Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedverified 2026-09-24 (gpt-6-sol)
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 tauberian prime number theorem

Example

For f(t)=e−tψ(et)−1, its transform is −ζ′(s+1)(s+1)ζ(s+1)−1s. Newman's theorem and monotone desmoothing recover ψ(x)∼x without a quantitative zero-free region.

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Dirichlet character chebyshev laplace transform: Fix a Dirichlet character χ modulo q≥1. Put Ψχ(x)=∑n≤xχ(n)Λ(n) and δχ=1 for the principal character, zero otherwise. The bounded, locally integrable function fχ(t)=e−tΨχ(et)−δχ has Laplace transform gχ(s)=−L′(s+1,χ)(s+1)L(s+1,χ)−δχs(Re⁡s>0). After the removable value at zero is filled in, this extends holomorphically to an open neighborhood of the closed right half-plane.

[F2]

Newman zagier tauberian theorem: Let f:[0,∞)→C be bounded and locally Lebesgue integrable. If g(z)=∫0∞f(t)e−zt dt, initially defined for Re⁡z>0, extends holomorphically to an open set containing {Re⁡z≥0}, then lim⁡T→∞∫0Tf(t) dt=g(0).

[F3]

Monotone chebyshev tauberian desmoothing: Let A:[1,∞)→[0,∞) be nondecreasing and locally integrable, with A(x)=O(x), and let a≥0. If ∫1∞(A(x)−ax)x−2 dx converges, then A(x)/x→a.

Verification

1.1F1

The character-transform lemma at q=1 proves boundedness, local integrability, the displayed transform and its continuation through the closed boundary, including the cancellation at s=0.

2.1F2F3step 1.1∎

Newman gives convergence of ∫0∞f(t) dt=∫1∞(ψ(x)−x)x−2 dx. Since psi is nonnegative, nondecreasing and O(x), desmoothing with a=1 yields ψ(x)/x→1. This is the qualitative assertion; no estimate of a uniform continuation width was needed.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

20 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