Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\pi(x)$ be the number of primes smaller than $x$. Do there exist unconditionally universal constants $c > d$ such that $$ \lim_{x \rightarrow \infty} \frac{\pi(x + \log^c x) - \pi(x)}{\log^{c-d} x} \geq 1 $$

We know that by Maier Theorem, it is not possible that $c = d+1$.

By Selberg theorem, for any function $y(x)$ grows faster than $\log^2 x$, it holds that $$ \lim_{x \rightarrow \infty} \frac{\pi(x + y) - \pi(x)}{y/\log x} = 1 $$ for \emph{almost} $x$ (assuming the Riemann hypothesis). Does it hold for \emph{all} $x$ if $y(x) = \log^c x$ for some constant $c$ (with Riemann hypothesis)?

share|improve this question
add comment

2 Answers

A weaker question is to ask for which functions $f$ the interval $[x,x+f(x)]$ contains a prime for all sufficiently large $x$. The sharpest uncoditional result is then that $f(x)\geq x^{0.525}$ is sufficient. We are therefore a long way from being able to prove results about $f(x)=\log^c x$.

share|improve this answer
    
And even then we don't know how large "sufficiently large" is. The best result I know with an explicit "sufficiently large" is $[x,x+x/(25\log^2x)]$ for x > 396 738 due to Dusart 2010. –  Charles Mar 29 '12 at 21:55
    
But it seems that OP is asking for results conditional on RH. –  Gerry Myerson Mar 29 '12 at 23:20
add comment

Assuming the Riemann Hypothesis, I believe the best known result is due to Cramer (I cannot figure out how to add the accent of the e) and it says the following:

There is a constant $C > 0$ such that if if the Riemann Hypothesis is true, then for every $x \ge2$ the interval $(x, x + C \sqrt{x} \log x)$ contains at least $\sqrt{x}$ prime numbers.

This is Theorem 13.3 in Montgomery and Vaughan's Multiplicative Number Theory.

Translating things from $\pi(x)$ to $\psi(x)$, exercise 2, pp. 430-431 of the same book outlines a proof that the Riemann Hypothesis implies that

$$\psi(x+y)-\psi(x)=y+O\left(\sqrt{x} \log x \log\left(\frac{2y}{\sqrt{x} \log x}\right) \right).$$

Thus an asymptotic holds as soon as $\frac{y}{\sqrt{x} \log x} \to \infty$. This formula simultaneously implies both Cramer's result and von Koch's well-known result that $$ \psi(x) = x + O(\sqrt{x}\log^2 x) \quad \text{equivalently } \quad \pi(x) = \int_2^x \frac{dt}{\log t} + O(\sqrt{x}\log x) $$ assuming the Riemann Hypothesis.

share|improve this answer
    
And improving this slightly, down to intervals $(x,x+C\sqrt{x\log x})$ assumes the Pair Correlation Conjecture of Montgomery (this was done by Heath-Brown). –  Dimitris Koukoulopoulos Apr 6 '12 at 14:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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