A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH? Building on this question scaling the imaginary part of $\rho$s in infinite products, I like to conjecture that:
$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{1-\mu_n} \right)\left(1- \frac{s}{\overline{\mu_n}} \right) \left(1- \frac{s}{\overline{1-\mu_n}} \right)$$
with $\mu_n = a + \Im(\rho_n)x i$ and $a,x \in \mathbb{R},x \ne 0, s \in \mathbb{C}$ and $\rho_n$ the n-th non-trivial zero of $\zeta(z)$, 
has the following closed form:
$$\displaystyle H(s,a,x):= \frac{\xi(\frac12 - \frac{a}{x} + \frac{s}{x})}{\xi(\frac12 - \frac{a}{x})} \frac{\xi(\frac12 - \frac{a}{x} + \frac{1}{x} - \frac{s}{x})}{\xi(\frac12 - \frac{a}{x}+ \frac{1}{x})}$$
where $\xi(z) = \frac12 z(z-1) \pi^{-\frac{z}{2}} \Gamma(\frac{z}{2}) \zeta(z)$ is the Riemann xi-function.
If this formula is correct, the 'constructed' zeros $\mu_n$ can be stretched/condensed vertically via $x$ on the imaginary axis and shifted left/right on the real line via $a$. In all cases they would yield an entire function expressed by this closed form (think of it as a reversed application of the Weierstrass factorization theorem, i.e. starting with products of 'constructed' zeros).
Further factorization also seems possible with:
$$\frac{\xi(\frac12 - \frac{a}{x} + \frac{s}{x})}{\xi(\frac12 - \frac{a}{x})} = \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1-\frac{s}{\overline{\mu_n}} \right)$$
and
$$\frac{\xi(\frac12 - \frac{a}{x} + \frac{1}{x} - \frac{s}{x})}{\xi(\frac12 - \frac{a}{x}+ \frac{1}{x})} = \prod_{n=1}^\infty \left(1- \frac{s}{1-\mu_n} \right) \left(1- \frac{s}{\overline{1-\mu_n}} \right)$$
When $a=\frac12$ and $x=1$, the formula correctly reduces to:
$$\frac{\xi(s)}{\xi(0)} = \prod_{\rho} \left(1- \frac{s}{\rho} \right) \left(1- \frac{s}{1-\rho} \right)$$
from which the known Hadamard product for $\zeta(s)$ can been derived.
Unfortunately I do not have a proof for this formula, however rigorously checked it against many 'brute force' calculations using the first 2mln $\rho$s (all correct results, but accurate up to 5 decimals max). I manufactured the formula by replicating the symmetry of the closed form for $\mu_n = a + n x i$ (i.e. running through the integers rather than $\Im(\rho_n)$, see the linked question). Since until today, all non-trivial zeros appear to be lying on the critical line, I have used $\frac12$ as the "source" for all zeros for different $a$'s i.e.: $\frac12 - \frac{a}{x} + \frac{s}{x}$ just inserts $\frac12$ when $\Re(s)=a$. I guess I have thereby implicitly assumed the RH in constructing the formula.
My questions:


*

*Is this a known closed form?

*Does a proof of this closed form imply the RH, i.e. does it "force" the Hadamard product into a "straight jacket" that only allows it to be valid when all $a=\Re(\rho_n)=\frac12$ ?  

UPDATE:
Assuming RH is true, I believe that I
  have found a nice proof for the
  equation in the OP. Since "the
  comments section is too small to
  contain it", I decided to put it as an
  answer to my own question to round it up.

