Define the function $F(s)$ as the Dirichlet series $$ F(s) = \sum_{n=1}^\infty \frac{1}{(n+1)n^{s-1}}, $$ which converges for $\operatorname{Re}(s)>1$.

Has anyone seen/studied this function before? It has the property that $$\zeta(s) = F(s)+F(s+1), \label{1} \tag{1} $$ where $\zeta(s)$ is the Riemann zeta function. This is trivial to check: $$ F(s)+F(s+1) = \sum_{n=1}^\infty \frac{n}{(n+1)n^s} + \sum_{n=1}^\infty \frac{1}{(n+1)n^s} = \sum_{n=1}^\infty \frac{n+1}{n+1} \cdot \frac{1}{n^s} = \zeta(s). $$ It also has several other properties which make it seem interesting. I'll start by giving some motivation and then describe the properties I noticed, and ask some related questions.

Motivation for studying $F(s)$:

  1. If we denote $G(s) = F(2s)$, a trivial coordinate change, then \eqref{1} becomes $\zeta(s)=G(s/2)+G((s+1)/2)$. This can be interpreted as the statement that $\zeta(s)$ is the image of $G(s)$ under the linear operator $T$ that sends an analytic function $g(s)$ to the function $T(g):=g\left(\frac{s}{2}\right) + g\left(\frac{s+1}{2}\right)$.

    This linear operator has some very nice properties. Notably, it is "Riemann-hypothesis preserving", in the following sense: if $g(s)$ is a complex polynomial with all its zeros on the line $\operatorname{Re}(s)=1/2$, then $T(f)$ is also such a polynomial. (This is a reformulation, via a simple coordinate change, of the statement that the operator that maps a polynomial $h(z)$ to $h(z-i/4)+h(z+i/4)$ is "hyperbolicity preserving", that is, sends polynomials with only real zeros to polynomials with only real zeros -- a special case of a more general result that I think is due to Polya).

    This hyperbolicity/RH preservation property also extends to certain entire functions of genus 0 or 1 (usually assumed to satisfy the functional equation $f(1-s)=f(s)$). See section 10.23 of Titschmarsh's book "The Theory of the Riemann Zeta Function", 2nd ed. This was used in some of Polya's failed (though still very interesting) attacks on the Riemann hypothesis. I've seen it used in some other places, for example in the paper "The Riemann hypothesis for certain integrals of Eisenstein series" by Lagarias and Suzuki.

    Now, $\zeta(s)$ is not an entire function, and doesn't exactly satisfy $\zeta(1-s)=\zeta(s)$, so the relevance of these properties of the linear operator $T$ to RH-type questions isn't entirely clear. But the equation $\zeta(s)=F(s)+F(s+1)$ is still suggestive of the possibility that by studying $F(s)$ and its zeros we could learn something about $\zeta(s)$ and its zeros, using these sorts of Polya-style ideas.

  2. The same linear operator $T$ (or rather, $\frac{1}{2}T$) is also the transfer operator for the doubling map $x\mapsto 2x \bmod 1$ from ergodic theory.

  3. The Bernoulli polynomials $B_n(x)$ (which are themselves related to $\zeta(s)$ in all sorts of ways) are the eigenfunctions of this transfer operator, satisfying the equation $$ T [B_n] = \frac{1}{2^{n-1}} B_n, $$ and they also have the symmetry $B_n(1-x)=B_n(x)$.

Properties of $F(s)$:

I've observed the following facts about $F(s)$:

  1. By rewriting \eqref{1} as $F(s) = \zeta(s) - F(s+1)$, one sees that this gives a way to analytically continue $F(s)$ to the region $\operatorname{Re}(s)>0$, then inductively to the region $\operatorname{Re}(s)>-1$, $\operatorname{Re}(s)>-2$, etc, using the formulas \begin{align*} F(s) &= \zeta(s) - F(s+1) \\ &= \zeta(s) - \zeta(s+1) + F(s+2) \\ &= \zeta(s) - \zeta(s+1) + \zeta(s+2) - F(s+3) = \ldots \\ &= \zeta(s) - \zeta(s+1) + \zeta(s+2) - \zeta(s+3) + \ldots + (-1)^k \zeta(s+k) + (-1)^{k+1} F(s+k+1) \end{align*} This shows that $F(s)$ can be analytically continued to a meromorphic function on $\mathbb{C}$, with poles at $s=1, 0, -1, -2, \ldots$. The pole at $s=n$ for integer $n\le 1$ is a simple pole with residue $(-1)^{n-1}$.

