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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 $G(x)$.

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Titchmarsh has a very well-written chapter on the Prime Number Theorem. This is hardly a needle in a haystack. –  Micah Milinovich Aug 6 '10 at 4:15

1 Answer 1

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.

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Unless I'm missing something, an "if and only if" dependence is more than an exercise. –  Kevin O'Bryant May 24 '10 at 20:48

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