Alphabeta Math
ExampleConstruction: AI-generatedVerification: 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 tauberian prime number theorem

Example

For f(t)=etψ(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 q1. Put Ψχ(x)=nxχ(n)Λ(n) and δχ=1 for the principal character, zero otherwise. The bounded, locally integrable function fχ(t)=etΨχ(et)δχ has Laplace transform gχ(s)=L(s+1,χ)(s+1)L(s+1,χ)δχs(Res>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)=0f(t)eztdt, initially defined for Rez>0, extends holomorphically to an open set containing {Rez0}, then limT0Tf(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 a0. If 1(A(x)ax)x2dx converges, then A(x)/xa.

Verification

1.1

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.

F1
2.1

Newman gives convergence of 0f(t)dt=1(ψ(x)x)x2dx. Since psi is nonnegative, nondecreasing and O(x), desmoothing with a=1 yields ψ(x)/x1. This is the qualitative assertion; no estimate of a uniform continuation width was needed.

F2F3step 1.1

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