0
$\begingroup$

In a response to one of my earlier questions which I believe was related to Evolution of Zeta Zeros from Fourier Transform of $e^{-t/2}\left(\psi'[e^t]-1\right)$, it was suggested I instead focus on the Fourier transform of $\frac{\psi[e^u]-e^u}{e^{u(1/2+\epsilon)}}$ where $\psi[x]$ is the second Chebyshev function. Evaluation of this Fourier transform results in terms which contain Dirac delta functions with complex arguments such as the following.
$$\text{FourierTransform}\left[\frac{-e^u}{e^{u \left(\frac{1}{2}+\epsilon \right)}},u,y\right]=-2\ \sqrt{2\ \pi}\ \delta\ [-i+2\ y+2\ i\ \epsilon]$$

Question 1: What is the meaning of a Dirac delta function with a complex argument? In what direction does the integral of a Dirac delta function with a complex argument evaluate to a non-zero result?

Evaluation of the suggested Fourier transform using the Fourier series representation of $\psi[x]$ (see Illustration of Fourier Series for Prime Counting Functions) results in additional terms which contain Dirac delta functions with complex arguments. I believe I can get some, but not all, of these terms which contain Dirac delta functions with complex arguments to cancel each other out.

Question 2: Since the Fourier transform is being evaluated over $y\in Reals$, can I assume all terms containing Dirac delta functions with complex arguments can safely be ignored?

$\endgroup$
1
$\begingroup$

Everything is explained in the books on the Riemann zeta function (Titchmarsh, Apostol, Edwards, Montgomery, Iwaniec..). And I don't understand what you did, so in short there is what you need to know :

From $\frac{\zeta'(s)}{\zeta(s)} = -s\int_1^\infty \psi(x)x^{-s-1}dx$, by inverse Mellin transform and the residue theorem, we have the Riemann explicit formula $$\psi(x) = \dfrac{1}{2\pi i}\int_{\sigma-i \infty}^{\sigma-i \infty}\left(-\dfrac{\zeta'(s)}{\zeta(s)}\right)\dfrac{x^s}{s}ds=x-\sum_\rho\frac{x^\rho}{\rho} - \log(2\pi) -\dfrac{1}{2}\log(1-x^{-2})$$

If the RH is true, with $\rho = 1/2+i\lambda$ it means that $$f(u) = \frac{\psi(e^u)-e^u+ \log(2\pi)+\dfrac{1}{2}\log(1-e^{-2u})}{e^{u/2}} = -\sum_\lambda \frac{e^{i \lambda u}}{1/2+i\lambda} = \int_{-\infty}^\infty \hat{f}(\omega)e^{i \omega u}d\omega$$ where $\hat{f}(\omega) = -\sum_\lambda \frac{\delta(\omega-\lambda)}{1/2+i\lambda} $ is the Fourier transform (in the sense of distributions) of $f(u)$.

Note that both $f'(u)$ and its Fourier transform is (up to a very simple term) a series of Dirac deltas, those kind of distributions being very rare (lattices, crystals, quasicrystals)

$\endgroup$
  • $\begingroup$ Were the results I illustrated in my question above significantly different than what you expected? For example, did you expect indications of a Dirac delta function instead of a minimum at values of y corresponding to the magnitude of the imaginary part of a zeta zero? $\endgroup$ – Steven Clark Jan 27 '17 at 2:05
  • $\begingroup$ I have Edward's book, so I'd appreciate it if you could point me to the specific section of this book which covers the consequences of the RH being true which you state in the last half of your response above. $\endgroup$ – Steven Clark Jan 27 '17 at 4:59
  • $\begingroup$ @SteveC. I think what I wrote is very clear. Did you start reading some books : complex analysis, Fourier analysis, distribution theory, texts on $\zeta(s)$ ? I wrote many times "the Fourier transform in the sense of distributions", did you read anything about that ? And the residue theorem ? $\endgroup$ – reuns Jan 27 '17 at 16:11
  • $\begingroup$ I'm confused because the results using the Fourier series representation of $\psi[x]$ do not agree with the consequences of the RH above. Based on your answer, I also investigated the Fourier transform of $f'(u)$ and obtained similar results to what I illustrated for the Fourier transform of $\frac{\psi[e^u]-e^u}{e^{u(1/2+\epsilon)}}$ in my question above. Both Fourier transforms seem to evaluate to a minimum (versus indications of a Dirac delta function) at values of y corresponding to the magnitude of the imaginary part of a zeta zero. $\endgroup$ – Steven Clark Jan 27 '17 at 16:28
  • $\begingroup$ I'm not sure what Fourier parameters you're using, but I get $\frac{\sqrt{2\ \pi}\ \delta (\omega+\lambda)}{1/2+i\ \lambda}$ versus $\frac{\delta (\omega-\lambda)}{1/2+i\ \lambda}$ for the Fourier transform of $\frac{e^{i\ \lambda\ u}}{\frac{1}{2}+i\ \lambda }$ using the default Fourier parameters $\{0,\ 1\}$, but this is a minor point and this result is consistent with your point on "Fourier transforms in the sense of distributions". $\endgroup$ – Steven Clark Jan 27 '17 at 16:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.