Bombieri-Vinogradov up to smaller moduli?

Bombieri-Vinogradov theorem (taken from Wikipedia) states: Let $x$ and $Q$ be any two positive real numbers with $x^{1/2}\log^{-A}x\leq Q\leq x^{1/2}.$ Then $$\sum_{q\leq Q}\max_{y<x}\max_{1\le a\le q\atop (a,q)=1}\left|\psi(y;q,a)-{y\over\varphi(q)}\right|=O\left(x^{1/2}Q(\log x)^5\right).$$

Here $\varphi(q)$ is the Euler totient function, and $$\psi(x;q,a)=\sum_{n\le x\atop n\equiv a\bmod q}\Lambda(n),$$ where $\Lambda$ denotes the von Mangoldt function.

My question is: Is it possible to have the statement (slightly weaker statement ok too) with smaller $Q$ ? Would it be possible to have similar statements with $Q= X^{\delta}$ for some small $\delta > 0$ (possibly adding smooth weights if necessary)?

• I don't understand the question. The Bombieri-Vinogradov theorem as stated works for any positive number $Q$, producing a non-trivial bound for $Q \leq X^{1/2 - \delta}$, say. What do you mean by choosing a 'smaller' $Q$? – Stanley Yao Xiao Jun 19 '17 at 1:44
• In the statement of the Wikipedia it says I have to take $Q \geq x^{1/2} \log^{-A}x$... – Johnny T. Jun 19 '17 at 1:45
• To answer your question: you get a small enough saving for the sum on the right hand side for any $Q \ll x^{1/2} (\log x)^{-A}$; but the right hand side might need to be replaced by $O(x (\log x)^{-B(A)})$ for some $B(A) > 0$ depending only on $A$. In other words, the level of distribution can be taken to be $1/2$. The Elliot-Halberstam conjecture asserts that the level of distribution can be taken to be $1$. Any established level of distribution implies all smaller values are admissible for a level of distribution. – Stanley Yao Xiao Jun 19 '17 at 1:53
• Any argument that proves the Bombieri-Vinogradov theorem will also prove it for any smaller level of distribution. You can read Friedlander and Iwaneic's book "Opera de Cribro" for a detailed account. – Stanley Yao Xiao Jun 19 '17 at 2:32
• You should probably say what you'd like to do with BV – reuns Jun 19 '17 at 7:15