TITLE: Distribution of primes in small intervals
QUESTION [6 upvotes]: 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  $ \textit{almost all}\ \ x$ (assuming the Riemann hypothesis). Does it hold for  $ \textit{all}\ \ x $
if $y(x) = \log^c x$ for some constant $c$ (with Riemann hypothesis)?

REPLY [6 votes]: 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.