# Estimate on the prime-counting function $\psi(x)$.

There is an elementary statement that I believe I have read somewhere, but I can't remember where. I'd like to know if the statement is correct (in which case it is surely standard) and if so, where I can find a proof of it.

The statement is about the prime-counting function $\psi(x) = \sum_{ p^n < x } \log p$. It is well-known that GRH is equivalent to $\psi(x) = x + O(x^{1/2} \log^2 x)$. The statement in question is:

If $1/2< \alpha < 1$ is a real number, and $\zeta(s)$ has no zero on $\Re s> \alpha$, then $\psi(x) = x + O(x^{\alpha} \log x)$?

The point here is that we get to replace $\log^2 x$ by $\log x$. Thanks for any reference (or disproof, if this is just a dream I made).

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Grosswald, Émile. "Sur l'ordre de grandeur des différences $\psi(x)-x$ et $\pi(x)-\ell i(x)$." (French) C. R. Acad. Sci. Paris 260 (1965), 3813–3816.
It seems (according to the Math Review) that for $\alpha$ fixed, the error is $O(x^\alpha)$. So, two powers of the logarithm can be saved.
Intuitively, the reason for the saving comes from the fact that it is known (from zero density estimates) that only a tiny fraction of the zeroes of zeta can have real part anywhere close to $\alpha$, so their net contribution to the explicit formula $\psi(x) = x - \sum_\rho \frac{x^\rho}{\rho} + \ldots$ is small enough that their portion of sum $\sum_\rho 1/\rho$ is absolutely convergent, thus producing no logarithmic factor. –  Terry Tao May 23 '13 at 16:37