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, 2017 at 21:15

1 Answer 1


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.