Building on this question [scaling the imaginary part of $\rho$s in infinite products](https://mathoverflow.net/questions/115128/what-happens-to-zetas-when-all-its-im-rho-n-are-scaled-linearly), 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:


1. Is this a known closed form?
2. 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 closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH?
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.


