13
$\begingroup$

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 :$\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}}$

$\endgroup$
  • 1
    $\begingroup$ Since $$\sum_{k\geq 2}\zeta(k)\frac{x^k}{k} = -\gamma x+\log\Gamma(1-x) $$ we have $$\sum_{k\geq 2}\frac{\zeta(k)}{k^2}=-\gamma+\int_{0}^{1}\frac{\log\Gamma(1-x)}{x}\,dx\stackrel{\text{IBP}}{=}-\gamma+\int_{0}^{1}\psi(x)\log(1-x)\,dx $$ so it might be useful to have closed forms for $$\int_{0}^{1}x^m \psi(x)\,dx.$$ $\endgroup$ – Jack D'Aurizio Mar 5 '20 at 16:19
  • $\begingroup$ @Jack D'Aurizio I've thought about that, but the integral is even more complicated (see my answer (A.3)). $\endgroup$ – Dr. Wolfgang Hintze Mar 5 '20 at 17:04
7
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.