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Statement

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

g ( s ) = ∑ n = 1 ∞ a ( n ) n s {\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 ℜ ( s ) > σ {\displaystyle \Re (s)>\sigma } . Then Perron's formula is

A ( x ) = ∑ n ≤ x ′ a ( n ) = 1 2 π i ∫ c − i ∞ c + i ∞ g ( z ) x z z d z . {\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.

Proof

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

g ( s ) = ∑ n = 1 ∞ a ( n ) n s = s ∫ 1 ∞ A ( x ) x − ( s + 1 ) d x . {\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 x = e t . {\displaystyle x=e^{t}.} Inverting it one gets Perron's formula.

Examples

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:

ζ ( s ) = s ∫ 1 ∞ ⌊ x ⌋ x s + 1 d x {\displaystyle \zeta (s)=s\int _{1}^{\infty }{\frac {\lfloor x\rfloor }{x^{s+1}}}\,dx}

and a similar formula for Dirichlet L-functions:

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

where

A ( x ) = ∑ n ≤ x χ ( n ) {\displaystyle A(x)=\sum _{n\leq x}\chi (n)}

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

Generalizations

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

∑ n = 1 ∞ a ( n ) f ( n / x ) = 1 2 π i ∫ c − i ∞ c + i ∞ F ( s ) G ( s ) x s d s {\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

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

and

F ( s ) = ∫ 0 ∞ f ( x ) x s − 1 d x {\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 f ( 1 / x ) = θ ( x − 1 ) , {\displaystyle f(1/x)=\theta (x-1),} for θ ( x ) {\displaystyle \theta (x)} the Heaviside step function.

• Page 243 of Apostol, Tom M. (1976), Introduction to analytic number theory, Undergraduate Texts in Mathematics, New York-Heidelberg: Springer-Verlag, ISBN 978-0-387-90163-3, MR 0434929, Zbl 0335.10001
• Weisstein, Eric W. "Perron's formula". MathWorld.
• Tenenbaum, Gérald (1995). Introduction to analytic and probabilistic number theory. Cambridge Studies in Advanced Mathematics. Vol. 46. Translated by C.B. Thomas. Cambridge: Cambridge University Press. ISBN 0-521-41261-7. Zbl 0831.11001.