Take the following two infinite products that have closed forms.

Assume: $\gamma_n > 0 \in \mathbb{R};s,a,x \in \mathbb{C}; x \ne 0,a \pm ix\gamma_n \ne 0$

The first product:

$$\displaystyle H_{int}(s,a,x) := \prod_{n=1}^\infty \left(1- \frac{s}{a + i x \gamma_n} \right) \left(1- \frac{s}{{a - i x \gamma_n}} \right) = \frac{\xi_{int}(0 -\frac{a}{x}+\frac{s}{x})}{\xi_{int}(0-\frac{a}{x})}$$

has $\gamma_n = n$, so runs through the integers with: $\xi_{int}(s) = \frac{\sinh(\pi s)}{s}$.

The second product, for which the closed form can be derived assuming RH is true,

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

has $\gamma_n = \Im(\rho_n)$, so runs through the non-trivial zeros $\rho_n$ with: $\xi_{rie}(s)= \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$.

The products clearly have a comparable structure and share the following characteristics:

1) They are both reflexive: $H(s,a,x) = H(2a-s,a,x)$ (since this is fully independent of $\gamma_n$).

2) Their $\xi(s)$-functions are reflexive as well:



3) They both have a similar 'base' Hadamard product (for $a=0$ and $a=\frac12$ respectively):

$$\displaystyle \xi_{int}(s) = \xi_{int}(0) \prod_{n=1}^\infty \left(1- \frac{s}{0+i n} \right) \left(1- \frac{s}{{0- i n}} \right) \rightarrow \xi_{int}(0)=\pi$$

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

4) They are both entire functions and the "undesired" zeros/poles from their meromorphic components $\sinh(s)$ and $\zeta(s)$ are annihilated via $s$ and $\frac12 s(s-1), \Gamma(\frac{s}{2})$ respectively. Both meromorphic elements can be expressed as an infinite product as well as an infinite sum (over a certain domain):

$$\sinh(s) = s \prod_{k=1}^\infty \left(1+ \frac{s^2}{k^2\pi^2)} \right)=\sum_{n=1}^\infty \frac{s^{2n-1}}{\Gamma(2n)}$$

$$\zeta(s) = \prod_{p} \left(\frac{1}{1-p^{-s}} \right)=\sum_{n=1}^\infty \frac{1}{n^s}$$

5) And both of the above can be analytically continued throughout the entire complex plane via:

$$\sinh(0-s) = -\sinh(s)$$

$$\zeta(1-s) = \chi(s)\zeta(s)$$

My questions:

1) Is there any other possible (or known) set of values for $\gamma_n$ that could yield yet another closed form? Or is this all there is, i.e. are the ("via parameter $x$ linearly scalable") integers $\gamma_n=n$ at $a=0$ and $\gamma_n =\Im(\rho_n)$ at $a=\frac12$, the only possible choices?

2) Since the $\rho_n$'s contain information about the (distribution of) primes and the primes in turn can induce the integers via unique multiplication, could there be a connection made between (the closed forms of) these two products?


  • $\begingroup$ Very interesting observation! $\endgroup$ Feb 23, 2013 at 0:59

1 Answer 1


In my quest for similar closed forms as the above, I did find:

$$\displaystyle H_{int^2}(s,a,x) := \prod_{n=1}^\infty \left(1- \frac{s}{a + i x \gamma_n} \right) \left(1- \frac{s}{{a - i x \gamma_n}} \right) = \frac{\xi_{int^2}(0 -\frac{a}{x}+\frac{s}{x})}{\xi_{int^2}(0-\frac{a}{x})}$$

with $\gamma_n = n^2$, running through the squared integers and $\xi_{int^2} = {\frac {\sin \left( \left(\frac12-\frac12i \right) \sqrt{2s} \ \pi \right) \sin \left( \left(\frac12+\frac12i \right) \sqrt{2s} \ \pi \right)}{s}}$

I do believe more closed forms exist, however only for $\gamma_n = n^{2k}$ with ($k= 1,2,3...)$ and would like to conjecture that for the 'base' Hadamard product it is true that:

$$\displaystyle \xi_{int^{2k}}(s) = \xi_{int^{2k}}(0) \prod_{n=1}^\infty \left(1- \frac{s}{0+i n^{2k}} \right) \left(1- \frac{s}{{0- i n^{2k}}} \right) \rightarrow \xi_{int^{2k}}(0)=\pi^{2k}$$

However, my (maybe too big) dream was to find a closed form for $\gamma_n = p_n$ with $p_n$ being the n-th prime. Via brute force calculations, I did find a few (all explainable) 'leads' for:

$$\displaystyle H_{prime}(s,a,x) := \prod_{n=1}^\infty \left(1- \frac{s}{a + i x p_n} \right) \left(1- \frac{s}{{a - i x p_n}} \right)$$

$$H_{prime}(1,1,1) = \frac{\pi^2}{15} = \frac{\zeta(4)}{\zeta(2)}$$ $$H_{prime}(1,\frac12,1) = 1$$ $$H_{prime}(i,i,1) = \frac{\pi^2}{6} = \zeta(2)$$

and the general rules, that also apply to the above: $$H_{prime}(s,a,x) = H_{prime}(2a-s,a,x)$$ $$H_{prime}(s,a,x) = \frac{1}{H_{prime}(s,1-a,x)}$$

I can also prove that a closed (entire) form must have the shape: $\dfrac{\xi_{prime}(\beta -\frac{a}{x}+\frac{s}{x})}{\xi_{prime}(\beta-\frac{a}{x})}$, however the key snag is that the meromorphic component of $\xi_{prime}(s)$ that generates the zeros (similar to $\zeta(s)$), must do so only at the primes, whilst all extra "undesired and maybe more trivial" zeros need to annihilated by other components of the entire function $\xi_{prime}(s)$.

This brought me immediately to Wilson's formula, where I used the $\Gamma$-function instead of the factorial and replaced the $mod$ by the $cos$ to produce a function that generates zeros only at integers that are prime:


however it unfortunately also generates a tremendous amount of 'undesired, but hopefully more trivial' (non-integer) zeros that I need to annihilate. Is there any known function (other than using the floor function) that could accomplish this? Any other ideas\steers?



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