Fourier Transform of the Riemann zeros (Dirac comb)?

Question

Lets assume RH and $\rho_i, i\in\Bbb N$ be the imaginary parts of the non-trivial zeros of the Riemann $\zeta$ function: $\zeta(\frac{1}{2}\pm\imath \rho_i)=0$, $(\forall i)$.

Does anonye know if anything (in case what) is known on the (real) Fourier-Transform of a "zeta-zero-Dirac-comb": $$ \mathcal{F}\left \{ \sum_{i=1}^{\infty} \delta(t - \rho_i ) + \delta(t + \rho_i)\right \}[s] $$

A very naive suggestion would be that its the primes. Though I think its probably not, because the connection seems to be more hidden and only revealed regarding complex functions. Nevertheless or especially for that, I think it could be an interesting question (though I expect the answer is known to specialists).

Answer 1

See Riemann-Weil's Explicit Formula under the RH the Fourier transform of the tempered distribution

$$f(u)= \sum_{n=1}^\infty \frac{\Lambda(n)}{n^{1/2}} (\delta(u-\ln n)+\delta(u+\ln n))$$ is

$$F(v) = \sum_{t \in \text{ imaginary parts of non-trivial zeros}} \delta(v-t)\quad +\quad G(1/2+iv)+G(1/2-iv)$$

where $$G(s) = \frac{1}{1-s}+\frac12 (\ln 2\pi + \gamma)+ \sum_{k=1}^\infty (\frac{1}{s+2k}- \frac1{2k})$$

Answer 2

Well this is a fairly simple Fourier Transform, that yields:

$$\mathcal{F}\left \{ \sum_{i=1}^{\infty} \delta(t - \rho_i ) + \delta(t + \rho_i)\right \} = 2 \sum_{i=1}^{\infty} \cos(2\pi \rho_i s)$$

So now your question reduces to "What does an infinite sum of cosinusoids, whose frequencies are the imaginary parts of the non-trivial roots of the Reimann-Zeta function, look like?"

We know for sure that it is an even function.

I suppose the RHS is easy enough to simulate in MatLab/Octave for the first $n$ roots.

Answer 3

Not an answer, but the function plot for the first 200 zeros:

The numbers are about 0.11, 0.175, ...