# Fourier Transform of the Riemann zeros (Dirac comb)?

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).

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

• Are you suggesting we see spikes at $\ln p_i^n$? – Rudi_Birnbaum Feb 9 '19 at 22:20
• @Rudi_Birnbaum Yes, spikes of relative amplitude $\frac{\log p}{p^{k/2}}$ at $\pm k\log p$ – reuns Feb 9 '19 at 22:23
• It seems to me its rather $\frac{ln p_i^n}{2 \pi}$ – Rudi_Birnbaum Feb 9 '19 at 22:27
• @Rudi_Birnbaum up to your Fourier normalization. And it is a Fourier transform in the sense of distributions, so you need to regularize it to make it more visible, eg. looking at $\mathcal{F}^{-1}[\sum_t \delta(v-t) e^{-t^2/n^2}]$ – reuns Feb 9 '19 at 22:29
• I'm trying to sort through some of the errors in the Wikipedia article at the link in the answer above (see en.wikipedia.org/wiki/Talk:Explicit_formulae_(L-function) ) as well as errors in the answer above itself. Assuming the definition $F(v)=\int\limits_{-\infty}^\infty f(u)\,e^{-i\,v\,u}\,du$, with respect to the answer above I believe $F(v)=-\left(2\,\pi\sum\limits_\rho\delta(v-\Im(\rho))+G(1/2+i\,v)+G(1/2-i\,v)\right)$ is closer to being correct, but this formula seems to evaluate with a slight offset, so perhaps the definition of $G(s)$ is not exactly correct either. – Steven Clark Feb 28 '19 at 23:03

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.

• In deed! I'll check it numerically. – Rudi_Birnbaum Feb 8 '19 at 21:55

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

The numbers are about 0.11, 0.175, ...