Closed form of Euler-type sum over zeta functions $\sum _{k=2}^{\infty } \frac{\zeta (k)}{k^2}$?

Question

Revisiting the question on the integral over the harmonic number I stumbled over the nice formula

$$\sum_{k\ge2} (-1)^{k+1}\frac{\zeta(k)}{k} = \gamma\tag{1}$$

where $\zeta(z)$ is the Riemann zeta function and $\gamma$ is Euler's gamma.

Searching SE I found solutions to related but even more complicated problems (see below) so I dropped $(1)$ and propose here instead the problem asked in the heading, viz. to find a closed expression for

$$s=\sum_{k\ge2} \frac{\zeta(k)}{k^2} \simeq 0.835998 \tag{2}$$

I tried several approaches but still couldn't find a closed expression. So I would consider this a tough sum.

More generally we can ask for sums of the form

$$s_q=\sum_{k\ge2} \frac{\zeta(k)}{k^q} \tag{3}$$

A similar tough sums is

$$s_{-1}=\sum_{k\ge2} \frac{\zeta(k)}{k(k-1)} = ? \tag{4}$$

Whereas for sums of the type $\sum_{k\ge2} \frac{\zeta(k)}{k(k+1)}$, $\sum_{k\ge2} \frac{\zeta(k)}{k(k+1)(k+2)}$, or generally

$$s_{p(m)}=\sum_{k\ge2} \frac{\zeta(k)}{(k)_m}=\text{closed expression} \tag{5}$$

where $(k)_m=k(k+1)(k+2)\ldots (k+m-1)$ is the Pochhammer symbol, closed expressions for any integer $m\ge2$ can be found using CAS (e.g. Mathematica).

My effort so far

Because of the length of these developments I have put them into a (preliminary) self answer.

Related problems

[1] A Tough Series $\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1}=-\gamma+\log(2)$
[2] Closed form for a zeta series :$\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}}$

Answer 1

This is a preliminary incomplete answer showing my effort to solve the problem.

What I did so far to find a closed expression to $(2)$ is mainly a reformulation. Maybe someone recognises one of these expressions.

It turned out that different approaches sometimes lead to the same result. I have therefore indicated "really" different formulas by putting them in a box.

0) Just be reassured not to miss trivial things I consulted the online-encyclopedia of integer sequences with the first few digits of $N(s)$.

Nothing relevant was found but the first 5 digits appear somewhere in several funny numbers, like https://oeis.org/A019694, Decimal expansion of 2*Pi/5.

1) Expanding zeta in a series and changing the order of summation, leaves another sum

$$s=\sum _{k=2}^{\infty } \frac{\zeta (k)}{k^2}=\sum _{k=1}^{\infty } \frac{1}{k^2}\sum _{m=1}^{\infty } \frac{1}{m^k} \\ =\sum _{m=1}^{\infty } \left(\sum _{k=2}^{\infty } \frac{1}{k^2 m^k}\right) \\ \boxed{s=\sum _{m=1}^{\infty } \left(\operatorname{Li}_2\left(\frac{1}{m}\right)-\frac{1}{m}\right)}\tag{A.1}$$

2) Replacing inverse power $\frac{1}{k^2}$ by an integral and doing the sum, leaves a nice compact integral

In fact,

$$\int_0^1 x^{k-1} \log \left(\frac{1}{x}\right) \, dx=\frac{1}{k^2}\tag{A.2.1}$$

and using the Taylor expansion of the harmonic number

$$\sum _{k=2}^{\infty } x^{k-1} \zeta (k)=-H_{-x}\tag{A.2.2}$$

we have

$$s=\sum _{k=2}^{\infty } \zeta (k) \int_0^1 x^{k-1} \log \left(\frac{1}{x}\right) \, dx \\ =\int_0^1 \log \left(\frac{1}{x}\right) \sum _{k=2}^{\infty } x^{k-1} \zeta (k) \, dx=\int_0^1 \left(-H_{-x}\right) \log \left(\frac{1}{x}\right) \, dx \\ \boxed{s=\int_0^1 H_{-x} \log (x) \, dx}\tag{A.2.3}$$

3) Exploring the integral $\int_0^1 H_{-x} \log (x) \, dx$

EDIT 06.03.20 begin

Using the basic relation $H_n=H_{n-1}+\frac{1}{n}$ and letting $n=1-x$ we can replace $H_{-x} \to H_{1-x} -\frac{1}{1-x}$ which, observing $\int_0^1\frac{\log(x)}{1-x}\,dx = -\zeta(2)$, leads to the possibly more pleasant form

$$s = \zeta(2) + \int_0^1 \log(1-x) H_{x}\,dx\tag{A.3.0}$$

EDIT end

Integrating by parts, $\int H_{-x} \, dx=\gamma x-\text{log$\Gamma $}(1-x)$, gives

$$s=\int_0^1 H_{-x} \log (x) \, dx=\int_0^1 \frac{(\operatorname{\log\Gamma}(1-x)-x\gamma) }{x} \, dx\tag{A.3}$$

Here it might appear helpful to have the generating integral

$$s(\xi)=\int_0^1 x^\xi H_{-x} \, dx\tag{A.3.1}$$

so that we can generate the $\log$ by the derivative with respect to $\xi$. But that integral is divergent at $x=1$.

