Can $\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}$, with $H_n$ the $n$-th harmonic number, be written in terms of $\zeta$ values?

Question

The Euler sums are given by $$S_{p,q} = \sum_{n = 1}^{\infty} \frac{H_{n}^{(p)}}{n^q},$$ where $$H_{n}^{(p)} = \sum_{j = 1}^{n} \frac{1}{j^p}.$$ According to Wolfram, Eq. (19), the following special case holds: $$S_{1,3} \equiv \sum_{n = 1}^{\infty} \frac{H_{n}}{n^3} = \frac{5}{4}\zeta(4),$$ where $H_n = H_n^{(1)}$ is the $n$-th harmonic number. I am interested in the following "modified Euler sum": $$\sum_{n = 1}^{\infty} \frac{H_{2n}}{(2n)^3} \equiv \frac{1}{8} \sum_{n = 1}^{\infty} \frac{H_{2n}}{n^3}.$$ Can it as well be expressed in terms of $\zeta$-values? I have not been able to locate any relevant literature.

Update I: For what it may be worth (it is not even strictly related to the original question of this post, although it does feature sums containing $H_{2n}$), I have been able to derive the following identity: $$\frac{124}{5}\zeta(5) = 2\pi^{2}\ln2 + \frac{16}{\pi^{2}} \sum_{n=1}^{\infty}[\frac{6H_{n}}{(2n+1)^{4}} + \frac{H_{n}^{(2)}}{(2n+1)^{3}} - \frac{12H_{2n}}{(2n+1)^{4}} - \frac{4H_{2n}^{(2)}}{(2n+1)^{3}}],$$ where $H_{n}^{(2)} \equiv \sum_{k=1}^{n} 1/k^{2}$, as standardly. This has been done by considering a contour integral at infinity (as a limit, of course) of the function $$f(s) = \frac{\Psi(s)}{s^{3}\cos(\pi s)^{2}},$$ the corresponding sum of all residues having to vanish. However, that identity leaves me no less stuck.

Update II: Using the following family of functions: $$F_{k}(s) = \frac{\Psi(s)}{s^{k}\sin(\pi s)}[1 + \cos(\pi s)],$$ where $k$ is a natural number, I have been able to establish the following identity [note the condition on $k$]: $$\sum_{n=1}^{\infty}\frac{H_{2n}}{(2n)^{k}} = 2^{-k} \gamma \zeta(k) + 2^{-(k+2)}(k+1)\zeta(k+1) - \frac{\pi}{4} \rm{Res}(F_{k};0),\text{ for } k \text{ even},$$ where $\rm{Res}(F_{k};0)$ denotes, of course, the residue of $F(s)$ at $s = 0$. It implies the following specific cases [note the presence of $n^{k}$ in the denominators of the sums, not $(2n)^{k}$ as above] \begin{align} \sum_{n=1}^{\infty}\frac{H_{2n}}{n^{2}} &= \frac{11 \zeta(3)}{4},\\ \sum_{n=1}^{\infty}\frac{H_{2n}}{n^{4}} &= \frac{37 \zeta(5)}{4} - 4\zeta(2)\zeta(3),\\ \sum_{n=1}^{\infty}\frac{H_{2n}}{n^{6}} &= \frac{135 \zeta(7)}{4} - 16\zeta(2)\zeta(5) - 4\zeta(4)\zeta(3). \end{align} Unfortunately, the method fails for k odd, in the sense that the resulting condition (of the sum of residues having to vanish) does not contain any harmonic numbers.

Answer 1

An attempt to get an expression through the integrals.

$$\sum_{n \geq 1}\frac{H_{2n}}{n^3}=\sum_{n \geq 1}\frac{1}{n^3} \int_0^1 \frac{1-x^{2n}}{1-x}dx=$$

$$\int_0^1 \frac{dx}{1-x} \left(\zeta(3)-\sum_{n \geq 1}\frac{x^{2n}}{n^3} \right)=\int_0^1 \frac{dx}{1-x} \left(\zeta(3)-\text{Li}_3 (x^2)\right)=$$

Let's change the variable to $x=e^{-t}$, then we have:

$$=\int_0^\infty \frac{ dt}{e^t-1} \left(\zeta(3)-\text{Li}_3 (e^{-2t})\right)=$$

There are known integral expressions:

$$\zeta(3)=\frac{1}{2} \int_0^\infty \frac{u^2 du}{e^u-1}$$

$$\text{Li}_3 (e^{-2t})=\frac{1}{2} \int_0^\infty \frac{u^2 du}{e^{2t+u}-1}$$

So we can write:

$$=\frac{1}{2} \int_0^\infty \int_0^\infty\frac{ dt}{e^t-1}\left( \frac{1}{e^u-1}-\frac{1}{e^{2t+u}-1} \right) u^2 du=$$

$$=\frac{1}{2} \int_0^\infty \int_0^\infty\frac{ dt}{e^t-1} \frac{e^u (e^{2t}-1)u^2 du}{(e^u-1)(e^{2t+u}-1)} = \frac{1}{2} \int_0^\infty \int_0^\infty \frac{e^u (e^t+1)u^2 du~dt }{(e^u-1)(e^{2t+u}-1)}$$

