The Hadamard Product for the Riemann zeta function is given by
$\zeta(s) = \frac{e^{[\ln(2\pi)-1-\gamma/2]s}}{2(s-1)\Gamma(1+\frac{s}{2})}\prod_{\rho}(1-\frac{s}{\rho})\exp(\frac{s}{\rho}),$
compare http://mathworld.wolfram.com/HadamardProduct.html, where the infinite product is taken over all nontrivial zeros, counting their multiplicity. From this it trivially follows that
$\zeta(s+\frac{1}{2}) = \frac{e^{[\ln(2\pi)-1-\gamma/2](s+\frac{1}{2})}}{2(s-\frac{1}{2})\Gamma(1+\frac{s+\frac{1}{2}}{2})}\prod_{\rho}(1-\frac{s+\frac{1}{2}}{\rho})\exp(\frac{s+\frac{1}{2}}{\rho})$.
But as the nontrivial zeros of $\zeta(s+\frac{1}{2})$ correspond one-to-one to the nontrivial zeros of $\zeta(s)$, the former ones being shifted $\frac{1}{2}$ to the left with respect to the latter ones, it may also, I guess, be written as
$\zeta(s+\frac{1}{2}) = \phi(s)\prod_{\rho}(1-\frac{s}{\rho-\frac{1}{2}})\exp(\frac{s}{\rho-\frac{1}{2}}),$
for some function $\phi(s)$, with the infinite product being taken over the exact same zeros as before. My question is now: What is $\phi(s)$ given by?, as a closed expression, i.e., without any infinite product.
Update [as a result of the comments below]: To find $\phi(s)$, I need to evaluate the following infinite product:
$\prod_{\rho} \frac{(1-\frac{s+\frac{1}{2}}{\rho})\exp(\frac{s+\frac{1}{2}}{\rho})}{(1-\frac{s}{\rho-\frac{1}{2}})\exp(\frac{s}{\rho-\frac{1}{2}})} = \prod_{\rho} (1-\frac{1}{2\rho})\times\prod_{\rho}\exp(\frac{2\rho-2s-1}{2\rho(2\rho-1)}),$
here generally assuming that the ratio of two infinite products is the infinite product of ratios. The first product is some constant, but the second product is not. How can I evaluate that latter one?
I have now determined the function $\phi(s)$. It is given by
$\phi(s) = -\frac{\zeta(\frac{1}{2})\pi^{1+\frac{s}{2}}\sqrt{2}}{8(s-\frac{1}{2})\Gamma(\frac{3}{4})\Gamma(\frac{5}{4}+\frac{s}{2})}.$
The calculations leading to this result do not strike me as particularly trivial, as one of the commenters of my original question asserted.