TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04

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Dirichlet series from arithmetic functions admit the Abel-summation integral formula

Statement

Let $θ \in R$ , let $(a_{n})_{n \geq 1}$ be complex coefficients, and put $A (x) : = \sum_{1 \leq n \leq x} a_{n}$ . If $A (x) = O (x^{θ})$ , then for every $s$ with $ℜ s > θ$ ,

$\sum_{n \geq 1} a_{n} n^{- s} = s \int_{1}^{\infty} A (x) x^{- s - 1} d x .$

For every integer $N \geq 1$ one has the endpoint formula

$\sum_{1 \leq n \leq N} a_{n} n^{- s} = A (N) N^{- s} + s \int_{1}^{N} A (x) x^{- s - 1} d x .$

Facts & Assumptions

Given: A real number $θ$ , complex coefficients $(a_{n})_{n \geq 1}$ , their summatory function $A (x) = \sum_{1 \leq n \leq x} a_{n}$ , and a complex number $s$ with $ℜ s > θ$ .

[L1]

Abel summation for complex coefficients expresses finite weighted sums through their partial sums (Abel summation by parts for complex coefficients and their partial sums).

[L2]

The growth bound means $∣ A (x) ∣ \leq C x^{θ}$ for large $x$ .

Proof

technique · direct

1.1

Fix an integer $N \geq 1$ . Extend the coefficients by $a_{0} : = 0$ , set $b_{0} : = 0$ , and put $b_{n} : = n^{- s}$ for $1 \leq n \leq N$ . The partial sums $S_{n} : = \sum_{k = 0}^{n} a_{k}$ in [L1] then satisfy $S_{0} = 0$ and $S_{n} = A (n)$ for $1 \leq n \leq N$ . Applying the tail identity in [L1] with $p = 1$ and $q = N$ gives $\sum_{n = 1}^{N} a_{n} n^{- s} = A (N) N^{- s} + \sum_{n = 1}^{N - 1} A (n) (n^{- s} - (n + 1)^{- s}) .$ Since $n^{- s} - (n + 1)^{- s} = s \int_{n}^{n + 1} x^{- s - 1} d x,$ the finite sum is exactly $s \int_{1}^{N} A (x) x^{- s - 1} d x$ , because $A (x)$ is constant on each interval $[n, n + 1)$ .

L1givenalgebra

2.1

By [L2], there are $C > 0$ and $x_{0} \geq 1$ such that, for $x \geq x_{0}$ , $∣ A (x) x^{- s - 1} ∣ \leq C x^{θ - ℜ s - 1} .$ The exponent is strictly less than $- 1$ , so the integral over $[x_{0}, \infty)$ converges absolutely. On $[1, x_{0}]$ , the function $A$ is a bounded step function and $x^{- s - 1}$ is continuous, so the integral there also exists. For all sufficiently large $N$ , the boundary term satisfies $∣ A (N) N^{- s} ∣ \leq C N^{θ - ℜ s} ⟶ 0.$ Letting $N \to \infty$ in step 1.1 therefore proves that the Dirichlet-series partial sums converge to the stated improper integral.

L2step 1.1algebra∎

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Sources

Leonard Tomczak, Analytic Number Theory, Theorem 3.5 (standard reference, not scraped)
Kiran S. Kedlaya, Notes on Analytic Number Theory, Chapter 1 (standard reference, not scraped)