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Wikipedia states under the entry for the von Mangoldt function:

The Fourier transform of the von Mangoldt function gives a spectrum with spikes at ordinates equal to imaginary part of the Riemann zeta function zeros.

(I believe "ordinates" should be changed to "abscissas".)

First, what does taking the Fourier transform of the von Mangoldt function mean?

Second, if meaningful, is it true?

Third, if true, how might this be related to the sum of the exponentials $e^{iImg(z_n)x}$ over the non-trivial zeros $z_n$ above and below the real axis, assuming the RH is true?

(There is some history behind this statement in an older MO-Q, but the analysis there is not clear to me either.)

Tom Copeland
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  • I suggested once that maybe the real part of the non trivial zeros of $\zeta$ could arise as the Fourier transform of the trivial zeros and their imaginary part as the Fourier transform of (powers of) primes or of the logarithms thereof. But I think we lack the relevant theoretical framework to adress these issues rigorously. – Sylvain JULIEN Sep 20 '19 at 06:08
  • a very extensive MO posting that addresses these issues: https://mathoverflow.net/q/162076/11260 – Carlo Beenakker Sep 20 '19 at 06:12
  • @CarloBeenakker , yes, I found that, but the arguments and conclusions there are not clear to me. – Tom Copeland Sep 20 '19 at 13:19
  • @SylvainJULIEN , it appears Mangoldt was able to work out rigorously the basic relation in 1895, but Riemann, of course, had understood already. See Wikipedia "Explicit formulae for L functions" https://en.m.wikipedia.org/wiki/Explicit_formulae_for_L-functions and "The Riemann hypothesis" by Conrey https://www.ams.org/notices/200303/fea-conrey-web.pdf – Tom Copeland Sep 23 '19 at 21:07

1 Answers1

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Start with the explicit formula

$$\sum_{n \le x}\Lambda(n) =\frac1{2i\pi} \int_{2-i\infty}^{2+i\infty} \frac{-\zeta'(s)}{\zeta(s)}\frac{x^s}s ds=1_{x > 1}\sum Res(\frac{-\zeta'(s)}{\zeta(s)}\frac{x^s}s)$$ $$=1_{x > 1}( x - \sum_\rho \frac{x^\rho}{\rho} - \frac12 \log 2\pi - \sum_{k=1}^\infty \frac{x^{-2k}}{-2k})$$

Since $\sum_\rho \frac1{|\rho|^2}<\infty$ the RHS converges in $L^1_{loc}$ thus we can differentiate both sides in the sense of distributions, the RHS being continuous at $1$ we get

$$\sum_n \Lambda(n) \delta(x-n) =1_{x > 1} - 1_{x > 1}\sum_\rho x^{\rho-1} +\frac{d}{dx} 1_{x > 1}\log(1-x^{-2})$$

If the RH is true, letting $x =e^u$ and multiplying both side by $e^{u/2}$, since $e^{u/2}\delta(e^u-n) = \frac{\delta(u-\log n)}{n^{1/2}}$

$$\sum_n \frac{\Lambda(n)}{n^{1/2}}\delta(u-\log n) =1_{u > 0}e^{u/2} - 1_{u > 0}\sum_t e^{itu} +e^{-u/2}\frac{d}{du}1_{u > 0} \log(1-e^{-2u}))$$

Making both side even $$\sum_n \frac{\Lambda(n)}{n^{1/2}} (\delta(u-\log n)+\delta(u+\log n))=e^{|u|/2} - \sum_t e^{itu} -e^{-|u|/2}(\frac{d}{d|u|} \log(1-e^{-2|u|})$$

Which means that we have the Fourier transform in the sense of distributions $$\mathcal{F}^{-1}\left[2 \pi \sum_t \delta(\omega-t)\right] = e^{|u|/2}-e^{-|u|/2}(\frac{d}{d|u|} \log(1-e^{-2|u|})-\sum_n \frac{\Lambda(n)}{n^{1/2}} (\delta(u-\log n)+\delta(u+\log n)) $$

If the RH is not true then those things are true only in the sense of analytic functionals, for example they hold when using Gaussians as test functions.

reuns
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  • Please identify $t$ ($\rho$ or $\rho-1$), $\omega$, the variables and normalization of the Fourier transform, $1_{u>0}$ (Heaviside step fct?), etc. If you take a FT over finite limits, i.e., multply the RHS by a rect fct. centered at the origin and apply the FT, then, if your analysis is correct, the delta fcts. will become convolved with a sinc fct (the FT of a rectangle fct.), or equivalently replaced by a sinc fct, which you can then plot. You should have sinc fcts normalized and distributed according to the LHS. – Tom Copeland Sep 20 '19 at 19:14
  • I see that you are borrrowing some of the notation in the analysis of https://en.m.wikipedia.org/wiki/Chebyshev_function – Tom Copeland Sep 20 '19 at 19:56
  • I'm not sure of what you are saying. Under RH $\sum_t$ is over the imaginary parts of non-trivial zeros. In the yellow part the normalization is $\mathcal{F}f(u) = \int_{-\infty}^\infty f(u) e^{-i \omega u} du$. If $\phi$ is Schwartz then $\int_{-\infty}^\infty e^{-i u \omega}\phi(u) ( e^{|u|/2}-e^{-|u|/2}(\frac{d}{d|u|} \log(1-e^{-2|u|})-\sum_n \frac{\Lambda(n)}{n^{1/2}} (\delta(u-\log n)+\delta(u+\log n)))du$ $ = 2\pi\sum_t \mathcal{F}\phi$ – reuns Sep 20 '19 at 20:01
  • I plotted here the numerical version https://math.stackexchange.com/a/3122844/276986 – reuns Sep 20 '19 at 20:11
  • Thanks, it'll take s little time to digest it. – Tom Copeland Sep 21 '19 at 15:43
  • Ok, I've mostly worked through the analysis (a little differently) for the Mangoldt and Chebyshev functions and arrived at essentially the same answer. I'm in the process of writing it up in my blog, where I'll try to sketch the trail through MSE, MO, and n-Category Cafe for others as well at some point. – Tom Copeland Sep 23 '19 at 20:57