Optimality of the Riemann Hypothesis

Question

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.

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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:

1. 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)

2. 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.

Answer 1

See Chapter 15 ("Oscillation of error terms") in Montgomery-Vaughan: Multiplicative number theory I. See especially Theorems 15.2-15.3 and 15.11.

Added by Steven Clark and GH from MO. For convenience, we copied below the relevant theorems and some additional text from the book. As usual, $$M(x):=\sum\limits_{n\le x}\mu(n)$$ is the Mertens function.

Theorem 15.2. Let $\Theta$ denote the supremum of the real parts of the zeros of the zeta function. Then for every $\varepsilon>0$, $$\psi(x)-x=\Omega_\pm(x^{\Theta-\varepsilon})\tag{15.1}$$ and $$\pi(x)-\mathrm{li}(x)=\Omega_\pm(x^{\Theta-\varepsilon})\tag{15.2}$$ as $x\to\infty$.

Theorem 15.3. Suppose that $\Theta$ is the supremum of the real parts of $\zeta(s)$, and there is a zero $\rho$ with $\Re\rho=\Theta$, say $\rho=\Theta+i\gamma$. Then $$\underset{x\to\infty}{\text{lim sup}}\ \frac{\psi(x)-x}{x^{\Theta}}\geq \frac{1}{| \rho |}\tag{15.4}$$ and $$\underset{x\to\infty}{\text{lim inf}}\ \frac{\psi(x)-x}{x^{\Theta}}\leq -\frac{1}{| \rho |}.\tag{15.5}$$

Theorem 15.11. As $x\to\infty$, $$\psi(x) -x = \Omega_{\pm} \bigl(x^{1/2} \log \log \log x\bigr),$$ and $$\pi(x) - \mathrm{li}(x) = \Omega_{\pm} \bigl(x^{1/2}(\log x)^{-1}\log \log \log x\bigr).$$

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Then in the manner of the proof of Theorem 15.3, we find that if $\Theta+i\gamma$ is a zero of $\zeta(s)$, then

$$\underset{x\to\infty}{\text{lim sup}}\ \frac{M(x)}{x^{\Theta}}\geq \frac{1}{| \rho\,\zeta'(\rho) |},\tag{15.11}$$

and

$$\underset{x\to\infty}{\text{lim inf}}\ \frac{M(x)}{x^{\Theta}}\leq -\frac{1}{| \rho\,\zeta'(\rho) |}.\tag{15.12}$$