  2. $F(s)$ has the special values \begin{align*} F(2) &= 1, \\ F(3) &= -1 + \zeta(2), \\ F(4) &= 1 - \zeta(2) + \zeta(3), \\ &\ \ \vdots \\ F(n) &= (-1)^n + \sum_{k=2}^{n-1} (-1)^{n+k+1} \zeta(k). \end{align*} (Proof: the evaluation of $F(2)$ is a trivial telescoping series, and the other ones follow from it by induction using \eqref{1}.)

  3. $F(s)$ has no zeros in the region $\operatorname{Re}(s)>2$.

    Proof: if $\operatorname{Re}(s)>2$ then it is easily checked that the first term $\frac{1}{2}$ in the series defining $F(s)$ dominates the sum of the remaining terms.

  4. Numerically using Mathematica I found some zeros of $F(s)$, at these points: \begin{align*} Z_1 &\approx 0.901294 + 14.11648 i, \\ Z_2 & \approx 0.85788022 + 21.0356764 i, \\ Z_3 & \approx 0.8389893 + 24.982853 i, \\ Z_4 & \approx 0.821207 + 30.4765 i, \\ Z_5 & \approx 0.812678 + 32.8777 i, \\ Z_6 & \approx 0.0987755 + 1.27788 i, \\ Z_7 & \approx -1.47031 + 1.65906 i. \end{align*}

  5. The functional equation for $\zeta(s)$ can be reformulated in terms of an auxiliary function related to $F(s)$. More precisely, define \begin{align*} q(s) &= \pi^{-s/2} \Gamma\left(\frac{s}{2}\right), \\ H(s) &= q(s) F(s) - q(1-s) F(2-s). \end{align*} Then $H(s)$ is a meromorphic function. With this notation, the functional equation for $\zeta(s)$ $$q(s)\zeta(s) = q(1-s)\zeta(1-s), \label{2} \tag{2}$$ can be rewritten as $$ q(s) F(s) + q(s) F(s+1) = q(1-s) F(1-s) + q(1-s) F(2-s), $$ or, rearranging terms, $$ q(s) F(s) - q(1-s) F(2-s) = q(1-s) F(1-s) - q(s) F(s+1), $$ This is simply the statement that $$ H(1-s) = H(s). \label{3} \tag{3} $$ So we see that \eqref{2} and \eqref{3} are equivalent.

  6. $F(s)$ has the Mellin transform representation $$ F(s) = \frac{1}{\Gamma(s-1)} \int_0^{\infty} \left(- e^x \log(1-e^{-x}) - 1\right) x^{s-2} dx \qquad (\operatorname{Re}(s)>2). $$ (Proof: expand the integrand in a series in powers of $e^{-x}$ and integrate termwise.)

My questions

  1. Has either of the functions $F(s)$ or $H(s)$ been studied before?

  2. Can someone see a way to derive the functional equation $H(1-s)=H(s)$ directly, without going through the functional equation for $\zeta(s)$? (This would give a new proof for the functional equation of $\zeta(s)$.)

  3. Do the functions $F(s)$ and $H(s)$ have any significance? Can they be used to prove anything interesting about $\zeta(s)$, or about any number theoretic quantities? Do they come up in any natural context?

  4. What can be said about the location of the zeros of the functions $F(s)$ and $H(s)$? Does the location of the zeros have any importance for anything?

