TheoremStatement: Literature-sourcedProof: 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 Dirichlet-type transfer criterion for divergence

Statement

Let $u$ be continuous on $[a, \infty)$ and suppose $\int_{a}^{\infty} u$ diverges. Let $v > 0$ be differentiable, satisfy $v (x) \to \infty$ , and suppose $\int_{a}^{\infty} \frac{∣ v^{'} (x) ∣}{v (x)^{2}} d x$ converges. Then $\int_{a}^{\infty} u (x) v (x) d x$ diverges.

Facts & Assumptions

Given: Functions $u, v$ satisfying the statement.

[L1]

The reciprocal rule gives $(1 / v)^{'} = - v^{'} / v^{2}$ (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L2]

The differentiable-multiplier clause of Dirichlet's test applies to a continuous function with bounded truncation primitive (Dirichlet's test for improper integrals).

[L3]

Convergence of an improper integral bounds its truncation primitive near infinity, while on the remaining compact interval the integral function is Lipschitz and hence bounded (Improper integrals over unbounded intervals, The integral function of a bounded integrable $f$ is Lipschitz, hence uniformly continuous).

Proof

technique · contradiction

1.1

Suppose for contradiction that $\int_{a}^{\infty} u v$ converges. Then its truncation primitive is bounded by [L3], and $u v$ is continuous. Put $g = 1 / v$ . Positivity and $v \to \infty$ give $g \to 0$ : for $ε > 0$ , eventually $v > 1 / ε$ , hence $0 < g < ε$ . Also [L1] and the hypothesis give absolute convergence of $\int g^{'}$ .

L3L1assume-contra

2.1

Apply [L2] with the continuous function $u v$ and multiplier $g$ . It yields convergence of $\int (u v) g = \int u$ , contradicting the hypothesis. Hence $\int u v$ diverges.

step 1.1L2discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

55 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

William F. Trench, Introduction to Real Analysis, Theorem 3.4.11 (standard reference, not scraped)