Perron's formula

Let ${\displaystyle \{a(n)\}}$ be an arithmetic function, and let

${\displaystyle g(s)=\sum _{n=1}^{\infty }{\frac {a(n)}{n^{s}}}}$

be the corresponding Dirichlet series. Presume the Dirichlet series to be uniformly convergent for ${\displaystyle \Re (s)>\sigma }$. Then Perron's formula is

${\displaystyle A(x)={\sum _{n\leq x}}'a(n)={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }g(z){\frac {x^{z}}{z}}\,dz.}$

Here, the prime on the summation indicates that the last term of the sum must be multiplied by 1/2 when x is an integer. The integral is not a convergent Lebesgue integral; it is understood as the Cauchy principal value. The formula requires that c > 0, c > σ, and x > 0.

An easy sketch of the proof comes from taking Abel's sum formula

${\displaystyle g(s)=\sum _{n=1}^{\infty }{\frac {a(n)}{n^{s}}}=s\int _{1}^{\infty }A(x)x^{-(s+1)}dx.}$

This is nothing but a Laplace transform under the variable change ${\displaystyle x=e^{t}.}$ Inverting it one gets Perron's formula.

Proofs of Perron's formula have been published by Tom M. Apostol^[2] and by Gérald Tenenbaum.^[3]

Because of its general relationship to Dirichlet series, the formula is commonly applied to many number-theoretic sums. Thus, for example, one has the famous integral representation for the Riemann zeta function:

${\displaystyle \zeta (s)=s\int _{1}^{\infty }{\frac {\lfloor x\rfloor }{x^{s+1}}}\,dx}$

and a similar formula for Dirichlet L-functions:

${\displaystyle L(s,\chi )=s\int _{1}^{\infty }{\frac {A(x)}{x^{s+1}}}\,dx}$

where

${\displaystyle A(x)=\sum _{n\leq x}\chi (n)}$

and ${\displaystyle \chi (n)}$ is a Dirichlet character. Other examples appear in the articles on the Mertens function and the von Mangoldt function.

Perron's formula is just a special case of the formula

${\displaystyle \sum _{n=1}^{\infty }a(n)f(n/x)={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }F(s)G(s)x^{s}ds}$

where

${\displaystyle G(s)=\sum _{n=1}^{\infty }{\frac {a(n)}{n^{s}}}}$

and

${\displaystyle F(s)=\int _{0}^{\infty }f(x)x^{s-1}dx}$

the Mellin transform. The Perron formula is just the special case of the test function ${\displaystyle f(1/x)=\theta (x-1),}$ for ${\displaystyle \theta (x)}$ the Heaviside step function.