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Introduction:

In this 1992 paper, J.B. Keiper (an amazing person, who tragically died way too young), derives several power series expansions of the Riemann $\xi$-function that involve infinite sums of powers of the non-trivial zeros $\rho$ (taken in pairs).

$$\sigma_k= \sum_{\rho} \frac{1}{\rho^k}$$

Analogy:

We know that the non-trivial zeros $\rho$ of $\zeta(s)$ drive the oscillating (error) term in the prime counting function(s). Analogously, the zeros $\mu = 0 \pm 2\pi n i$ induce the oscillating (sawtooth) behaviour in the integer counting function. The entire function associated with the $\mu$'s is $\xi_i(s)=\frac{2}{s}\sinh\left(\frac{s}{2}\right)$. Now also define ($\mu$'s taken in pairs):

$$\hat{\sigma}_k= \sum_{\mu} \frac{1}{\mu^k} = \frac{\zeta(1-k)}{\Gamma(k)}$$

Comparison:

I found that all the expansions for $\xi(s)$ from Keiper's paper can be replicated for $\xi_i(s)$ and the results/relations are fully equivalent except for these three:

\begin{align} \sum_{k=1}^\infty \frac{\sigma_k}{k} &=0 \qquad \tag{Kei 16}\\ \sigma_1 &= -\sum_{k=1}^\infty \sigma_k \qquad \tag{Kei 17}\\ \sigma_{j+1} &= (-1)^{j+1} \sum_{k=1}^\infty \binom{k-1}{j}\sigma_k \qquad \tag{Kei 18} \end{align} where in the latter series a $\sigma$ could be derived recursively. This recursion disappears in the equivalent series with the $\hat{\sigma}$'s: \begin{align} \sum_{k=1}^\infty \frac{\hat{\sigma}_k}{k} &=\ln\left(\frac{1}{{\rm e}-1}\right) \\ \frac{1}{{\rm e}-1} &= -\sum_{k=1}^\infty \hat{\sigma}_k \\ -1 + \frac{1}{\Gamma(j+1)} \sum_{k=0}^{j} \Gamma(j+1-k){j+1 \brace j+1-k} ({\rm e}-1)^{k-j-1} &= (-1)^{j+1} \sum_{k=1}^\infty \binom{k-1}{j}\hat{\sigma}_k \\ \end{align} where ${a \brace b}$ is a Stirling number of the second kind. Question:

Where could the differences in recursion between these series of $\sigma$ and $\hat{\sigma}$ originate from? I could conceive of two possible explanations:

  • the functional equation $\xi(s)=\xi(1-s)$ versus $\xi_i(s) = \xi_i(-s)$ or
  • some connection in the recursion of $\sigma$ to the fundamental theorem of arithmetic?

Or maybe there is a simpler rule at work here? Appreciate any ideas.

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  • $\begingroup$ What do you mean by '$\sigma$ can be derived from series of $\sigma$'? Is this just (Kei 18) restated? $\endgroup$
    – Stopple
    Aug 11, 2021 at 15:39
  • $\begingroup$ @Stopple Yes it is. $\endgroup$
    – Agno
    Aug 11, 2021 at 18:03
  • $\begingroup$ It might be clearer to simply say that is is a recursion for the $\sigma$ $\endgroup$
    – Stopple
    Aug 11, 2021 at 22:12

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