Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11
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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.

Comparison tests for improper integrals

Statement

Suppose 0≤f≤g eventually toward a singular end. If the improper integral of g converges there, then the integral of f converges. If instead ∣f∣≤g eventually and ∫g converges, then ∫f converges absolutely and hence converges.

The same assertions hold separately at +∞, at −∞, and at either finite singular endpoint.

Facts & Assumptions

Given: The stated eventual pointwise bounds and local Riemann integrability.

[L1]

Proper integration preserves pointwise order (If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a)).

[L2]

A nonnegative improper integral converges exactly when its truncation integrals are bounded (A nonnegative improper integral converges iff its truncated integrals are bounded).

[L3]

Absolute convergence implies convergence (Absolute convergence implies improper convergence).

[L4]

Finite initial pieces do not affect convergence (Improper convergence is independent of finite truncations and split points).

Proof

technique · direct
1.1L4L1L2

Discard the finite portion before the eventual inequality using [L4]. On every remaining compact truncation, [L1] gives 0≤∫f≤∫g. Convergence of ∫g bounds the latter truncations, so [L2] gives convergence of ∫f.

2.1L3step 1.1∎

If ∣f∣≤g, step 1.1 applied to ∣f∣ proves absolute convergence; [L3] then proves convergence of f. The argument depends only on the direction of truncation and therefore proves every endpoint form.

Depends on

Dependency tree · 2 levels

Sources