We already know what the Prime Number Theorem says — \pi(x) \sim x/\ln x — and we've seen the one-line slogan that it comes down to the zeta function having no zeros on the line \Re(s)=1. This page is the real thing: the full Hadamard–de la Vallée Poussin proof of 1896, assembled piece by piece. By the end you'll see exactly how a statement about counting primes gets converted into a contour integral, how the answer x falls out of a single pole, and where the genuinely hard analysis actually lives (spoiler: not where a first glance suggests).

The whole argument is a machine with five moving parts. Get the parts in your head first, then we'll service each one:

Add the last two together: \psi(x) = x + o(x), i.e. \psi(x)\sim x. A short summation-by-parts step turns that into \pi(x)\sim x/\ln x, and we are done.

Move 1 — the generating series

We do not count primes directly. Instead we weight each prime power by a logarithm using the von Mangoldt function

and study its running total, the Chebyshev function \psi(x) = \sum_{n\le x}\Lambda(n). The pay-off is that this weighting is exactly the one the zeta function produces when you take its logarithmic derivative. Starting from the Euler product \zeta(s)=\prod_p (1-p^{-s})^{-1}, taking \log and then -\,d/ds gives the identity developed on the logarithmic derivative of zeta:

This is the hinge of the entire proof. On the left is a single analytic object whose poles we can chase around the complex plane; on the right is the arithmetic we want to understand. A pole of -\zeta'/\zeta appears wherever \zeta has a pole (at s=1) or a zero (with a sign flip). That is why the location of zeta's zeros will end up dictating the size of \psi(x).

Move 2 — Perron's formula turns the series into an integral

Perron's formula is the dictionary between a Dirichlet series and the partial sums of its coefficients. Applied to our series with coefficients \Lambda(n), it reads

The vertical line of integration sits at some fixed c>1, safely inside the region where the series converges. The factor x^{s}/s is doing the counting: as an operator it acts like a smoothed step function that is 1 for n and 0 for n>x, so the integral selects exactly the terms with n\le x. All the arithmetic is now hidden inside one complex integral. The rest of the proof is a single question: what happens to this integral as we slide the line of integration to the left?

Moves 3 & 4 — shift the contour and collect the pole

Cauchy's residue theorem says that sliding a contour across an analytic region changes the integral only by the residues of whatever singularities you cross. We move the line from \Re(s)=c leftward toward and just past \Re(s)=1. On the way we cross exactly one singularity: the pole of the integrand at s=1. Schematically:

The integrand \left(-\zeta'/\zeta\right)x^{s}/s has a simple pole at s=1, inherited from the simple pole of \zeta itself. Crossing it contributes its residue to \psi(x), and — this is the punchline of the whole subject — that residue is exactly x. So

The first term is the main term. The second — the integral along the new, shifted contour — is what we must show is small. Let's nail the residue first, because it is the easy and beautiful part.

Worked example — why the main term is exactly x

Near s=1, the zeta function has a simple pole with residue 1: its Laurent expansion is \zeta(s) = \dfrac{1}{s-1} + \gamma + O(s-1). Taking the logarithmic derivative of a simple pole is mechanical. If \zeta(s) \approx \dfrac{1}{s-1} near s=1, then \ln\zeta(s) \approx -\ln(s-1), so

In other words -\zeta'/\zeta also has a simple pole at s=1, and crucially its residue there is +1 — a pole of \zeta of order one becomes a pole of -\zeta'/\zeta with residue equal to that order. Now multiply by the smooth factor x^{s}/s, which at s=1 equals x^{1}/1 = x. For a simple pole the residue is just (residue of the pole factor) times (value of the smooth factor):

There it is. The residue is x because zeta's pole at s=1 has residue 1, and because x^{s}/s evaluates to x at s=1. The famous asymptotic \psi(x)\sim x is, at heart, the statement "\zeta has a simple pole of residue 1 at s=1". Everything else in the proof is the work of showing that this main term is not swamped by the leftover integral.

Move 5 — the zero-free region makes the rest o(x)

We could not have shifted the contour to the left at all if the integrand had singularities blocking the way. Its singularities to the left of s=1 are precisely the zeros of \zeta (each zero of \zeta is a pole of -\zeta'/\zeta). So the shift is legal exactly to the extent that we have a region, just left of the line, that is free of zeros. This is the theorem that took Hadamard and de la Vallée Poussin 37 years after Riemann to reach:

Inside this region we also need bounds on how large \zeta'/\zeta can get, so that the integral over the shifted contour is genuinely controlled. The standard estimate is polynomial in the height:

With the contour pushed to \Re(s)=\sigma = 1 - c/\ln(|t|+2), the size of the integrand along it is governed by |x^{s}| = x^{\sigma} = x\cdot x^{-c/\ln(|t|+2)}. The factor x^{-c/\ln(\cdots)} is what beats the main term down: it is strictly less than 1, so the shifted integral is smaller than x by a factor that tends to 0. Hence the leftover is o(x), and \psi(x) = x + o(x).

Truncation at height T, and balancing T against x

There is a practical wrinkle we glossed over. The Perron integral runs all the way up the infinite vertical line, but you cannot shift an infinitely long contour and control it for free. The rigorous version uses the truncated Perron formula: integrate only between heights -T and +T, and pay an explicit error for the missing tails,

Now two competing effects must be balanced by the single free parameter T:

You cannot make both terms tiny at once, so you choose T as a function of x to make the two errors comparable. Pushing the algebra through — set T so that \ln T \asymp \sqrt{\ln x} — the optimum yields the classical de la Vallée Poussin error term

which is comfortably x + o(x). The \exp(-c\sqrt{\ln x}) savings is exactly the imprint of a zero-free region of width \sim 1/\ln T: widen the region and the error shrinks; the Riemann Hypothesis, a zero-free region all the way to \Re(s)=\tfrac12, would upgrade this to an error of size \sqrt{x}\,(\ln x)^2.

From \psi(x)\sim x to \pi(x)\sim x/\ln x

The final step is bookkeeping, not analysis. The three Chebyshev counts are asymptotically equal, \psi(x)\sim\vartheta(x)\sim x (the prime powers p^k,\,k\ge2 contribute only O(\sqrt{x}\ln x), which is negligible), and a summation by parts converts the log-weighted count \vartheta(x)=\sum_{p\le x}\ln p into the plain count \pi(x)=\sum_{p\le x}1:

And there is the Prime Number Theorem, delivered by a contour integral. The engine ran: a series, a pole, a shift, a residue of x, and a zero-free region to sweep the remainder under o(x).