}{\ln^{3}\pars{x}\ln\pars{1 - x} \over x}\,\dd x}
_{\ds{\mc{I}_{1}}}\ +\
{1 \over 6}\
\underbrace{\int_{0}^{1}{\ln^{3}\pars{1 - x}\ln\pars{x} \over x}\,\dd x}
_{\ds{\mc{I}_{2}}}
\\[5mm] & = {\mc{I}_{1} + \mc{I}_{2} \over 6}\label{1}\tag{1}
\end{align}

$\ds{\Large \mc{I}_{1} = ?}$.
\begin{align}
\mc{I}_{1} & =
-\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{3}\pars{x}\,\dd x =
3\int_{0}^{1}\mrm{Li}_{3}'\pars{x}\ln^{2}\pars{x}\,\dd x =
-6\int_{0}^{1}\mrm{Li}_{4}'\pars{x}\ln\pars{x}\,\dd x
\\[5mm] & =
6\int_{0}^{1}\mrm{Li}_{5}'\pars{x}\,\dd x =
6\,\mrm{Li}_{5}\pars{1} \implies
\bbx{\large \mc{I}_{1} = 6\,\zeta\pars{5}}\label{2}\tag{2}
\end{align}

$\ds{\Large \mc{I}_{2} = ?}$.
\begin{align}
\mc{I}_{2} & =
\left.\partiald[3]{}{\mu}\partiald{}{\nu}
\int_{0}^{1}{\bracks{\pars{1 - x}^{\mu} - 1}x^{\nu} \over x}\,\dd x
\,\right\vert_{\ \mu = 0\,,\ \nu\ =\ 0}
\\[5mm] & =
\left.\partiald[3]{}{\mu}\partiald{}{\nu}
\int_{0}^{1}\bracks{\pars{1 - x}^{\mu}x^{\nu - 1} - x^{\nu - 1}}\dd x
\,\right\vert_{\ \mu = 0\,,\ \nu\ =\ 0}
\\[5mm] & =
\partiald[3]{}{\mu}\partiald{}{\nu}
\bracks{{\Gamma\pars{\mu + 1}\Gamma\pars{\nu} \over \Gamma\pars{\mu + \nu + 1}} - {1 \over \nu}}_{\ \mu = 0\,,\ \nu\ =\ 0}
\\[5mm] & =
\partiald[3]{}{\mu}\partiald{}{\nu}
\braces{{\Gamma\pars{\mu + 1}\bracks{\Gamma\pars{\nu + 1}/\nu} \over \Gamma\pars{\mu + \nu + 1}} - {1 \over \nu}}_{\ \mu = 0\,,\ \nu\ =\ 0}
\\[5mm] & =
\partiald[3]{}{\mu}\partiald{}{\nu}
\bracks{{\Gamma\pars{\mu + 1}\Gamma\pars{\nu + 1} -
\Gamma\pars{\mu + \nu + 1} \over
\nu\,\Gamma\pars{\mu + \nu + 1}}}_{\ \mu = 0\,,\ \nu\ =\ 0}
\end{align}
\begin{equation}
\mbox{This limit is a 'cumbersome task'. Its value is}\
\bbx{\large\mc{I}_{2} = 12\,\zeta\pars{5} - \pi^{2}\zeta\pars{3}}
\label{3}\tag{3}
\end{equation}