 A: The proposed formula is not true if RH is not true.
Let $\Theta$ be the upper bound of the real parts of the zeros of $\zeta(s)$. 
Your product has zeros at $\mu$, $1-\mu$, $\overline{\mu}$ and $\overline{1-\mu}$
with $\mu=a+\gamma x i$ where $\rho=\beta+i\gamma$ runs through the non trivial zeros of $\zeta(s)$
so that the supremum of the real parts of the zeros of this product is $\sup(a,1-a)$.
But your function 
$$H(s,a,x):=\frac{\xi\left(\frac12-\frac{a}{x}+\frac{s}{x}\right)}
{\xi\left(\frac12-\frac{a}{x}\right)}\frac{\xi\left(\frac12-\frac{a}{x}+\frac{1-s}{x}\right)}{\xi\left(\frac12-\frac{a}{x}+\frac{1}{x}\right)}$$
has zeros at $\beta x-\frac{x}{2}+a+i\gamma x$ and $1-\beta x+\frac{x}{2}-a-i\gamma x$.
Therefore assuming $x>0$ the supremum of the zeros of $H(a,s,x)$ is the greater of
$\Theta x-\frac{x}{2}+a$ and $1-(1-\Theta)x+\frac{x}{2}-a=1+\Theta x-\frac{x}{2}-a$,
Or $\Theta x-\frac{x}{2}+\sup(a,1-a)$
Your equality, for $x>0$, would imply therefore 
$$\Theta x-\frac{x}{2}+\sup(a,1-a)=\sup(a,1-a).$$
That is $\Theta x-\frac{x}{2}=0$, and therefore $\Theta=1/2$.
This is RH.  
The case $x<0$ is similar.
A: The conjecture above:
$H(s,a,x) =\prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{1-\mu_n} \right)\left(1- \frac{s}{\overline{\mu_n}} \right) \left(1- \frac{s}{\overline{1-\mu_n}} \right)= \frac{\xi(\frac12 - \frac{a}{x} + \frac{s}{x})}{\xi(\frac12 - \frac{a}{x})} \frac{\xi(\frac12 - \frac{a}{x} + \frac{1}{x} - \frac{s}{x})}{\xi(\frac12 - \frac{a}{x}+ \frac{1}{x})}$
with $\mu_n = a + \Im(\rho_n)x i$ might be expanded even further. 
When the domains for $a$ and $x$ are extended to $\mathbb{C}_{/(x=0)}$, the formula for $H(s,a,x)$ still appears to work (tested at 5 decimals accuracy) and for instance induces the following result (assume $\gamma_n = \Im(\rho_n)$):
$H(s,0,i) =\prod_{n=1}^\infty \left(1 + \frac{s}{\gamma_n} \right) \left(1- \frac{s}{\gamma_n} \right)\left(1- \frac{s}{{1+\gamma_n}} \right) \left(1- \frac{s}{{1-\gamma_n}} \right)= \frac{\xi(\frac12 + \frac{s}{i})}{\xi(\frac12)} \frac{\xi(\frac12 + \frac{1}{i} - \frac{s}{i})}{\xi(\frac12 + \frac{1}{i})}$
i.e. all imaginary parts of $\rho_n$ become real and their infinite product still has a closed form  that, as expected, induces a zero at e.g. $H(\gamma_n,0,i)$.
Or, when we take one of the two sub-factors from the OP, this gives:
$$\displaystyle \xi\left(\frac12+\frac{s}{i}\right) = \xi\left(\frac12 \right)\prod_{n=1}^\infty \left(1 - \frac{s^2}{\gamma_n^2} \right)$$
and for $s=i$, this result emerges:
$$\displaystyle \xi\left(\frac32\right) = \xi\left(\frac12 \right)\prod_{n=1}^\infty \left(1 + \frac{1}{\gamma_n^2} \right)$$
and $s=\frac{-i}{2}$ or $s=\frac{i}{2}$  yields this quite interesting outcome:
$$\displaystyle \xi\left(\frac12 \right) = \frac12 \prod_{n=1}^\infty \left(\frac{1}{1 + \frac{1}{(2 \gamma_n)^2} }\right) = \Xi(0)$$
So that:
$$\Xi\left(\frac12\right) = \frac12 \prod_{n=1}^\infty \left(\frac{(2 \gamma_n)^2 -1} {(2 \gamma_n)^2 +1}\right)$$
Here's my 'Overflow' follow-up question: I have tested the conjecture in Maple up to $\gamma_{2.000.000}$, but would really appreciate it when somebody could replicate and confirm the outcomes (I do not possess Mathematica, but do know the $\rho$s are already coming with the package so not much 'programming' should be required). 
Many thanks!
A: Below is a proposed proof, that assuming the RH, the following equation is true:
