TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Newman zagier tauberian theorem

Statement

Let $f : [0, \infty) \to C$ be bounded and locally Lebesgue integrable. If $g (z) = \int_{0}^{\infty} f (t) e^{- z t} d t$ , initially defined for $Re z > 0$ , extends holomorphically to an open set containing ${Re z \geq 0}$ , then $\lim_{T \to \infty} \int_{0}^{T} f (t) d t = g (0) .$

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Newman damped contour estimates: Let $f : [0, \infty) \to C$ be locally integrable with $∣ f ∣ \leq B$ , let $g (z) = \int_{0}^{\infty} f (t) e^{- z t} d t$ for $Re z > 0$ , and $g_{T} (z) = \int_{0}^{T} f (t) e^{- z t} d t$ . For $R > 0, T \geq 0$ , set $K_{R} (z) = (1 + z^{2} / R^{2}) / z$ . On the right and left semicircles $C_{+}, C_{-}$ of radius R, $∣ \int_{C_{+}} (g - g_{T}) e^{T z} K_{R} (z) d z ∣ \leq \frac{2 π B}{R}, ∣ \int_{C_{-}} g_{T} e^{T z} K_{R} (z) d z ∣ \leq \frac{2 π B}{R} .$ Integrals at the imaginary endpoints are interpreted as improper limits when needed.

[F2]

The residue theorem for a null-homologous cycle: Let $Ω \subseteq C$ be open, let $f$ be meromorphic on $Ω$ with pole set $S$ , and let $Γ$ be admissible for the residue theorem in $Ω$ . Then $\int_{Γ} f (z) d z = 2 π i \sum_{a \in S} n (Γ, a) Res (f, a),$ where only finitely many terms are nonzero.

[F3]

Dominated convergence: Let $f$ and $(f_{n})$ be measurable complex-valued functions such that $f_{n} \to f$ almost everywhere and $∣ f_{n} ∣ \leq g$ almost everywhere for a single nonnegative measurable function $g$ with $\int g d μ < + \infty$ . Then $f \in L^{1} (μ)$ , $\int ∣ f_{n} - f ∣ d μ ⟶ 0,$ and hence $\int f_{n} d μ ⟶ \int f d μ .$

Proof

1.1

Choose a bound $B \geq 0$ for $∣ f ∣$ and fix $R > 0$ . The finite transform $g_{T}$ is entire: on compact z-sets its difference quotients and derivatives are dominated by integrable constants times $∣ f (t) ∣$ on [0,T]. Compactness of the imaginary segment permits $0 < δ < R$ such that the closed region ${∣ z ∣ \leq R, Re z \geq - δ}$ and a neighborhood are in the continuation domain. Its positively oriented boundary C has a right semicircle and a left path staying strictly left except at its two endpoints.

F3given

2.1

Apply the residue theorem to $(g - g_{T}) e^{T z} K_{R} (z)$ on C. Its sole possible pole is zero, with residue $g (0) - g_{T} (0)$ . Split the contour into the right arc, the g left-path integral, and minus the $g_{T}$ left-path integral. Deform the last integral to the left semicircle: $g_{T} e^{T z} K_{R} (z)$ is holomorphic in the region between these two left paths, which does not contain zero.

F2step 1.1

3.1

After division by $2 π$ , the right-arc and left-semicircle absolute contributions are each at most $B / R$ . On the fixed left path the g integrand is bounded independently of T, since the path misses zero, and tends to zero except at the endpoints. Dominated convergence makes that integral tend to zero. This argument applies to every sequence of real T tending to infinity, hence to the full limit. Thus ${lim sup}_{T \to \infty} ∣ g (0) - g_{T} (0) ∣ \leq 2 B / R$ .

F1F3step 2.1

4.1

The radius R can be arbitrarily large; for each radius only its own positive strip width is needed. Letting R tend to infinity gives $g_{T} (0) \to g (0)$ , which is precisely convergence of the asserted improper integral. If B=0 the assertion is immediate from the same estimates.

step 3.1algebra∎

Depends on

Used by

Newman tauberian prime number theorem Example
Prime number theorem arithmetic progressions Theorem

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

§1.4, Theorem 1.8 and its complete proof (standard reference, not scraped)