\eqref{1}, \eqref{2} and \eqref{3} lead to

$$
\bbx{\sum_{k = 1}^{\infty}{H_{k} \over k^{4}} =
3\,\zeta\pars{5} - {1 \over 6}\,\pi^{2}\,\zeta\pars{3}}
$$
A: I wonder if there is not a problem somewhere.
I agree that $$\sum_{k=1}^\infty \frac{H_k}{k^4}=-\zeta(2)\zeta(3)+\text{something}$$ Setting $$\text{something}=\int_0^1 \frac{\text{Li}_2^2(x)}{x}dx$$ may be not correct since the numerical integration leads to $0.843825$ while, numerically
$$\sum_{k=1}^\infty \frac{H_k}{k^4}=1.13348$$ making $$\text{something}=3.11078$$ which is identified as $3\zeta(5)$.
It could be good that you describe the intermediate steps.
A: The explicit values of all the Euler sums $\sum_{n\geq 1}\frac{H_n}{n^s}$ are well known and related to convolutions of $\zeta$ values. Flajolet and Salvy showed how to derive them from the residue theorem, for instance.
Here I will outline  a different technique, less efficient but more elementary. Let us assume that $a,b$ are distinct positive real numbers and recall that $\sum_{n\geq 1}H_n z^n = -\frac{\log(1-z)}{1-z}$. 
By partial fraction decomposition
$$ \frac{1}{(n+a)(n+b)}=\frac{\frac{1}{b-a}}{n+a}+\frac{\frac{1}{a-b}}{n+b}$$
hence
$$ \sum_{n\geq 1}\frac{H_n}{(n+a)(n+b)} = -\int_{0}^{1}\frac{\log(1-x)}{1-x}\left[\frac{x^{a-1}}{b-a}+\frac{x^{b-1}}{a-b}\right]\,dx$$
where the resulting integral can be computed by integration by parts and by differentiating Euler's Beta function. The outcome is:
$$\sum_{n\geq 1}\frac{H_n}{(n+a)(n+b)} = \frac{H_{a-1}^2-H_{b-1}^2-\psi'(a)-\psi'(b)}{2(a-b)}$$
hence by considering the limit as $b\to a$:
$$ \sum_{n\geq 1}\frac{H_n}{(n+a)^2}=H_{a-1}\psi'(a)-\tfrac{1}{2}\psi''(a). $$
In order to compute $\sum_{n\geq 1}\frac{H_n}{n^s}$, it is enough to apply $\left(\lim_{a\to 0^+}\frac{d^{s-2}}{da^{s-2}}\right)$ to both sides of the previous line. In the $s=4$ case we get:
$$ \sum_{n\geq 1}\frac{H_n}{n^4}=\frac{1}{6}\lim_{a\to 0^+}\left[3\,\psi'(a)\,\psi''(a)+H_{a-1}\,\psi'''(a)-\tfrac{1}{2}\psi^{IV}(a)\right]=\color{blue}{3\,\zeta(5)-\zeta(2)\,\zeta(3)}. $$
A: Since $H_k=H_{k-1}+\frac1k$, we have $$\sum^\infty_{k=1}\frac{H_k}{k^4}=\sum^\infty_{k=1}\frac{H_{k-1}}{k^4}+\sum^\infty_{k=1}\frac1{k^5}
=\sum^\infty_{k=1}\frac{H_{k-1}}{k^4}+\zeta(5).\tag1$$
If we define $$\sigma_h(s,t)=\sum^\infty_{n=1}\frac1{n^t}\sum^{n-1}_{k=1}\frac1{k^s},$$ already Euler expressed those sums by special values of the Riemann Zeta Function. He proved rigorously
$$2\sigma_h(1,m)=m\,\zeta(m+1)-\sum^{m-2}_{k=1}\zeta(m-k)\,\zeta(k+1)$$ for $m\ge2$ (that's equation (3) in this paper). So the sum of the RHS of (1) is $\sigma_h(1,4)$, and the above formula easily gives $\sigma_h(1,4)=2\,\zeta(5)-\zeta(2)\,\zeta(3)$, and (1) gives the final result
$$\sum^\infty_{k=1}\frac{H_k}{k^4}=3\,\zeta(5)-\zeta(2)\,\zeta(3).$$
A: To find your Euler sum I will make use of the following result which can be found here
$$\sum_{k=1}^{\infty} \dfrac{H_k^{(p)}}{k^q}=\frac{(-1)^{q+1}}{\Gamma(q)} \int_{0}^{1} \frac{\ln^{q - 1} (u) \text{Li}_{p}(u)}{u (1 - u)}{du}.$$
Here $H^{(p)}_k$ are the generalised harmonic numbers with $H^{(1)}_k$ corresponding to the ordinary harmonic numbers $H_k$. It should be noted that the above result is proved using real methods only.
So when $p = 1$ and $q = 4$ we have
$$\sum^\infty_{k = 1} \frac{H_k}{k^4} = -\frac{1}{\Gamma (4)} \int^1_0 \frac{\ln^3 (u) \, \text{Li}_1 (u)}{u (1 - u)} \, du = \frac{1}{6} \int^1_0 \frac{\ln^3 (u) \ln (1 - u)}{u (1 - u)} \, du,$$
where $\text{Li}_1 (u) = - \ln (1 - u)$ has been used.
The resulting integral can be found by reducing it to a double limit of the derivative of the beta function $\text{B}(x,y)$ as follows. As
$$\text{B}(x,y) = \int^1_0 t^{x - 1} (1 - t)^{y - 1} \, dt,$$
we have
$$\lim_{x \to 0^+} \lim_{y \to 0^+} \partial^3_x \partial_y \text{B}(x,y) = \int^1_0 \frac{\ln^3 (u) \ln (1 - u)}{u (1 - u)} \, du.$$
Thus
\begin{align*}
\sum^\infty_{k = 1} \frac{H_k}{k^4} &= \frac{1}{6} \lim_{x \to 0^+} \lim_{y \to 0^+} \partial^3_x \partial_y \text{B}(x,y) = \frac{1}{6} \left [-\frac{3}{4} \psi^{(4)} (1) + 3 \psi^{(2)}(1) \psi^{(1)}(1) \right ].
\end{align*}
Here $\psi^{(m)}(x)$ is the polygamma function of order $m$. Values for the polygamma function when their arguments are equal to unity are well known. They are:
$$\psi^{(4)}(1) = - 24 \zeta(5), \,\, \psi^{(2)}(1) = -2 \zeta (3), \,\,\psi^{(1)}(1) = \zeta (2),$$
and yields
$$\sum^\infty_{k = 1} \frac{H_k}{k^4} = 3 \zeta (5) - \zeta (2) \zeta (3).$$
