# A Hadamard product representation for Keiper's $\tau$-function?

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:

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:

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

$$\frac{\xi'(s)}{\xi(s)}=\sum_\rho \frac1{s-\rho}+\frac1{\rho} = -\sum_{j\ge 1} s^j\rho^{-j-1}$$ 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.

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