7
$\begingroup$

I was wondering about the following, and I was hoping that some expert here could answer, rather than me indulging in a search for a needle in the haystack of formulas in books like Titchmarsch.

Notation:

  • $\zeta(s)$ is the Riemann zeta function.
  • $f : \mathbb R^+ \rightarrow (0,1/2)$ is such that $\zeta(s)$ does not vanish between $s = 1+it$ and $s=1 - f(t) + it$.
  • $\pi(x)$, $Li(x)$ as in wikipedia.

Assuming the above data, suppose the version of the prime number theorem that can be proven is:

$$ \pi(x) = Li(x) + O\left(G(x)\right) $$

Question:

Can G(x) be given a closed form expression showing its precise(if and only if) dependence on $f(t)$?

Heuristics: When $f = 0$, $G(x) = x \mathrm{e}^{-a\sqrt{\ln x}}$ and when $f = 1/2$, $G(x) = \sqrt x \ln x$. So possibly there would be a term like $x^{1-f(x)}$ in a putative expression for $G(x)$.

$\endgroup$
1
  • 1
    $\begingroup$ Titchmarsh has a very well-written chapter on the Prime Number Theorem. This is hardly a needle in a haystack. $\endgroup$ Aug 6, 2010 at 4:15

1 Answer 1

7
$\begingroup$

Your heuristic is wrong: $G(x)=x\exp{(-a\sqrt{\log{x}}})$ follows from $f=\frac{c}{\log{(|t|+3)}}$ for some fixed real $c>0$.

I really don't want to tell you the answer, because this is a great exercise! A big hint: use the "approximate explicit formula"

$\psi(x)=x-\sum_{|\rho|\leq T} \frac{x^{\rho}}{\rho}+O(T^{-1} x \log^2{x}),$

bound the sum over zeros trivially given what you know about $f$, and then choose $T$ so that the two error terms balance.

$\endgroup$
1
  • 2
    $\begingroup$ Unless I'm missing something, an "if and only if" dependence is more than an exercise. $\endgroup$ May 24, 2010 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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