3a) Inserting the definition of $H$ as an integral leaves another integral

$$s=\int_0^1 \left(\int_0^1 \frac{\left(1-z^{-x}\right) \log (x)}{1-z} \, dz\right) \, dx \\ =\int_0^1 \left(\int_0^1 \frac{\left(1-z^{-x}\right) \log (x)}{1-z} \, dx\right) \, dz \\ =\int_0^1 \frac{-\log (z)+\log (\log (z))+\Gamma (0,\log (z))+\gamma }{\log (z)-z \log (z)} \, dz \\ \boxed{s=\int_0^{\infty } \frac{t+\log (-t)+\Gamma (0,-t)+\gamma }{t \left(1-e^{t}\right)} \, dt}\tag{A.3.2}$$

3b) Inserting the definition of $H$ as an infinite sum, leaves another infinite sum

$$s=\int_0^1 \log (x) \sum _{m=1}^{\infty } \left(\frac{1}{m}-\frac{1}{m-x}\right) \, dx \\ =\sum _{m=1}^{\infty } \int_0^1 \left(\frac{1}{m}-\frac{1}{m-x}\right) \log (x) \, dx=\sum _{m=1}^{\infty } c(m)\tag{A.3.3}$$

with

$$c(1)=\frac{1}{6} \left(\pi ^2-6\right)\tag{A.3.4}$$

and

$$c(m\gt1)=\int_0^1 \left(\frac{1}{m}-\frac{1}{m-x}\right) \log (x) \, dx \\ =-\operatorname{Li}_2\left(\frac{m-1}{m}\right)-\frac{1}{m}-\log ^2(m)+\log (m-1) \log (m)+\frac{\pi ^2}{6}\tag{A.3.5}$$

This can be simplified appreciable using the transformation formula

$$\text{Li}_2\left(\frac{m-1}{m}\right)=-\text{Li}_2\left(\frac{1}{m}\right)-\log \left(\frac{1}{m}\right) \log \left(\frac{m-1}{m}\right)+\frac{\pi ^2}{6}$$

to give

$$c(m) = \text{Li}_2\left(\frac{1}{m}\right)-\frac{1}{m}$$

so that we have found a complicated way to regain exactly $(A.1)$.

4) Replace zeta by an integral, leaves another integral

We have

$$\zeta (k)=\frac{1}{\Gamma (k)}\int_0^{\infty } \frac{t^{k-1}}{e^t-1} \, dt\tag{A.4.1},$$

so that our sum becomes

$$s=\sum_{k\ge2} \frac{1}{k^2}\frac{1}{\Gamma (k)}\int_0^{\infty } \frac{t^{k-1}}{e^t-1} \, dt=\int_0^{\infty } \frac{1}{e^t-1}\left( \sum_{k\ge2}\frac{1}{k^2}\frac{t^{k-1}}{\Gamma (k)}\right)\, dt \\ =\int_0^{\infty } \frac{-\log (-t)-\Gamma (0,-t)-e^t \Gamma (2,t)-\gamma +1}{t \left(e^t-1\right)} \, dt\tag{A.4.2}$$

We can simplify the integrand.

The incomplete gamma function is defined as

$$\Gamma (r,y)=\int_y^{\infty }x^{r-1} \exp (-x)\, dx\tag{A.4.3}$$

This gives

$$\Gamma (2,t) =e^{-t} (t+1)\tag{A.4.3a} $$

and we can see (e.g. by plotting) that the combination

$$-\log (-t)-\Gamma (0,-t)\tag{A.4.3b}$$

is real for all real $t$. I don't know a name for this expression.

This gives finally

$$s = \int_0^{\infty } \frac{1}{t(1-e^t)} \left(t+\log (-t)+\Gamma (0,-t)+\gamma \right)\, dt\tag{A.4.4}$$

which coincides with the last formula of $(A.3.2)$.

5) Generating functions

Defining the generating functions analogous to $(3)$

$$g(q,z) =\sum_{k\ge2} \frac{z^k}{k^q}\zeta(k)\tag{A.5.1}$$

we have

$$g(0,z) = -z (\psi ^{(0)}(1-z)+\gamma ) = - z H_{-z}\tag{A.5.2}$$

and the sequence

$$g(q,z)=\int_{0}^z \frac{g(q-1,y)}{y}\,dy, q=1,2,\ldots \tag{A.5.3}$$

Giving

$$g(1,z)=\int_0^z H_{-y} \, dy=\gamma z-\operatorname{\log\Gamma}(1-z)\tag{A.5.4}$$

and the g.f. we are looking for

$$g(2,z)=\gamma z-\int_0^z \frac{1}{y}\operatorname{\log\Gamma}(1-y) \, dy=\text{?}\tag{A.5.5}$$

This one we have encountered already in $(A.3)$.

Notice that interestingly $\lim_{z\to -1} \, g(0,z)=1$ in spite of the fact, that the series diverges. In fact, there is no limit but two partial sums with even and odd parity tend to $\frac{1}{2}$ and $\frac{3}{2}$, respectively, i.e. the sequence has two accumulation points, and their arithmetic mean is $=1$.

6) Complex contour integral

I am not sure if this approach could lead to a closed expression but it might be interesting.

Representing the infinite series as a complex contour integral with the "kernel function" $H_{-z}$ and a path coming from $i+\infty$, going to $i+\frac{3}{2}$, to $-i+\frac{3}{2}$, and then back to $-i+\infty$, then bending the path around we arrive at the following representation of our sum $s$:

$$s=2-\gamma -\frac{1}{2 \pi }\int_{\frac{1}{2}-i\infty }^{\frac{1}{2}+i \infty } \frac{H_{-z} \zeta \left(z\right)}{z^2} \, dz\tag{6.1}$$

where the terms before the integral are the residue of the integrand at $z=1$:

$$\text{Res}\left(\frac{H_{-z} \zeta (z)}{z^2}\right)|_{z=1} =-2 + \gamma\tag{6.2}$$

Notice that the integral is taken on the critical strip where the zeta function has its non-trivial zeroes (if Riemann was right).