9

Usually, the Riemann hypothesis is introduced along the following lines:
(1.1) Geometric progressions were known for forever
(1.2) Euler factorization links a product of primes to a sum of natural numbers
(1.3) The harmonic series diverges, thus there are infinitely many primes
(1.4) Riemann expanded the zeta function to the complex plane and conjectured that all non-trivial zeroes have $\Re=1/2$

So far, so good. Nice and steady. How he came to this conclusion isn't obvious, but this is why the RH is an open problem. Suddenly ...

(2.1) Non-trivial zeroes of the zeta function impose constraints on the distribution of prime numbers including the accuracy of the prime counting function distribution of (a) gaps between primes, (b) twin primes ...

Indeed, (1.2) linked all natural numbers to primes. But why is the distribution of zeroes relevant? Why not a Fourier transform of the zeta function, not the distribution of ones?

In other words, I understand why a certain property of the zeta function is relevant to primes. But why does this specific property happen to be the distribution of zeroes?

Edit: I've read this What is so interesting about the zeroes of the Riemann $\zeta$ function? but roughly around this point

"This formula says that the zeros of the Riemann zeta function control the oscillations of primes around their 'expected' positions."

"Roughly speaking, the explicit formula says the Fourier transform of the zeros of the zeta function is the set of prime powers plus some elementary factors."

it gets hard to follow.

Philippe Knecht
  • 541
sixtytrees
  • 409
  • @Peter Sheldrick, Rigorous proofs are tough, but I am asking about the line of a sketch at a "5K basic complex analysis" level. – sixtytrees Jul 12 '16 at 16:28
  • 1
    the Abel summation formula gives $\ln \zeta(s) = \sum_{p^k} \frac{p^{-sk}}{k} = s \int_2^\infty J(x) x^{-s-1}dx$ where $J(x) = \sum_{p^k < x} \frac{1}{k} = \pi(x)+ \mathcal{O}(\sqrt{x})$. With the change of variable $x = e^u$ you get $\frac{\ln \zeta(s)}{s} = \int_{\ln 2}^\infty J(e^u) e^{-su} du$ i.e. the Laplace/Fourier transform of $J(e^u)$ – reuns Sep 11 '16 at 08:34
  • 3
    And Riemann found that the functional equation meant $\xi(s) = s (s-1) \pi^{-s/2} \Gamma(s/2) \zeta(s)$ obeys to $\xi(s) = \xi(1-s)$. Together with $\xi(s) = \overline{\xi(\overline{s})}$ it means $\Xi(t) = \xi(1/2+it)$ is real. Proving it changes of sign infinitely many times is not difficult, so that it has infinitely many zeros $\implies \zeta(s)$ has infinitely many zeros on $Re(s) = 1/2$. This is the main reason why he made his hypothesis. – reuns Sep 11 '16 at 13:07

1 Answers1

6

  • with $\psi(x) = \sum_{p^k < x} \ln p$ you have the Riemann explicit formula $$\psi(x) = x - \sum_\rho \frac{x^\rho}{\rho} + \mathcal{O}(1) $$ ($\rho$ being the non trivial zeros of $\zeta(s)$, obtained from the residue theorem applied to $\frac{-\zeta'(s)}{\zeta(s)} = \sum_{p^k} p^{-sk}\ln p $ $= s \int_2^\infty \psi(x) x^{-s-1}dx$ $ \implies \psi(x) = \frac{1}{2i\pi} \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{-\zeta'(s)}{\zeta(s)} \frac{x^s}{s}ds$ by inverse Laplace/Mellin transform)

  • then the Riemann hypothesis is that $$\frac{\psi(e^u) - e^u}{e^{u/2}} = - \sum_\rho \frac{e^{i u \text{ Im}(\rho)}}{\rho} + \mathcal{O}(e^{-u/2})$$ i.e. $\frac{\psi(e^u) - e^u}{e^{u/2}}$ is almost periodic, having an expansion that is "almost" a Fourier series (you doubt it would have a huge impact on the distribution of prime numbers)

reuns
  • 79,880
  • I missed the connection to domain of almost periodic functions. gr8Answer. – sixtytrees Sep 20 '16 at 23:56
  • In his blog, I say Mumford's blog, he provide us an explanation/heuristic of that (this is the title of such entry in his blog) The lowest zeros of Riemann's zeta are in front of your eyes. Then I've added, this time, this comment here as companion of your answer. –  Sep 23 '16 at 18:06