Bounds on error term in prime number theorem directly from exponential sum estimates Most improvements on the zero-free region for $\zeta(s)$ go through bounds on the exponential sums $$\sum_{n\sim N} n^{it}$$
for $N$ in certain ranges depending on $|t|$. Is there any way to directly show the prime number theorem from such exponential sum bounds (with the improved error terms the resulting zero-free regions yield), without going through $\zeta(s)$?
 A: There is not a commonly agreed definition on "without going through $\zeta(s)$," but it is possible to wrap up all the $\zeta$ processes between the exponential sum and the remainder of the PNT.
According to Titchmarsh's The theory of the Riemann zeta function, it is not difficult to convert growth condition of $\zeta(s)$ into a zero-free region:
Lemma A (see §3.10 of Titchmarsh): Let $\phi(t)$ and $1/\theta(t)$ be positive nondecreasing functions defined on $t\ge0$ such that

*

*$\zeta(s)\ll\exp\phi(t)$ in $1-\theta(t)\le\sigma\le2$ and $s=\sigma+it$

*$\theta(t)\le1$

*$\phi(t)\to+\infty$ as $t\to+\infty$

*$\phi(t)/\theta(t)=o(\exp\phi(t))$
Then there is a constant $c_0>0$ such that $\zeta(s)$ is free of zeros whenever
$$
\sigma\ge1-c_0{\theta(2|t|+1)\over\phi(2|t|+1)}
$$
To convert the zero-free region into the remainder of the PNT, we may consider the explicit formula provided by Montgomery & Vaughan's Multiplicative Number Theory I: Classical Theory:
Lemma B (see §12.1 of Montgomery & Vaughan): For $T\le x$ and $\psi(x)=\sum_{n\le x}\Lambda(n)$, we have
$$
\psi(x)=x-\sum_{|\Im\rho|\le T}{x^\rho\over\rho}+\mathcal O\left(x\log x\over T\right)
$$
where $\rho$ denotes the nontrivial zeros of $\zeta(s)$.
Under Lemma B, we see that to obtain remainder for the PNT, all we need is to estimate the sum over nontrivial zeros. For convenience, we write
$$
\Theta_T=1-c_0{\theta(2T+1)\over\phi(2T+1)}
$$
Then we have
$$
\left|\sum_{|\Im\rho|\le T}{x^\rho\over\rho}\right|\le x^{\Theta_T}\sum_{|\Im\rho|\le T}{1\over|\Im\rho|}
$$
To estimate the reciprocal sum, we quote Riemann-von Mangoldt formula:
Lemma C (see §9.4 of Titchmarsh): Let $N(T)$ denote the number of nontrivial zeros of $\zeta(s)$ with imaginary parts lying between zero and $T$, then
$$
N(T)={T\over2\pi}\log{T\over2\pi}-\log{T\over2\pi}+\mathcal O(\log T)
$$
This indicates that
$$
\sum_{|\Im\rho|\le T}{1\over|\Im\rho|}=2\int_0^T{\mathrm dN(u)\over u}\ll(\log T)^2
$$
Plugging these results into Lemma B, we get
Theorem (relationship between $\zeta(s)$'s growth condition and the remainder of the PNT): Let $\theta(t)$ and $\phi(t)$ be functions satisfying the conditions of Lemma A. Then there exists a constant $c_0>0$ such that for any $T\le x$, we have
$$
\psi(x)=x+\mathcal O\left\{x(\log x)^2\exp\left(-{c_0\theta(2T+1)\log x\over\phi(2T+1)}\right)+{x\log x\over T}\right\}
$$
