Vtgyjfy

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.

The explicit values of all the Euler sums $sum_{ngeq 1}frac{H_n}{n^s}$ 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 $sum_{ngeq 1}H_n z^n = -frac{log(1-z)}{1-z}$.
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
$$ sum_{ngeq 1}frac{H_n}{(n+a)(n+b)} = -int_{0}^{1}frac{log(1-x)}{1-x}left[frac{x^{a-1}}{b-a}+frac{x^{b-1}}{a-b}right],dx$$
where the resulting integral can be computed by integration by parts and by differentiating Euler's Beta function. The outcome is:
$$sum_{ngeq 1}frac{H_n}{(n+a)(n+b)} = frac{H_{a-1}^2-H_{b-1}^2-psi'(a)-psi'(b)}{2(a-b)}$$
hence by considering the limit as $bto a$:
$$ sum_{ngeq 1}frac{H_n}{(n+a)^2}=H_{a-1}psi'(a)-tfrac{1}{2}psi''(a). $$
In order to compute $sum_{ngeq 1}frac{H_n}{n^s}$, it is enough to apply $left(lim_{ato 0^+}frac{d^{s-2}}{da^{s-2}}right)$ to both sides of the previous line. In the $s=4$ case we get:
$$ sum_{ngeq 1}frac{H_n}{n^4}=frac{1}{6}lim_{ato 0^+}left[3,psi'(a),psi''(a)+H_{a-1},psi'''(a)-tfrac{1}{2}psi^{IV}(a)right]=color{blue}{3,zeta(5)-zeta(2),zeta(3)}. $$

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 $3zeta(5)$.

It could be good that you describe the intermediate steps.

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
$$2sigma_h(1,m)=m,zeta(m+1)-sum^{m-2}_{k=1}zeta(m-k),zeta(k+1)$$ for $mge2$ (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
$$sum^infty_{k=1}frac{H_k}{k^4}=3,zeta(5)-zeta(2),zeta(3).$$

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
$$sum^infty_{k = 1} frac{H_k}{k^4} = 3 zeta (5) - zeta (2) zeta (3).$$

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newcommand{ds}[1]{displaystyle{#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{-,{lnpars{1 - x} over 1 - x}}} ,{dd x over x}
\[5mm] & =
{1 over 6}int_{0}^{1}{ln^{3}pars{x}lnpars{1 - x} over xpars{1 - x}}
,dd x
\[5mm] & =
{1 over 6}int_{0}^{1}{ln^{3}pars{x}lnpars{1 - x} over x},dd x +
{1 over 6}int_{0}^{1}{ln^{3}pars{x}lnpars{1 - x} over 1 - x},dd x
\[5mm] & =
{1 over 6}
underbrace{int_{0}^{1}{ln^{3}pars{x}lnpars{1 - x} over x},dd x}
_{ds{mc{I}_{1}}} +
{1 over 6}
underbrace{int_{0}^{1}{ln^{3}pars{1 - x}lnpars{x} over x},dd x}
_{ds{mc{I}_{2}}}
\[5mm] & = {mc{I}_{1} + mc{I}_{2} over 6}label{1}tag{1}
end{align}

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

$$
bbx{sum_{k = 1}^{infty}{H_{k} over k^{4}} =
3,zetapars{5} - {1 over 6},pi^{2},zetapars{3}}
$$