$$\pi = 3\prod_{\zeta(1/2+it) = 0}\frac{9+4t^2}{1+4t^2}\iff\text{RH is true}.$$

15$\begingroup$ I'd suggest you expand on this a bit, tell us your reasoning for asking the question, what you mean by $\mathfrak{I}(r)$, why you believe or disbelieve it, etc. $\endgroup$ – Keith Millar Nov 5 '17 at 18:18

2$\begingroup$ ℑ(r) denotes the imaginary part. If RH is true then the product equals Pi. My question relates to whether it is true vice versa. That is, if the product is equal to Pi then and RH is true. r is a zeta zero. $\endgroup$ – Dimitris Valianatos Nov 5 '17 at 18:51

17$\begingroup$ Despite appearances, this question makes sense (and the equivalence is probably correct). But all it amounts to is writing down the Hadamard formula for the completed Riemann $\xi$ function evaluated at $2$. The product is presumably over positive ordinates $t$. $\endgroup$ – Lucia Nov 5 '17 at 18:52

3$\begingroup$ Posted a year ago here: Riemann Hypothesis and Prime summation $\endgroup$ – jeq Nov 6 '17 at 0:01

6$\begingroup$ Next time, please ask a question. In the above example, you could have asked: is this statement true and has it been asked before? $\endgroup$ – GH from MO Nov 6 '17 at 1:39
Yes, this is equivalent to RH (but not in any significant way). Recall the completed Riemann $\xi$function
$$
\xi(s) = s(s1) \pi^{s/2} \Gamma(s/2) \zeta(s),
$$
which, by Hadamard's factorization formula can be written as
$$
e^{A+Bs} \prod_{\rho}\Big(1\frac{s}{\rho}\Big) e^{s/\rho},
$$
where the product is over all nontrivial zeros $\rho$ of $\zeta(s)$.
Now one can check that $A=0$ (plug in $s=0$) and that $B= \sum_{\rho}\text{Re }(1/\rho)$. It follows that
$$
\xi(s) = \prod_{\rho: \text{Im}(\rho) >0} \Big \frac{s\rho}{\rho}\Big^2,
$$
by grouping complex conjugate zeros (and the product now converges). Now evaluate this at $s=2$: thus
$$
\xi(2) =2 \times 1 \times \pi^{1} \times \Gamma(1) \times \zeta(2) = \frac{\pi}{3}
$$
equals
$$
\prod_{\rho: \text{Im}(\rho) >0} \Big \frac{2\rho}{\rho}\Big^2.
$$
Split the product over zeros into two factors: the first one from zeros on the critical line, and the second one over zeros not on the critical line (if any). The first factor is simply $$ \prod_{\substack{\rho = 1/2 +i\gamma \\ \gamma >0}} \frac{(3/2)^2+\gamma^2}{(1/2)^2 +\gamma^2} = \prod_{\substack{\rho = 1/2 +i\gamma \\ \gamma >0}} \frac{9+4\gamma^2}{1+4\gamma^2}. $$ If RH is true then the second factor is $1$. If RH is false, then note that the contribution of the zeros $\beta+i\gamma$ and $1\beta+i\gamma$ together to the second product is $$ \frac{(2\beta)^2+\gamma^2}{\beta^2+\gamma^2} \frac{(1+\beta)^2+\gamma^2}{(1\beta)^2 +\gamma^2} > 1; $$ (both factors are $\ge 1$ since $0\le \beta \le 1$, and at least one of them must be strictly larger than $1$).
There's a little bit more fun to be had with this problem. It can be used to show easily that if $\gamma_0$ is the first ordinate of a zero (not necessarily on the critical line) of $\zeta(s)$ then $$ \frac{\pi}{3} \ge \frac{9+4\gamma_0^2}{1+4\gamma_0^2}, $$ and since $\pi$ is so close to $3$, one can extract from this the fairly good bound that $\gamma_0 \ge 6.49\ldots$. This general idea is of course well known, but I thought this particular choice was pretty!