Assuming the RH and $s \in \mathbb{C}, \rho_n =\frac12 \pm i\gamma_n$, the following (altered) Hadamard product:

$$\displaystyle \displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{\frac12+ (-1)^n i \gamma_n} \right) \left(1- \frac{s}{\frac12+ (-1)^{n+1} i \gamma_n} \right) = \frac{\xi_{rie}(s)}{\xi_{rie}(0)}$$

runs through the alternating non-trivial zeros $\rho_n$ with $\xi_{rie}(s)= \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right) \zeta(s)$.

Contrary to the factors of the original Hadamard product, these alternating factors do converge.

This question suggest that many similarities exist between infinite (Hadamard/Weierstrass) products using $\gamma_n=n$ and $\gamma_n=\Im(\rho_n)$, and this question shows that a closed form for alternating factors using $\gamma_n=n$ does exist. I therefore like to conjecture that also a closed form exists for the alternating formula above.

Let's call the closed form for each factor $A_-$ and $A_+$ and it is easy to see that:

$$\displaystyle A_-A_+=\frac{\xi_{rie}(s)}{\xi_{rie}(0)}=s(s-1) \pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right) \zeta(s)$$

is the entire function to be split into two factors.

Splitting this function is easy to do for the Gamma-part: $G_-G_+=s(s-1) \pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right)$, that for instance (there are more ways) could be factored into:

$$\displaystyle G_-=s\Gamma\left(\frac{s}{4}\right) \pi^{-\frac{s}{4}}2^{\frac{s}{4}-\frac32} \text{ and } G_+=(s-1)\Gamma \left(\frac{s}{4}+\frac12\right) \pi^{-(\frac{s}{4}+\frac12)}2^{\frac{s}{4}+\frac12}$$

But what to do with $\zeta(s)$?

The poles of $G_-$ (-4,-8,...) and $G_+$ (-2,-6,...) might provide some hints, since they need to be annihilated by the zeros of the 'to be found' $\zeta(s)$-factors. It is also clear that a $\zeta(s)$-factor must now induce alternating non-trivial zeros only i.e.: $\frac12+14.134...i,\frac12-21.022...i,\frac12+25.010...i, \dots$ (and its complement). Dividing the infinite product factors $A_-$ en $A_+$ (using n=699) by $G_-$ and $G_+$ respectively, one gets the following graphs of what the (absolute) $\zeta(s)$-factors might look like:

Question (apologies for the long intro):

The graphs of the two potential factors for $\zeta(s)$ above, could both be seen as analytically continued functions across $\mathbb{C}/1$, that have been derived "bottom up" from their alternating zeros. This would imply that in the domain $\Re(s)>1$ the two factors must multiply into:

$$\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s} = \prod _{p \in \mathbb{P}}(1-p^{-s})^{-1}$$

Hence my question: are there any ways to split, the known analytically "discontinued" expressions for $\zeta(s)$ in the domain $\Re(s)>1$, into two factors that each can be analytically continued again?


(1) I have f.i. tried splitting the Euler product into its $(p \mod 4 = 1)$ and $(p \mod 4 = 3)$ factors, however did not see any way to analytically continue these.

(2) Also hoped to find some 'natural' connection with the alternating Zeta: $\eta(s)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s}$ that is valid over the domain $\Re(s)>0$, however so far have been unsuccessful in factoring it further.

  • $\begingroup$ You (or Martin Sleziak) wrote "en". Is that Dutch for "and" ? $\endgroup$ May 12 '17 at 21:15

Haven't found a full answer yet, however did discover a nice way (very different from the approach in the question) to split $\zeta(s)$ into two factors that are valid in $\mathbb{C}{/1}$ and that each induce the alternating conjugates of the $\rho$s.

With $\displaystyle \chi(s)=\pi^{-\frac{s}{2}}\,\Gamma\left(\frac{s}{2}\right)$ and $K(s)=\Psi\left(\frac{s}{2}\right)-\ln(\pi)$, with $\Psi\left(s\right)$ the Digamma function, then:

$$\zeta(s):= -\dfrac{2}{\chi(s)}\cdot\dfrac{\chi(1-s)\cdot\zeta'(1-s)+\chi(s)\cdot\zeta'(s)}{K(1-s)+K(s)}\qquad(1)$$

This allows the $\zeta(s)$ function to be split into (assuming principal branches for the squared roots):

$$\zeta_1(s):= i\cdot \sqrt{\dfrac{2}{\chi(s)}}\cdot\dfrac{\sqrt{\chi(1-s)\cdot\zeta'(1-s)}+i\cdot\sqrt{\chi(s)\cdot\zeta'(s)}}{\sqrt{K(1-s)}+i\cdot\sqrt{K(s)}}$$


$$\zeta_2(s):= i\cdot\sqrt{\dfrac{2}{\chi(s)}}\cdot\dfrac{\sqrt{\chi(1-s)\cdot\zeta'(1-s)}-i\cdot\sqrt{\chi(s)\cdot\zeta'(s)}}{\sqrt{K(1-s)}-i\cdot\sqrt{K(s)}}$$

so that $\zeta(s)=\zeta_1(s)\cdot\zeta_2(s)$.

The graph below shows a plot of both functions and how the zeros alternate. The additional pair of roots induced by the numerator at $\frac12 \pm 6.28...$ is exactly cancelled by the denominator.

enter image description here

Note that multiplying both sides of (1) by $\frac12\, s(s-1)\, \chi(s)$ gives the Riemann $\xi(s)$ function as:


Of course this can also be split into factors with e.g. $\sqrt{s\,(s-1)}$ and I had hoped to then link these to each alternating factor of the Hadamard product in the question (but no connection found).

The main issue I encountered is that although the $\zeta_1(s)$ and $\zeta_2(s)$ do nicely behave in the complex plane, in the real domain there are discontinuities induced by the real roots of $\zeta'(s)$ in the numerator and by roots of $K(s)$ in the denominator, that don't nicely annihilate each other (since squared roots are taken).


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