So we have:

$$\sum_{n \geq 1}\frac{H_{2n}}{n^3}=\frac{1}{2} \int_0^\infty \int_0^\infty \frac{e^u (e^t+1)u^2 du~dt }{(e^u-1)(e^{2t+u}-1)}$$

We could set:

$$e^{-t}=x \\ e^{-u}=y$$

Then

$$\sum_{n \geq 1}\frac{H_{2n}}{n^3}=\frac{1}{2} \int_0^1 \int_0^1 \frac{ (1+x) \ln^2 y ~dx~dy }{(1-y)(1-x^2 y)}$$

Which could be separated into two distinct parts:

$$\sum_{n \geq 1}\frac{H_{2n}}{n^3}=\frac{1}{2} \int_0^1 \int_0^1 \frac{ \ln^2 y ~dx~dy }{(1-y)(1-x^2 y)}+\frac{1}{4} \int_0^1 \int_0^1 \frac{ \ln^2 y ~dx~dy }{(1-y)(1-x y)}=$$

$$=\frac{1}{2} \int_0^1 \int_0^1 \frac{ \ln^2 y ~dx~dy }{(1-y)(1-x^2 y)}-\frac{1}{4} \int_0^1 \frac{ \ln^2 y \ln(1-y) ~dy }{y(1-y)}$$

Mathematica gives for that last integral:

$$-\frac{1}{4} \int_0^1 \frac{ \ln^2 y \ln(1-y) ~dy }{y(1-y)}=\frac{\pi^4}{144}$$

So we can write (also taking the first integral w.r.t. $x$):

$$\sum_{n \geq 1}\frac{H_{2n}}{n^3}=\frac{1}{2} \int_0^1 \frac{ \tanh^{-1} \sqrt{y} \ln^2 y ~dy }{\sqrt{y}(1-y)}+\frac{\pi^4}{144}$$

The first integral can be simplified:

$$\sum_{n \geq 1}\frac{H_{2n}}{n^3}= 4\int_0^1 \frac{ \tanh^{-1} y \ln^2 y ~dy }{1-y^2}+\frac{\pi^4}{144}$$

No idea what to do with the last integral, Mathematica can't take it. However, the integrand is nice, there's not even any singularities.

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We could try integration by parts with $$u=\tanh^{-1} y \ln^2 y \\ dv=\frac{dy }{1-y^2}$$

$$I=\int_0^1 \frac{ \tanh^{-1} y \ln^2 y ~dy }{1-y^2}=(\tanh^{-1} y)^2 \ln^2 y \bigg|_0^1-I-2 \int_0^1 (\tanh^{-1} y)^2 \ln y ~\frac{dy}{y}$$

So:

$$I=-\int_0^1 (\tanh^{-1} y)^2 \ln y ~\frac{dy}{y}= \int_0^\infty t (\tanh^{-1} e^{-t})^2 dt$$

Not really a simplification, but also an interesting expression.

Moreover, it is very fitting for numerical computation of the integral, because for moderately large $t$ we have $$\tanh^{-1} e^{-t} \asymp e^{-t}$$ and so the integral becomes elementary.

Hence, we can write:

$$\sum_{n \geq 1}\frac{H_{2n}}{n^3} = \frac{\pi^4}{144}+4\int_0^\infty t (\tanh^{-1} e^{-t})^2 dt$$

$$\sum_{n \geq 1}\frac{H_{2n}}{n^3} \approx \frac{\pi^4}{144}+(1+2 \tau) e^{-2 \tau}+ 4\int_0^\tau t (\tanh^{-1} e^{-t})^2 dt$$

Where the last integral can be evaluated with any suitable numerical quadrature. $\tau=5$ already gives $6$ accurate digits for the series.

Answer 2

Tim Jameson's notes could be helpful: Polylogarithms, multiple zeta values, and the series of Hjortnaes and Comtet, https://www.maths.lancs.ac.uk/jameson/polylog.pdf

Answer 3

$$ T = \sum_{n\geq 1}\frac{H_n}{(n+1)^3}(-1)^{n+1} = \frac{1}{2}\int_{0}^{1}\frac{\log(1+x)\log^2(x)}{1+x}\,dx$$ can be seen to be equal to $$ \frac{1}{24} \left(-\pi^4-4 \pi^2 \log^2(2)+4\log^4(2)+96\,\text{Li}_4\left(\tfrac{1}{2}\right)+84\log(2)\zeta(3)\right) $$ due to the polylogarithms machinery (see De Doelder, 1991, also mentioned by Flajolet and Salvy in 1998). Of course $\sum_{n\geq 1}\frac{H_{2n}}{(2n)^3}$ is a linear combination of $T$,$\zeta(3)$ and $\zeta(4)$, hence it is inextricably related to the "ugly" terms $\log(2)\zeta(3)$ and $\text{Li}_4\left(\frac{1}{2}\right)=\sum_{n\geq 1}\frac{1}{2^n n^4}$.

