Perron's formula

Remarks

The proof rests on the discontinuous integral. Fix $c > 0$, and for $y > 0$ let $I(y) = \lim_{T \to \infty} \frac{1}{2\pi i} \int_{c - iT}^{c + iT} \frac{y^{s}}{s} \, ds$. Then $I(y) = 1$ for $y > 1$, $I(1) = 1/2$, and $I(y) = 0$ for $0 < y < 1$. For $y > 1$ one closes the path to the left and picks up the residue $1$ at $s = 0$; for $y < 1$ one closes it to the right, where there is no pole. One applies this with $y = x/n$ to each term $a(n) n^{-s}$ and exchanges sum and limit. The exchange uses the explicit error bound of the discontinuous integral and the absolute convergence of $F$ on $\operatorname{Re} s = c$. The limit cannot be replaced by an absolutely convergent integral in general. For $F = 1$, that is $a(1) = 1$ and $a(n) = 0$ otherwise, the integrand $x^{s}/s$ has modulus of order $1/\abs{s}$ on the line. Example: take $a = \Lambda$, the von Mangoldt function, so that $F(s) = -\zeta'(s)/\zeta(s)$, and take $c > 1$. Then the formula expresses the Chebyshev function $\psi(x)$ as a contour integral. Shifting the contour to the left and collecting the residue $x$ at $s = 1$ is the route to the prime number theorem.

Going deeper

For a student

The formula is an inverse Mellin transform: by the Mellin representation of Dirichlet series, $F(s)/s$ is the Mellin transform at $-s$ of the summatory function. In practice one uses a truncated version with an explicit error term in $T$ and $x$, then moves the line of integration to the left of $\operatorname{Re} s = 1$. The poles of $F(s) x^{s}/s$ then give the main terms, and bounds for $F$ in the critical strip control the error. This is how the prime number theorem and its explicit formulas are derived from properties of the zeta function.