The PNT is indeed equivalent to $\lim_{x\to\infty} \frac{\psi(x) -x}{x}=0$ which Von Mangoldt's formula and some trivial estimates reduce to proving $$ \lim_{x\to\infty} \sum_{ \zeta(\rho)=0} \frac{x^{\rho-1}}{\rho}=0.$$

However, this is not easy. It's easy to show that $\zeta(\rho)=0$ implies $\operatorname{Re}(\rho)\leq 1$. To get the PNT, you at least need to show that each term in the sum converges to $0$, which requires improving the inequality to $\operatorname{Re}(\rho)<1$. This requires its own argument.

However, that is by no means sufficient. Since the sum over zeroes is an infinite sum, one can't simply exchange the sum with the limit. Instead one needs a bound for the individual terms, showing they decay reasonably rapidly, and a bound for the number of terms. The first one is done by a zero-free region, i.e. by an even stronger bound on $\operatorname{Re}(\rho)$, while the second is done by zero-counting estimates. Each requires its own argument.

It's helpful to give a version of the von Mangoldt formula which involves a truncated sum over zeroes with an explicit error term, but I don't think this is necessary as there are other approaches.

Furthermore, to even make von Mangoldt's formula make sense, you need analytic continuation of $\zeta$, which is its own argument. There are some cheap proofs of that but you'll need one that also gives the functional equation since you have that $\ln(1-x^{-2})$ term. I don't know if that is contained in your reference.

The proof also depends on the whole machinery of complex analysis - certainly the Cauchy integral formula but probably also a bunch of other stuff.

If you included all this background material, I'm pretty sure it would be longer than the elementary proofs.

But what is meant by "elementary" doesn't have much to do with the total length, but more to do with the amount of background material, and specifically material that a priori seems to have nothing to do with the original problem - i.e. why to prove a result about counting whole numbers of a certain type should we have to introduce real numbers, complex numbers, functions on the complex numbers, analytic functions, zeroes and poles of meromorphic functions, and numerous other things?

Still the non-elementary proofs are, in fact, generally preferred, since they give better conceptual understanding of why the result is true and how it can be generalized and strengthened.

Having looked at the book, some additional comments: The book indeed skips many steps, even just in the proof of von Mangoldt's formula. In the first displayed equation on p. 201, the function on the left side is not actually integrable, since it decays only as $1/z$ in the imaginary direction and $\int 1/z =\infty$. The integral can be interpreted precisely as a limit of integrals over symmetric intervals. In a few lines this integral (or limit of integrals) is exchanged with an infinite sum. One can't actually always do that, and more justification is needed.

Most crucially, the residue theorem applies to *closed* curves, while the contour integral here is over an infinite line. To apply the residue theorem in this context one needs some estimates on the growth rate of the function (in this case, the logarithmic derivative of zeta) at infinity. To see why this is problematic, note that nothing obvious explains why we should sum over the zeroes to the left of the line $c+it$, instead of to the right (where there are no zeroes, suggesting the sum should be zero). To figure out why the left side works and the right side doesn't one needs to draw more complicated contours and estimate the integrals over those.

The necessity of proving non vanishing of $\zeta$ on the line $1+it$ is mentioned on p. 202. It's not mentioned that this on its own isn't enough to prove the PNT and more work is needed.

Some of the needed material for the proof on the analytic continuation and functional equation is contained in Appendix E. It also doesn't make sense to describe the simplicity of the proof without including this!

Riemannand notReimann. $\endgroup$14more comments