Explicit formula

Remarks

This is the explicit formula of Riemann and von Mangoldt. The proof starts from Perron's formula: for $c > 1$, $$\psi_0(x) = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} \left(-\frac{\zeta'(s)}{\zeta(s)}\right) \frac{x^s}{s} \, ds,$$ because $-\zeta'(s)/\zeta(s) = \sum_n \Lambda(n) n^{-s}$ with the von Mangoldt function $\Lambda$. Moving the line of integration to the left picks up residues. The pole at $s = 1$ gives $x$, and each nontrivial zero gives $-x^\rho/\rho$. The point $s = 0$ gives $-\zeta'(0)/\zeta(0) = -\log(2\pi)$. Each trivial zero $-2n$ gives $x^{-2n}/(2n)$, and these terms add up to $-\frac{1}{2}\log(1 - x^{-2})$. The sum over zeros converges only conditionally, so the order of summation matters. Heuristically, the prime number theorem $\psi(x) \sim x$ reflects that each term $x^\rho/\rho$ is $o(x)$, since $\operatorname{Re} \rho < 1$. A rigorous proof along these lines also needs a zero-free region to control the whole sum.

Going deeper

For a student

Under the Riemann hypothesis each term has size $\abs{x^\rho} = \sqrt{x}$, and a truncated form of the formula gives $\psi(x) = x + O(\sqrt{x} \log^2 x)$. Weil generalized the formula to a symmetric identity between a sum over the zeros of an L-function and a sum over prime powers, for a class of test functions. Similar formulas hold for Dirichlet L-functions, and they give the distribution of primes in arithmetic progressions.