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:

$$\xi_{int}(s)=\xi_{int}(2(0)-s)$$

$$\xi_{rie}(s)=\xi_{rie}(2(\frac12)-s)$$

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?

Thanks.