Answer 4

\begin{align} \sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}&=\frac12\sum_{n=1}^\infty\frac{H_n}{n^3}+\frac12\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}\\ &=\frac12\left(\frac54\zeta(4)\right)+\frac12S\\ &=\frac58\zeta(4)+\frac12S \end{align}

\begin{align} S&=\sum_{n=1}^\infty\frac{(-1)^n H_n}{n^3}=\frac12\int_0^1\frac{\ln^2x}{x}\sum_{n=1}^\infty H_n(-x)^n\ dx=-\frac12\underbrace{\int_0^1\frac{\ln^2x\ln(1+x)}{x(1+x)}\ dx}_{x=(1-y)/y}\\ &=\frac12\underbrace{\int_{1/2}^1\frac{\ln^2((1-x)/x)\ln(x)}{1-x}\ dx}_{x=1-y}=\frac12\int_0^{1/2}\frac{\ln^2(x/(1-x))\ln(1-x)}{x}\ dx\\ &=\frac12\left(\int_0^{1/2}\frac{\ln^2x\ln(1-x)}{x}\ dx+\int_0^{1/2}\frac{\ln^3(1-x)}{x}\ dx\right)-\int_0^{1/2}\frac{\ln x\ln^2(1-x)}{x}\ dx\\ &=\frac12\left(I_1+I_2\right)-I_3 \end{align} Applying IBP for the first integral by setting $dv=\ln x/x$ and $u=\ln(1-x)$ and letting $x=1-y$ for the second integral, we have:

\begin{align} I_1+I_2&=\frac13\ln^42+\frac13\int_0^{1/2}\frac{\ln^3x}{1-x}\ dx+\int_{1/2}^1\frac{\ln^3x}{1-x}\ dx\\ &=\frac13\ln^42+\int_0^1\frac{\ln^3x}{1-x}\ dx-\frac23\int_0^{1/2}\frac{\ln^3x}{1-x}\ dx\\ &=\frac13\ln^42-6\zeta(4)-\frac23\sum_{n=1}^\infty\int_0^{1/2}x^{n-1}\ln^3x\ dx\\ &=\frac13\ln^42-6\zeta(4)+\frac23\sum_{n=1}^\infty\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}\right)\\ &=4\operatorname{Li_4}\left(\frac12\right)+4\ln2\operatorname{Li_3}\left(\frac12\right)+2\ln^22\operatorname{Li_2}\left(\frac12\right)+\ln^42-6\zeta(4) \end{align} Applying IBP for the third integral by setting $dv=\ln x/x$ and $u=\ln^2(1-x)$, \begin{align} I_3=\int_0^{1/2}\frac{\ln x\ln^2(1-x)}{x}\ dx&=\frac12\ln^42+\underbrace{\int_0^{1/2}\frac{\ln^2x\ln(1-x)}{1-x}\ dx}_{x=1-y}\\ &=\frac12\ln^42+\int_{1/2}^1\frac{\ln x\ln^2(1-x)}{x}\ dx \end{align} By adding the third integral to both sides, we get: \begin{align} I_3&=\frac14\ln^42+\frac12\int_0^1\frac{\ln x\ln^2(1-x)}{x}\ dx\\ &=\frac14\ln^42+\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1 x^{n-1}\ln x\ dx\\ &=\frac14\ln^42+\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\left(-\frac{1}{n^2}\right)\\ &=\frac14\ln^42+\zeta(4)-\sum_{n=1}^\infty\frac{H_n}{n^3}\\ &=\frac14\ln^42-\frac14\zeta(4) \end{align} Grouping $I_1, I_2$ and $I_3$: \begin{align} S&=2\operatorname{Li_4}\left(\frac12\right)+2\ln2\operatorname{Li_3}\left(\frac12\right)+\ln^22\operatorname{Li_2}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac14\ln^42\\ &=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42 \end{align} note that we used $$\operatorname{Li_3}\left( \frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$ $$\operatorname{Li_2}\left( \frac12\right) =\frac12\zeta(2)-\frac12\ln^22$$ Finally $$\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}=\operatorname{Li_4}\left(\frac12\right)-\frac34\zeta(4)+\frac78\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac{1}{24}\ln^42$$