Bounds for PNT from Wiener-Ikehara? 
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*What sort of bounds on the error term in the Prime Number Theorem can one obtain through a Wiener-Ikehara approach?


*Same question, but for the Mertens function $M(x)=\sum_{n\leq x} \mu(n)$.
 A: The Wiener-Ikehara approach to PNT (as an asymptotic statement $\pi(x) \sim x/\log x$) only needs the nonvanishing of $\zeta(s)$ on ${\rm Re}(s) \geq 1$, nothing about its growth conditions. I don't think you can expect an error term in PNT just from the minimal assumption of no zeros on ${\rm Re}(s) \geq 1$.
If you allow some growth conditions on ${\rm Re}(s) \geq 1$ then error terms can be given. See the following three papers, with MathSciNet reference numbers at the end of each.

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*J. Čížek, On the proof of the prime number theorem, Časopis Pěst. Mat. 106 (1981), no. 4, 395–401, 436. (MR0637818)


*A. Grant, Fourier transforms and the P.N.T. error term,
Časopis Pěst. Mat. 112 (1987), no. 4, 337–347. (MR0921321)


*A. Grant, A further note on the P.N.T. error term,
Časopis Pěst. Mat. 112 (1987), no. 4, 348–350. (MR0921322)
For $\pi(x) - {\rm Li}(x)$, the first paper got the error term $O(x/(\log x)^n)$ for all positive integers $n$ using only information about the zeta-function on ${\rm Re}(s) \geq 1$: the nonvanishing and upper bounds on all $|\zeta^{(k)}(s)|$ and on $|1/\zeta(s)|$ away from a neighborhood of $s=1$. The second paper got error term $O(xe^{-g(x)})$, where $g(x) = (1/25)(\log \log x)^2/\log\log\log x$ The third paper got error term $O(xe^{-g(x)})$, where $g(x) = L(\log x)^{1/12}$ for some $L > 0$.
A: Per KConrad's suggestion, I am turning my comment into an answer.
The Wiener-Ikehara tauberian theorem states that if $S:[0,\infty)\to\mathbb{R}$ is a non-decreasing function and
$$F(s) = \int_1^{\infty}S(x)x^{-s}\frac{dx}{x}$$
converges for Re$(s)>1$ and that there exists $a\in\mathbb{C}$ such that
$$F(s)-\frac{a}{s-1}$$
continuously extends to Re$(s)\geq 1$, then
$$\lim_{x\to\infty}\frac{S(x)}{x}=a.$$
On the other hand, a version of the Ingham-Karamata tauberian theorem states that if $S:[0,\infty)\to\mathbb{R}$ is Lipschitz continuous on its domain and $F(s)$ analytically continues across the line Re$(s)=0$, then
$$\lim_{x\to\infty}S(x) = 0.$$
It is reasonable to ask if these two results can be interpolated.  That is, if $F$ has a pole at $s=1$ of order at most 1 with residue $a$ (equal to zero if no pole exists) and there exists a constant $\delta>0$ such that
$$F(s) - \frac{a}{s-1}$$
analytically continues to the region Re$(s)\geq 1-\delta$, do tauberian arguments alone ensure that there exists a constant $0\leq c(\delta)<1$ such that
$$S(x) = ax+O(x^{c(\delta)})?$$
The answer is no.  This was proved by Debruyne and Vindas.  One needs additional information on $F$, such as quantitative upper and lower bounds on $F$ and/or its derivatives in the region of analytic continuation.  Without such information, the rate of convergence in the above tauberian results is in fact optimal (which can be quantified as an $\Omega_{\pm}$ statement), and one can construct examples of $S$ and $F$ that show this.  See the follow-up paper by Debruyne, Vindas, and Broucke.
This means that if you want a nontrivial error term in the prime number theorem via tauberian methods, even if you assume RH, then you must have more information about $\zeta(s)$, probably in the form of quantitative upper and/or lower bounds in the zero-free region you use/assume.  I think that it is likely (but I do not know definitively) that any upper and/or lower bound that you insert into a tauberian proof would yield a better error term in the prime number theorem if instead you insert said bounds into an "explicit formula".  You could also proceed as in the largely elementary (and quite beautiful) proof of the prime number theorem on page 40 of Iwaniec and Kowalski's book Analytic Number Theory.  (I say "largely elementary" because they use Fourier inversion.  But everything in their proof does take place in the region Re$(s)>1$.)
