# Distribution of primes in small intervals

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)?

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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.

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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

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 $f(x)=\log^c x$.

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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