Is this equivalent to RH - Riemann hypothesis?

Question

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

Answer 1

Yes, this is equivalent to RH (but not in any significant way). Recall the completed Riemann $\xi$-function $$ \xi(s) = s(s-1) \pi^{-s/2} \Gamma(s/2) \zeta(s), $$ 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 non-trivial 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!

Answer 2

I begin by prefacing I have no formal mathematics education.

This answer mirrors the answer given by Lucia. We have $$ \frac{\pi}{6}=\xi(2)=\frac{1}{2}\prod_{\rho}\left(1-\frac{2}{\rho} \right).\label{a}\tag{1} $$ Here the product is taken over the roots of $\xi,$ and should be done so over conjugate pairs i.e., the factors for a pair of zeroes of the form $\rho$ and $1−\rho$ should be grouped together. After multiplying the LHS and RHS of $\ref{a}$ by $2$, we get: $$ \frac{\pi}{3}=\prod_{\rho}\left(1-\frac{2}{\rho} \right). $$ Now, assume Riemann's hypothesis, that is all non trivial zeros have real part equal to $\frac{1}{2}.$ Writing $\rho=\frac{1}{2}+i\gamma$ and $\rho^*=\frac{1}{2}-i\gamma,$ leads to: \begin{align} \frac{\pi}{3} &=\prod_{\gamma >0}\left(1-\frac{2}{1/2+i\gamma} \right)\left(1-\frac{2}{1/2-i\gamma} \right)\\ &=\prod_{\gamma >0}\frac{(-3+2i\gamma)(-3-2i\gamma)}{(1+2i\gamma)(1-2i\gamma)}\\ &=\prod_{\gamma >0}\frac{4\gamma^2+9}{4\gamma^2+1}. \end{align}