$$\displaystyle \frac{\xi(\frac12 - \frac{a}{x} + \frac{s}{x})}{\xi(\frac12 - \frac{a}{x})} = \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{\overline{\mu_n}} \right)$$
where $\mu_n = a + i x \gamma_n$ and $\gamma_n = \Im(\rho_n)$, with $\rho_n$ the n-th non-trivial zero of $\zeta(s)$. 
Take $t, \gamma_n \in \mathbb{R},a,x,s \in \mathbb{C}$ with $x \ne 0$ and $\gamma_n > 0$. 
Starting from Hadamard's proof that:
$$\xi(s) = \xi(0) \prod_{n=1}^\infty \left(1- \frac{s}{\rho_n} \right) \left(1- \frac{s}{1-\rho_n} \right) \qquad (1)$$
with $\xi(s) = \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$ being the Riemann xi-function.
Now we assume the RH and all $\Re(\rho_n)=\frac12$. Take $s=\dfrac{a+ i t}{x}$.
This gives:
$$\xi\left(\dfrac{a+ i t}{x}\right) = \xi(0) \prod_{n=1}^\infty \left(1- \frac{\frac{a}{x} +\frac{i t}{x}}{\frac12 + i \gamma_n} \right) \left(1- \frac{\frac{a}{x} +\frac{i t}{x}}{\frac12 - i \gamma_n} \right)$$
and can be expanded into:
$$\xi\left(\dfrac{a+ i t}{x}\right) = \xi(0) \prod_{n=1}^\infty \left(\frac{\frac12 - \frac{a}{x} + i \gamma_n -\frac{i t}{x}}{\frac12 + i \gamma_n} \right) \left(\frac{\frac12 - \frac{a}{x} - i \gamma_n - \frac{i t}{x}}{\frac12 - i \gamma_n} \right)$$
By multiplying each factor with $\dfrac{(\frac{a}{x} + i \gamma_n)}{(\frac{a}{x} + i \gamma_n)}$ and $\dfrac{(\frac{a}{x} - i \gamma_n)}{(\frac{a}{x} - i \gamma_n)}$ respectively, we get:
$$\displaystyle \xi\left(\dfrac{a+ i t}{x}\right) = \xi(0) \prod_{n=1}^\infty \left(\frac{\frac{a}{x} + i \gamma_n}{\frac12 + i \gamma_n} \right) \left(\frac{\frac{a}{x} - i \gamma_n}{\frac12 - i \gamma_n} \right) \prod_{n=1}^\infty \left(\frac{\frac12 - \frac{a}{x} + i \gamma_n -\frac{i t}{x}}{\frac{a}{x} + i \gamma_n} \right) \left(\frac{\frac12 - \frac{a}{x} - i \gamma_n - \frac{i t}{x}}{\frac{a}{x} - i \gamma_n} \right) $$
$$\displaystyle = \xi(0) \prod_{n=1}^\infty \left(1- \frac{\frac12 -\frac{a}{x}}{\frac12 + i \gamma_n)} \right) \left(1- \frac{\frac12-\frac{a}{x}}{\frac12 - i \gamma_n} \right) \prod_{n=1}^\infty \left(1- \frac{\frac{2a}{x}-\frac12+\frac{i t}{x}}{\frac{a}{x} + i \gamma_n)} \right) \left(1- \frac{\frac{2a}{x}-\frac12+\frac{i t}{x}}{\frac{a}{x}- i \gamma_n} \right)$$
By now injecting $\dfrac12 - \dfrac{a}{x}$ into equation (1), this can be simplified as:
$$\xi\left(\dfrac{a+ i t}{x}\right) = \xi\left(\frac12 - \frac{a}{x}\right) \prod_{n=1}^\infty \left(1- \frac{\frac{2a}{x}-\frac12+\frac{i t}{x}}{\frac{a}{x} + i \gamma_n}\right) \left(1- \frac{\frac{2a}{x}-\frac12+\frac{i t}{x}}{\frac{a}{x}- i \gamma_n} \right)$$
and since $\dfrac{2a}{x} + \dfrac{i t}{x} = s + \dfrac{a}{x}$ this can be rewritten into:
$$\xi\left(\dfrac{a+ i t}{x}\right) = \xi\left(\frac12 - \frac{a}{x}\right) \prod_{n=1}^\infty \left(1- \frac{x(s+\frac{a}{x}-\frac12)}{a+ i x \gamma_n}\right) \left(1- \frac{x (s+\frac{a}{x}-\frac12)}{a- i x \gamma_n} \right)$$
so that we can now say that:
$$\xi\left(\frac12 - \frac{a}{x} + \frac{s}{x}\right) = \xi\left(\frac12 - \frac{a}{x}\right) \prod_{n=1}^\infty \left(1- \frac{s}{a+ i x \gamma_n} \right) \left(1- \frac{s}{a- i x \gamma_n} \right)$$
and the desired result is obtained:
$$\displaystyle \frac{\xi(\frac12 - \frac{a}{x} + \frac{s}{x})}{\xi(\frac12 - \frac{a}{x})} = \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{\overline{\mu_n}} \right)$$
When starting from $s = \dfrac{1-(a+i t)}{x}$ the equivalent result is $\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{1- \mu_n} \right) \left(1- \frac{s}{\overline{1-\mu_n}} \right)$