  5. Can anyone see other interesting properties that $F(s)$ and $H(s)$ satisfy?

  • $\begingroup$ As noted in Property 1 of the original question, every function $f(s)$ is formally of the form $g(s) + g(s+1)$: just let $g(s) = f(s) - f(s+1) + f(s+2) - f(s+3) + ...$. If $f(s)$ is a Dirichlet series then one can check that the sum defining $g(s)$ converges. This suggests to me that the expression $\zeta(s) = F(s) + F(s+1)$ will not be useful. In particular, the fact that you have a formula for $F(s)$ is just a distraction. $\endgroup$ Aug 29, 2022 at 2:27
  • $\begingroup$ @DavidFarmer property 1 of which question? Are you seeing a link somewhere to another question? (This one, perhaps?) $\endgroup$
    – Dan Romik
    Aug 29, 2022 at 3:11
  • 2
    $\begingroup$ @DavidFarmer I do enjoy messing around with cool formulas whether they are useful or not. But as for your “not useful” remark, surely as a general principle it’s not always true that explicit formulas are a distraction and don’t add anything useful as compared to general/more abstract ways of looking at things? In other words, I’m not sure I share your intuition on the usefulness issue, though you may be right of course. $\endgroup$
    – Dan Romik
    Aug 29, 2022 at 3:14
  • $\begingroup$ @DanRomik There is a typo in the 3rd displayed equation: $n^{s-1}$ should be $n^s$ right before the "${}= \zeta(s)$". It's nothing important, but I thought you may want to fix it. $\endgroup$ Aug 29, 2022 at 4:48
  • $\begingroup$ @SalvoTringali fixed, thanks. $\endgroup$
    – Dan Romik
    Aug 29, 2022 at 6:49

1 Answer 1


This is not an answer but a quick remark about a possible direct proof of the functional equation for $F$.

First recall the identity $$\int_0^{\infty} x^{({s-3})/2}\exp(-n^2\pi x)\,dx=\pi^{-(s-1)/2}\ \Gamma\left(\frac{s-1}{2}\right) n^ {-(s-1)}\,\mathrm{for}\, \Re(s)>2.$$ Multiplying both sides by $1/(n+1)$ and summing for $n$ we get $$\int_0^{\infty} x^{({s-3})/2}Q(x)\,dx\ =\pi^{-(s-1)/2}\ \Gamma\left(\frac{s-1}{2}\right)F(s),$$ where $Q(x)=\sum_{n=1}^\infty\frac{\exp(-n^2\pi x)}{n+1}$.

Now the proof of the functional equation for $\zeta$ relies on the fact that the function $\psi(x)=\sum_{n=1}^\infty{\exp(-n^2\pi x)}$ satisfies $$2\psi(x)+1={\frac{1}{x^{1/2}}}\left[2\psi\left(\frac{1}{x}\right)+1\right].$$ So maybe there is some similar property for $Q$ that lead to a functional equation for $F$.

About the zeros, since $\zeta$ has infinitely many zeros on the line $\Re(s)=1/2$, the obvious thing is that if your equation holds for $\Re(s)>0$, then $F$ must have infinitely many zeros on the lines $\Re(s)=3/2$ and $\Re(s)=1/2$

So, if $\rho$ is a zero of $\zeta$ then $\rho$, $1-\rho$, $2-\rho$, and $1+\rho$ are also zeros of $F$.

  • $\begingroup$ Interesting, thanks. How does your suggested functional equation $q(s-1)F(s)=q(1-s)F(2-s)$ relate to the equation $q(s)(F(s)+F(s+1))=q(1-s)(F(1-s)+F(2-s))$? In any case, I tested your equation numerically and it does not seem to be satisfied unfortunately. $\endgroup$
    – Dan Romik
    Aug 28, 2022 at 18:18
  • $\begingroup$ My functional equation was pure speculation. The main idea is that you should try to study $Q(x)$ in order to find some identity like the one for $\psi$. If such identity exist (you could start with Poisson formula) you might have a proof for the functional equation of $F$ that looks like the proof for the functional equation of $\zeta$. $\endgroup$
    – Gui
    Aug 28, 2022 at 22:36
  • $\begingroup$ But the statement about the zero of $F$ remains valid if your equation holds for at least $re(s)=>1/2$. If not, I'm not sure how you can deduce anything for the zero of $\zeta$ starting from $F$ (the $=>$ means greater or equal). $\endgroup$
    – Gui
    Aug 28, 2022 at 22:50
  • 3
    $\begingroup$ @Guigui, wouldn't $>=$ be better for an inequality...? Or, just use MathJax/(La)TeX? :) $\endgroup$ Aug 28, 2022 at 23:06
  • $\begingroup$ It is, I'm not used to Latex sorry! $\endgroup$
    – Gui
    Aug 28, 2022 at 23:07

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