The Euler product tells us that the zeta function knows about the primes. But knowing is not the same as speaking: the product itself is a clumsy object to push analysis through. The single instrument that turns zeta's silent knowledge into an audible statement about primes is its logarithmic derivative,

This one function is the hinge of the whole theory. On one side it is a tidy Dirichlet series whose coefficients are the von Mangoldt function \Lambda(n) — pure prime data. On the other side, its poles sit exactly at the zeros of \zeta. Feed it into a contour integral and those two descriptions collide: the primes on one side, the zeros on the other. That collision is the explicit formula for \psi(x). Everything in this course flows through \zeta'/\zeta.

From the Euler product to a prime-powered series

Start where the primes live — the Euler product, valid for \Re(s) > 1:

A product is hard to differentiate, but a logarithm turns it into a sum. Take \log of both sides:

Now differentiate term by term with respect to s. Recall \frac{d}{ds}p^{-s} = -p^{-s}\log p. For a single prime,

Summing over all primes gives the logarithmic derivative directly, and it is cleaner to write with the overall minus sign moved to the front:

So -\zeta'/\zeta is the Dirichlet series of \Lambda. This is why \zeta'/\zeta carries the primes: its coefficients are supported entirely on prime powers, each weighted by \log p.

Worked example: expanding one prime's factor

Let's verify the last equality by expanding a single prime's contribution as a geometric series and checking that its coefficients really are \Lambda. Fix a prime p and look at

Because \Re(s) > 1 we have |p^{-s}| < 1, so \frac{1}{1 - p^{-s}} = \sum_{k\ge 0}p^{-ks}. Multiply through by p^{-s}\log p:

Every term is \log p divided by (p^{k})^{s} — exactly \Lambda(p^{k})/(p^{k})^{s}. So this single prime contributes the series terms at n = p, p^{2}, p^{3}, \dots, each with coefficient \log p, and nothing anywhere else. Take p = 2 for concreteness: the factor produces

Summing over all primes stitches these strands together — \Lambda(4) = \log 2, \Lambda(9) = \log 3, \Lambda(6) = 0 (since 6 is not a prime power) — reproducing \sum_n \Lambda(n)n^{-s} term for term.

The other face: poles at the zeros

The Dirichlet series above only lives in the half-plane \Re(s) > 1. But \zeta'/\zeta continues meromorphically to the whole plane, and there it wears a completely different face — a landscape of simple poles. The rule for reading them is elementary complex analysis: the logarithmic derivative of any meromorphic function has a simple pole at each zero and each pole of that function, with residue equal to the order.

The proof is one line: write f(s) = (s - s_{0})^{m}g(s) with g(s_{0}) \ne 0. Then \log f = m\log(s - s_{0}) + \log g, and differentiating gives f'/f = \dfrac{m}{s - s_{0}} + \dfrac{g'}{g}, where the second piece is analytic at s_{0}. Apply this to \zeta:

• At each zero \rho of \zeta (order m_{\rho}), the function \zeta'/\zeta has a simple pole of residue +m_{\rho} — the multiplicity of the zero.
• At s = 1, where \zeta has its simple pole (order 1), \zeta'/\zeta has a simple pole of residue -1.

Read the two faces together: on the right, coefficients that are prime data; across the plane, poles that are the zeros. \zeta'/\zeta is the translator between them.

Seeing the pole landscape

Every marked point below is a simple pole of \zeta'/\zeta. One special pole sits at s = 1 with residue -1 (inherited from zeta's pole). All the others sit at zeros of \zeta: the nontrivial zeros \rho = \tfrac12 + i\gamma in symmetric pairs up the critical line, and the trivial zeros at s = -2, -4, -6, \dots. Each carries residue equal to the multiplicity of that zero (+1 for a simple zero).

This picture is the machine. When we integrate -\dfrac{\zeta'(s)}{\zeta(s)}\dfrac{x^{s}}{s} around a large contour, the residue theorem harvests exactly these dots: the pole at s = 1 yields the main term, and each zero \rho yields an oscillating correction.

The partial-fraction expansion

There is a closed formula that lays all of these poles out at once. Applying the Hadamard product for \zeta (its factorisation over the zeros) and taking the logarithmic derivative produces the celebrated expansion:

Every ingredient is a pole you already recognise:

• -\dfrac{1}{s - 1} is the pole at s = 1 with residue -1 — the pole of \zeta showing through.
• each \dfrac{1}{s - \rho} is a pole of residue +1 at a nontrivial zero \rho.
• the term -\tfrac12\dfrac{\Gamma'}{\Gamma}\!\left(\tfrac{s}{2}+1\right) packages the Gamma factor from the functional equation; its own poles (from \Gamma'/\Gamma at s = -2, -4, \dots) are the trivial zeros of \zeta.
• the paired +\dfrac{1}{\rho} terms are convergence factors — they make the sum over \rho converge without changing where the poles are.

The whole content of "the zeros drive the primes" is compressed into this one line: it is \zeta'/\zeta written as a running total of its poles.

How this manufactures the explicit formula

Here is the payoff, in outline. The Chebyshev function \psi(x) = \sum_{n\le x}\Lambda(n) is the natural prime-counting sum (weighted by \log p). Perron's formula recovers such a partial sum from the Dirichlet series by a contour integral:

Now push the contour far to the left and collect residues. The integrand is -\zeta'/\zeta times x^{s}/s, so at every pole of \zeta'/\zeta we pick up a residue:

• the pole at s = 1 (residue of -\zeta'/\zeta is +1) times x^{s}/s = x^{1}/1 gives the main term x;
• each zero \rho (residue -1 from -\zeta'/\zeta) contributes -\dfrac{x^{\rho}}{\rho};
• the trivial zeros and the pole of x^{s}/s at s = 0 supply the remaining lower-order pieces.

Assembling the residues gives the explicit formula

Read it as sound: x is the steady main tone, and each zero \rho = \tfrac12 + i\gamma adds an oscillation of amplitude x^{1/2}/|\rho| and frequency \gamma. The primes' fluctuations are a chord struck by the zeros — and every note passed through \zeta'/\zeta to get here.