Riemann $\zeta$ and Chebyshev's estimates

Question

I would like to shred some light concerning the relations between properties of the Riemann $\zeta$ function and the weak version of the Prime Number Theorem proven by Chebyshev. More precisely, we have the following known facts :

• the pole at $s=1$ is equivalent to the infinitude of primes
• the non-existence of zeros on $\sigma = 1$ is equivalent to the PNT in the form $\pi(x) \sim x / \log x$
• the known (or conjectured) zero-free region implies estimates for the error term

However, the proofs I find concerning the Chebyshev's result $$\pi(x) \asymp \frac{x}{\log x}$$

do not use any analytic property of the Riemann zeta function but rather computations with orders of arithmetic functions (I do not know whether or not the proof I have in mind is Chebyshev's).

Is there any proof using analytic properties of $\zeta$?

My attention has been caught by a remark in Montgomery-Vaughan's book (Multiplicative Number Theory), stating that Chebyshev's estimates have been derived from the behavior of $\log \zeta(s)$ when $s \to 1^{+}$.

Answer 1

How about this for a start? It shows $\limsup_{x \to \infty} \frac{\pi(x)}{x/\log x} \ge 1$.

For $\sigma > 1$, $$\log \zeta(s) = \log \prod_p (1-p^{-s})^{-1} = \sum_p \sum_{m \ge 1} \frac{1}{mp^{ms}} = \sum_p \frac{1}{p^s}+O_{s \to 1}(1).$$ Now let $s=1+\epsilon$ for small $\epsilon > 0$. Then $$\log \frac{1}{\epsilon} \sim \log \zeta(s) = \sum_p \frac{1}{p^s} = \lim_{x \to \infty} \sum_{p \le x} \frac{1}{p^s} = \lim_{x \to \infty} \frac{\pi(x)}{x^s}+s\int_1^x \frac{\pi(t)}{t^{s+1}}dt = (1+\epsilon)\int_1^\infty \frac{\pi(t)}{t^{2+\epsilon}}dt.$$

Now if there were some $\delta > 0$ so that $\pi(t) \le (1-\delta)\frac{t}{\log t}$ for all large $t$, then $$1 = \lim_{\epsilon \downarrow 0} \frac{\log(1/\epsilon)}{\log(1/\epsilon)} = \lim_{\epsilon \downarrow 0} \frac{1}{\log(1/\epsilon)}\int_1^\infty \frac{\pi(t)}{t^{2+\epsilon}}dt \le \lim_{\epsilon \downarrow 0} \frac{1}{\log(1/\epsilon)}(1-\delta)\int_2^\infty \frac{1}{t^{1+\epsilon}\log(t)}dt = 1-\delta,$$ a contradiction.

Answer 2

A simple Google search for "proof of prime number theorem" turned up this, which appears to be what you want:

/https://people.mpim-bonn.mpg.de/zagier/files/doi/10.2307/2975232/fulltext.pdf

Here is the title and first paragraph:

Newman's Short Proof of the Prime Number Theorem

D. Zagier

Dedicated to the Prime Number Theorem on the occasion of its 100th birthday

The prime number theorem, that the number of primes < x is asymptotic to x/log x, was proved (independently) by Hadamard and de la Vallee Poussin in 1896. Their proof had two elements: showing that Riemann's zeta function $\zeta(s)$ has no zeros with Re(s) = 1, and deducing the prime number theorem from this. An ingenious short proof of the first assertion was found soon afterwards by the same authors and by Mertens and is reproduced here, but the deduction of the prime number theorem continued to involve difficult analysis. A proof that was elementary in a technical sense-it avoided the use of complex analysis-was found in 1949 by Selberg and Erdos, but this proof is very intricate and much less clearly motivated than the analytic one. A few years ago, however, D. J. Newman found a very simple version of the Tauberian argument needed for an analytic proof of the prime number theorem. We describe the resulting proof, which has a beautifully simple structure and uses hardly anything beyond Cauchy's theorem.