The Riemann hypothesis is equivalent to the assertion that the prime counting function $\pi(x) := \sum_{p \le x} 1$ deviates from the logarithmic integral $Li(x) = \int_2^x \frac{dx}{\log x}$ in the order $O(\sqrt{x} \log x)$. Since $\log x = O(x^\alpha)$ for any $\alpha>0$, the Riemann hypothesis implies that:
$$\forall \alpha > \frac12, |\pi(x) - Li(x)| = O(x^\alpha)$$
From what I understand, this is the best possible power bound available, meaning that for any $\alpha \le \frac12$, we *don't* have $|\pi(x) - Li(x)| = O(x^\alpha)$. (I'm not sure whether the bounds like $O(\sqrt{x} \log \log x)$ could be possible.)

Where can I read a proof that this indeed is the best possible power bound? The answer on this question claims that any textbook on analytic number theory will do, but I'd like to know an explicit textbook reference that I can look into.

More concretely, I followed the steps outlined in the question referenced above, and could not progress at one point. For the Chebyshev function $\psi(x) = \sum_{p^n \le x}\log p$, I derived that \begin{align*} \left| \psi(x) - x \right| = \left| \log 2\pi + \sum_{\rho} \frac{x^\rho}{\rho} \right| = \left| \log 2\pi + \frac12 \log (1- \frac1{x^2}) + \sum_{\rho \text{ ntv.}} \frac{x^\rho}{\rho} \right| \sim \left| \sum_{\rho \text{ ntv.}} \frac{x^\rho}{\rho} \right| \end{align*} where $f(x) \sim g(x)$ iff $\lim_{x \rightarrow \infty} f(x)/g(x) = 1$, the summation over $\rho$ is the summation over the zeroes of the Riemann zeta, and the qualifier "ntv." refers to counting only the nontrivial zeroes. Since $\psi(x) \sim \pi(x) \log x$, at this point I'd like to show two things:

The Riemann hypothesis is equivalent to the assertion that $|\psi(x) - x| < \sqrt{x} \log ^2 x$ for large enough $x$ (Schoenfield 1976, according to Wikipedia)

Regardless the Riemann hypothesis, one

*cannot*have $|\psi(x) - x| = O(x^\alpha)$ for any $\alpha \le \frac12$.

It looks as if both of the equivalences should follow straightforwardly from the expression $$|\psi(x) - x| \sim \left| \sum_{\rho \text{ ntv.}} \frac{x^\rho}{\rho} \right|$$ However I don't know how to work out the phase factors appearing in the following expression: $$\frac{x^\rho}{\rho} = \left( \frac{e^{it \log x} (\sigma - it)}{\sigma^2+t^2} \right) \cdot x^\sigma \text{, where $\rho = \sigma + it$}$$ to obtain a lower bound or an upper bound to the error term. Further analysis indeed depends on the distribution of the imaginary part of the nontrivial zeroes, which will control how much the weights $\frac1\rho$ contribute in the end.

notconverge absolutely and there is no easy way to to estimate it directly. You need to use atruncatedexplicit formula, summing over nontrivial zeros $\rho$ with $|{\rm Im}(\rho)| \leq T$ and adding an error term on the right side that depends on $x$ and $T$. When $T = x$ you'll get the standard error term $O(\sqrt{x}\log x)$. $\endgroup$ – KConrad Apr 24 at 1:233more comments