In this paper J.B. Keiper defined the following function:

$$\tau_k = \sum_{j=1}^k (-1)^j\,{k-1 \choose j-1} \sigma_{j+1} \qquad k \ge 1, k \in \mathbb{N} \tag{1}$$

where $\displaystyle\sigma_r = \sum_{\rho}\frac{1}{\rho^r}$, with $\rho$ a non-trivial zero of $\zeta(s)$ and taken in pairs $\rho, 1- \rho$.

Keiper also notes that $\tau_k = \lambda_{k+1} + \lambda_{k-1} -2\lambda_k$ where $\lambda_k$ is Li's constant ($=k\lambda_k^{Keiper})$. These $\lambda_k$ can be expressed in terms of the (paired) non-trivial zeros:

$$\lambda_k = \sum_{\rho,1-\rho} \left(1-\left(\frac{\rho}{\rho-1}\right)^k+1-\left(\frac{\rho-1}{\rho}\right)^k \right)$$

and from this it follows that $\lambda_{-k} = \lambda_{k}$ and $\tau_{-k} = \tau_{k}$.

Numerical evidence suggests this could be continued further into this entire function:

$$\tau_s = \sum_{j=1}^\infty\, (-1)^j\, {s-1 \choose j-1} \sigma_{j+1} \qquad s \in \mathbb{C} \tag{2} $$

Below is a plot to illustrate its behaviour. The RH implies the oscillations must stay (at integer values) within the two bounds and vice versa:

enter image description here

With $\tau_s$ being an entire function, it should have a Hadamard product factorisation into its zeros. After computing the first few of them, I found only a single complex zero $z=55.309... + 12.826...i$ (i.e. inducing a party of 4 complex zeros in total). All other zeros appear to be real and come in pairs $(\mu_{n},\mu_{-n})$, with $\mu_{1..24}$ shown below:

enter image description here

All the typical $e^{(.)}$-factors in the Hadamard product nicely cancel out and what remains is:

$$\tau_s = \tau_0\left(1-\frac{s^2}{z^2}\right)\left(1-\frac{s^2}{\overline{z^2}}\right)\prod_n\left(1-\frac{s^2}{\mu_n^2}\right)\left(1-\frac{s^2}{\overline{\mu_n^2}}\right) \tag{3}$$

where $\tau_0 = \sigma_1 = 1 + \dfrac{\gamma-\log(\pi)}{2}-\log(2)$. Numerical results look pretty good already at 24 real zeros.

Q: Is there anything more to learn from the existence of such a canonical Hadamard product? Could it say anything about the oscillating behaviour of $\tau_s$ or the distribution of its zeros?

Note 1:

$\tau_s$ could also be expressed in terms of $\rho$'s:

$$\tau_s = \sum_{\rho,1-\rho} \left(\frac{\left(\frac{\rho}{\rho-1}\right)^s}{\rho^2}+\frac{\left(\frac{\rho-1}{\rho}\right)^s}{(1-\rho)^2)}\right)\qquad s\in \mathbb{C}, \Re(s) \ge 0$$

$$\tau_s = \sum_{\rho,1-\rho} \left(\frac{\left(\frac{\rho}{\rho-1}\right)^s}{(1-\rho)^2}+\frac{\left(\frac{\rho-1}{\rho}\right)^s}{\rho^2}\right)\qquad s\in \mathbb{C}, \Re(s) \le 0 $$

Note 2:

Also found that $\tau_s$ could be split into a smooth first term and an oscillating second term:

$$\tau_s=\sum_{k=1}^{\infty} \left( {s+k-1\choose k-1}\cdot \frac{\zeta(k+1)}{(-2)^{k+1}}- {-s+k-1\choose k-1} \sigma_{k+1}^*\right)$$

with $\sigma_k^*$ defined here. The first term is asymptotic to $\dfrac{1-2^{-s}}{2s}$ for $\Re(s) \gt 0$.

Note 3:

It seems also possible to expand the domain of $\lambda_k$:

$$\lambda_s = \sum_{j=1}^{\infty} \, (-1)^{j+1}\, {s \choose j} \sigma_{j} \qquad s \in \mathbb{C} \tag{4}$$

and of $\lambda_k - \lambda_{k+1}$:

$$\lambda_s - \lambda_{s+1} = \sum_{j=1}^\infty\,{s + j - 1 \choose s} \sigma_{j} \qquad s \in \mathbb{C} \tag{5} $$

However, their Hadamard products turn out to be less elegantly compared to the one for $\tau_s$. Product (4) seems to have only complex zeros (which makes sense, since the positivity of $\lambda_k$ for all $k$ also implies the RH, hence no real zeros are expected to be found). Product (5) also appears to have a single real zero at $s=-\frac12$ and complex zeros otherwise.


1 Answer 1


$$\frac{\xi'(s)}{\xi(s)}=\sum_\rho \frac1{s-\rho}+\frac1{\rho} = -\sum_{j\ge 1} s^j\rho^{-j-1}$$ the Taylor series is valid for $|s|<|1/2+14.13i|$

Thus the coefficients have exponential decay which implies that $$ \sum_{j=0}^\infty\, (-1)^j\, {z-1 \choose j} \sigma_{j+2}$$ is entire.

  • $\begingroup$ Thanks! A proof like this is obviously much better than my numerical assessment. $\endgroup$
    – Agno
    Jan 11, 2021 at 12:20

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