A polynomial is completely pinned down by its roots: if you know that p(x) vanishes exactly at r_1, r_2, \dots, r_n, then up to a constant it is the product c\,(x-r_1)(x-r_2)\cdots(x-r_n). The roots and the function are two views of the same object. The most beautiful — and most consequential — idea in this corner of analytic number theory is that the same is very nearly true for the zeta function, an object with infinitely many zeros.

Hadamard's theorem lets us write the completed zeta function as an explicit product over its zeros. Do that, take a logarithmic derivative, and out falls a sum that runs over the non-trivial zeros \rho — the very sum that, matched against a sum running over the primes, becomes the explicit formula. This product is the hinge on which "zeros" turns into "primes". It is worth meeting carefully.

Entire functions have a factorisation too

For polynomials the story ends with the roots. For a general entire function (one holomorphic on the whole plane) it cannot: an entire function can have infinitely many zeros, and a naive infinite product \prod_\rho (1 - s/\rho) over them usually diverges. Weierstrass fixed the divergence by attaching a small correcting exponential to each factor; Hadamard sharpened this into a precise statement once you know the function's order.

The order \lambda of an entire function f measures how fast it can grow: it is the smallest number such that |f(s)| \le \exp(|s|^{\lambda + \varepsilon}) for large |s|. A polynomial has order 0; e^{s} and \sin s have order 1; e^{s^2} has order 2. The magic number for us is order exactly 1 — that is the class the completed zeta function belongs to.

Read the theorem as "polynomial factorisation, plus one exponential prefactor e^{a+bs}, plus one convergence factor per zero." Everything below is just this statement applied to the right function.

The right function is ξ, not ζ

We cannot apply Hadamard directly to \zeta: it has a pole at s=1, and its trivial zeros at -2,-4,-6,\dots clutter the picture. The cure is to bundle \zeta together with the Gamma factor into the completed zeta function \xi:

This \xi is a small marvel. The factor s(s-1) cancels the pole of \zeta at s=1 and the pole of \Gamma(s/2) at s=0; the remaining poles of the Gamma factor sit exactly on top of the trivial zeros of \zeta and annihilate them. What is left is entire, satisfies the clean functional equation

and — the fact we borrow — is of order exactly 1. Its only zeros are precisely the non-trivial zeros of \zeta, all lying in the critical strip 0 < \Re(s) < 1. It is the perfect input for Hadamard's theorem.

The Hadamard product for ξ

Since \xi is entire of order 1 with \xi(0)\neq 0, the theorem applies verbatim. Writing \rho for the non-trivial zeros:

Two symmetries organise the zeros. First, the functional equation \xi(s) = \xi(1-s) means that if \rho is a zero then so is 1-\rho. Second, because \xi is real on the real axis, zeros come in conjugate pairs \rho, \bar\rho. Combining both, the zeros form quadruples \{\rho,\ 1-\rho,\ \bar\rho,\ 1-\bar\rho\} (a symmetric pair \rho, 1-\rho if \rho is on the critical line, where 1-\rho = \bar\rho). This pairing is not decoration — it is what makes the constant B meaningful, as we will see.

Seeing the pairing

The horizontal axis is \Re(s), the vertical axis \Im(s). Take one hypothetical off-line zero \rho and watch its three forced companions appear: the conjugate \bar\rho (mirror across the real axis), the reflection 1-\rho (mirror across the centre line \Re(s)=\tfrac12), and 1-\bar\rho. In the infinite product each such partner supplies a factor, and it is the pairs summing together that keep \sum_\rho 1/\rho under control.

On the Riemann Hypothesis every zero sits on the centre line, where the quadruple collapses to the conjugate pair \rho, \bar\rho = 1-\rho. Either way, the zeros are never alone — they come with a partner whose 1/\rho can be added to theirs.

Why the pairing rescues the sum

Here is the subtle heart of the matter. The zeros are spread up the strip with imaginary parts \pm t growing without bound, and their density means

That is exactly the fingerprint of order 1: the exponent that separates divergence from convergence is 1 itself. (The number of zeros up to height T grows like \tfrac{T}{2\pi}\log T, so \sum 1/|\rho|^s behaves like \int \frac{\log t}{t^{s}}\,dt, which converges for s>1 and diverges at s=1.)

So \sum_\rho 1/\rho is not absolutely convergent — you cannot just add the terms in any order. But pair each \rho with its partner 1-\rho. Since \tfrac{1}{\rho} + \tfrac{1}{1-\rho} = \tfrac{1}{\rho(1-\rho)} and the denominator grows like |\rho|^2, the paired sum converges beautifully. This is why the constant B = -\sum_\rho \Re(1/\rho) only makes sense when the terms are grouped by their symmetric pairs. The convergence factors e^{s/\rho} in Hadamard's product are the mechanism that performs exactly this pairing bookkeeping automatically.

Worked example — logarithmic derivative to ζ'/ζ

The payoff of the product is what happens when we take a logarithmic derivative \frac{d}{ds}\log(\cdot) = \frac{(\cdot)'}{(\cdot)}, because a product turns into a sum. Start from the Hadamard product and take \log:

Differentiate term by term. The constant A dies, Bs gives B, and each bracket gives \frac{d}{ds}\big[\log(1-s/\rho) + s/\rho\big] = \frac{-1/\rho}{1 - s/\rho} + \frac{1}{\rho} = \frac{1}{s-\rho} + \frac{1}{\rho}:

Each zero \rho contributes a simple pole 1/(s-\rho) — the residue-1 signature of a simple zero, exactly what the argument principle would predict — plus the 1/\rho that pairs it into convergence. Now unwrap \xi back into \zeta. Taking the logarithmic derivative of \xi(s) = \tfrac12 s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s) turns the product of factors into a sum of their logarithmic derivatives:

Solving for \zeta'/\zeta and substituting the partial-fraction expansion above gives the identity that carries the zeros into the logarithmic derivative of zeta:

Read it aloud: the logarithmic derivative of \zeta is a constant, a pole -1/(s-1) from \zeta's own pole at s=1, a sum of simple poles one per non-trivial zero, and the smooth Gamma-factor terms that encode the trivial zeros. This is the bridge: on the left, an object we can also expand as -\sum_{n} \Lambda(n) n^{-s} over prime powers; on the right, a sum over zeros. Set the two expansions equal, integrate, and you obtain the explicit formula linking primes to zeros directly.

Reading off the constants

The two constants deserve a second look because they are so often misremembered. From the product, set s=0: every factor (1 - 0/\rho)e^{0} = 1, so \xi(0) = e^{A}, giving A = \log\xi(0) = \log\tfrac12. For B, differentiate \log\xi and evaluate at s=0: only the Bs term survives at s=0 among the pieces that depend on s linearly, and matching against \xi'/\xi(0) yields

That last equality uses the pairing: grouping \rho with \bar\rho turns \tfrac1\rho + \tfrac1{\bar\rho} = 2\Re\tfrac1\rho into real terms and makes the sum converge. Numerically B \approx -0.0230957 — a small negative number, which forces \sum_\rho \Re(1/\rho) > 0. Because every zero has 0 < \Re(\rho) < 1, each \Re(1/\rho) > 0, so the sign is exactly as it should be — a small consistency check that the whole edifice hangs together.