Let $\pi(x)$ be the prime counting function. The statement $\pi(x) = Li(x) + O(\ln^3(x) \sqrt x) $ is in "essence" weaker than what we can conclude from the RH if the RH is true.
I assume it is correct to say that this statement does not imply the "stronger" statement $|\pi(x) - Li(x)| < \ln(x) \sqrt x$ for $x>2656$ that follows from the RH ? If that is incorrect, let me know.
If $\pi(x) = Li(x) + O(\ln^3(x) \sqrt x) $ is true, and it is indeed weaker than the RH, then what does that say about the nontrivial Riemann zeta zero's ? Do they change in real part or distance or multiplicity ? And how does that work ?
edit : I used big O notation :
see https://en.wikipedia.org/wiki/Big_O_notation
edit 2 :
To clarify (or avoid confusion) I mentioned
$|\pi(x) - Li(x)| < \ln(x) \sqrt x$
and this follows from the truth of the Riemann Hypothesis,
But the Riemann Hypothesis implies a stronger statement :
and I am not sure if there is a stronger implication from RH or if it would be possible to find a stronger implication from RH , but anyways RH implies this :
$|\pi(x) - Li(x)| < \frac{\ln(x) \sqrt x}{8 \pi}$ for $x>2656$.
Also I am uncertain what is the sharpest bound that is still consistant with the RH.
For instance is
$|\pi(x) - Li(x)| < \frac{\sqrt{\ln(x) x}}{8 \pi}$ for $x>q$.
for some fixed $q$ still potentially consistant with RH ?
The following question and answers were clarifying :
Riemann Hypothesis and the prime counting function
copy answer :
Any bound $|\pi(x)-\rm {li}(x)|\le f(x)$ with $f(x)=O(x^{1/2 + \epsilon})$ implies the Riemann hypothesis, and RH is equivalent to the existence of a bound of that type. The point of this particular $f(x)$ is that it is known to be provable from the Riemann hypothesis, and the parameters in the proof have been worked out explicitly (and might be some of the best ones currently available), so that the end result is not "$f(x) = A\sqrt{x}(\log x)^B$ for $x > C$ for some $A,B$ and $C$", but $A = \frac{1}{8 \pi}$, $B=1$, $C = 2657$. The $O()$ notation signifies that the limit of $\frac{\log f(x)}{\log x}$ is $1/2$; if the limit had been $\rho \in [\frac{1}{2},1]$ the (non-trivial) zeros of the Riemann zeta would have real parts in the interval $[1 - \rho, \rho]$. The same bound with any other values of $A,B,C$ would also have been equivalent to RH.
$C$ can be eliminated by inflating $A$ to cover the first $2656$ cases and then one would get a bound valid for all $x \geq 1$. This would be less informative, because the important thing is to minimize the exponent $B$ of the logarithm (which is related to the vertical distribution of the zeros), then the constant $A$ (which measures some finer aspect of the zero distribution, but I don't know if it has been articulated what that is). The cutoff $C$ is much less significant, because it is not an asymptotic, large-$x$, quantity, and can be affected by moving a single zero along the critical line. (end copy)
But this still does not completely answer my question. It does show that the real parts are still $1/2$ though.
See also the related question :
where I ask for the case,
Is
$|\pi(x) - Li(x)| < \frac{\sqrt{\ln(x) x}}{8 \pi}$ for $x>q$.
for some fixed $q$ still potentially consistant with RH ?
If $\zeta(s)$ has a nontrivial zero $\rho = \sigma_0 + it_0$ with $\frac{1}{2} < \sigma_0$, then
$$ \pi(X) - \mathrm{Li}(X) = \Omega_{\pm}(X^{\sigma_0}).$$
As this is a less common Landau-type notation, this means that $\limsup ([\pi(X) - \mathrm{Li}(X)]/X^{\sigma_0}) > 0$ and $\liminf([\pi(X) - \mathrm{Li}(X)]/X^{\sigma_0}) < 0$. In words, $\pi(X) - \mathrm{Li}(X)$ grows at least as large as $X^{\sigma_0}$ both positively and negatively, infinitely often.
Thus $\pi(X) - \mathrm{Li}(X) = O(X^{\frac{1}{2} + \epsilon})$ for all $\epsilon > 0$ necessarily implies that there are no zeros with real part greater than $1/2$, and thus implies the Riemann Hypothesis.