Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11
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.

A positive continuous integrand can have finite integral while unbounded on every tail

Example

There is a positive continuous f:[0,∞)→R for which ∫0∞f converges although f is unbounded on every tail.

Facts & Assumptions

Given: For each positive integer k, put hk=2−k−2/k and let sk be the symmetric triangular function supported on [k−hk,k+hk], zero at the endpoints, and of height k at its center. Define f(x)=1(1+x)2+∑k=1∞sk(x).

[L1]

The supports of the sk are pairwise disjoint, and every compact interval meets only finitely many of them.

[L2]

A triangle of height k and half-width hk has integral khk=2−k−2.

Verification

technique · construction
1.1

By [L1], local finiteness and matching zero endpoint values make the spike sum continuous; adding the positive continuous baseline preserves positivity and continuity. Direct differentiation gives primitive 1−1/(1+x) for the baseline, whose improper integral is one.

L1
2.1

By [L2], additivity, and [L3], the total integral of all spikes is ∑k=1∞2−k−2<∞. Given ε>0, choose K so that both the geometric spike tail from K and the baseline tail 1/(1+K) are below ε/2. For K≤u<v, nonnegativity bounds the integral over every partial spike by the full spike area, so ∫uvf<ε. The Cauchy criterion therefore gives convergence of ∫0∞f.

step 1.1L2L3
3.1

At every positive integer k, f(k)≥sk(k)=k. Integers occur arbitrarily far out, so f is unbounded on every tail.

given∎

Depends on

Used by

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Dependency tree · two levels

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Sources