Riemann Hypothesis and Prime Count

Question

Let $\pi(a)$ be the number of primes below $a>0$. The prime number theorem states $\pi(a)\sim\frac{a}{\ln a}$. My question is trivial. Is $$\frac{a}{\ln a}\leq\pi(a)\leq\frac{a}{\ln a}+c\sqrt{a}\ln a$$ for some constant $c>0$ iff the Riemann Hypothesis holds true? Is there an easy proof?

Answer 1

The original question:

Let $\pi(a)$ be the number of primes below $a>0$. The prime number theorem states $\pi(a)\sim\frac{a}{\ln a}$. My question is trivial. Is $$\frac{a}{\ln a}\leq\pi(a)\leq\frac{a}{\ln a}+c\sqrt{a}\ln a$$ for some constant $c>0$ iff the Riemann Hypothesis holds true? Is there an easy proof?

This is an answer to the question as originally posted.

The question is whether it is true that $|\pi(x)-\frac{x}{\log x}|= O(\sqrt{x}\log x).$ It seemed you had seen this somewhere but the usual statement from Schoenfeld$^1$ assuming RH is that $|\pi(x)-Li(x)|=O(\sqrt{x}\log x).$ My point in the comment was that the suggestion

$$Li(x)\sim \frac{x}{\log x} \implies |\pi(x)-\frac{x}{\log x}|=O(\sqrt{x}\log x)$$

is wrong although the right-hand side might well be true.

Assume RH and so also from Schoenfeld we have $\vartheta(x)-x = O(\sqrt{x}\log^2 x).$ Then

$$|\pi(x)-\frac{x}{\log x}|= \frac{\vartheta(x)-x}{\log x}+\int\frac{\vartheta(t)}{t\log^2 t}dt =\int\frac{\vartheta(t)}{t\log^2 t}dt +\frac{O(\sqrt{x}\log^2 x)}{\log x}$$

$$=|\pi(x) -\frac{x}{\log x}|= O\left(\frac{x}{\log^2 x}\right)+ O(\sqrt{x}\log x) $$

EDIT: I overlooked the integral term in my original answer. The answer to the original question is that the claim is false.

Lowell Schoenfeld, Sharper Bounds for the Chebyshev Functions $\theta(x)$ and $\psi(x)$ (part II), Math. of Computation, vol. 30, no.134 (1976)

Answer 2

The other answer is completely false, unfortunately. RH is equivalent to \[\pi(x) - \mathrm{Li}(x) = O(\sqrt{x} \log x).\] Now it is not hard to show that \[\mathrm{Li}(x) = \frac{x}{\log x} \sum_{k=0}^{m - 1}{\frac{k!}{(\log x)^k}} + O\left(\frac{x}{(\log x)^{m + 1}}\right)\] for any $m \geq 0$ (just use the definition of $\mathrm{Li}(x)$ and repeated integration by parts). Thus \[\pi(x) = \frac{x}{\log x} \sum_{k=0}^{m - 1}{\frac{k!}{(\log x)^k}} + O\left(\frac{x}{(\log x)^{m + 1}}\right).\] In particular, taking $m = 3$ yields \[\pi(x) - \frac{x}{\log x} = \frac{x}{(\log x)^2} + \frac{2x}{(\log x)^3} + O\left(\frac{x}{(\log x)^4}\right).\] So unconditionally we have that \[\pi(x) \geq \frac{x}{\log x} + \frac{x}{(\log x)^2}\] for all sufficiently large $x$, which contradicts the claimed inequality.