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Absolute convergence implies improper convergence
Statement
Every absolutely convergent improper integral converges. Moreover, on a one-ended interval, For a mixed interval the same conclusion applies separately to each singular-end piece.
Facts & Assumptions
Given: Convergence of the improper integral of .
On every compact interval, (If are integrable on then so are , , , and , and ).
The improper Cauchy criterion characterizes convergence (Cauchy criterion for improper integrals).
Inequalities persist under limits of real sequences; integer and reciprocal truncation sequences approach the infinite and finite singular ends (Limits preserve non-strict inequalities, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Every complete ordered field is Archimedean, For every in a complete ordered field there is a natural with ).
Proof
By the Cauchy criterion [L2], remote tail integrals of are arbitrarily small. The proper inequality [L1] makes the corresponding tail integrals of no larger in absolute value. A second application of [L2] proves convergence of .
Apply [L1] on compact truncations. Along integer truncations at infinity, or reciprocal truncations at a finite endpoint, both sides converge to the corresponding improper values; [L3] passes the inequality to those sequence limits and gives the displayed bound.
For a mixed integral, absolute convergence is required on every piece. Steps 1.1–1.2 apply piecewise, and finite addition completes the claim.
Depends on
- Absolute and conditional convergence of improper integrals
- Cauchy criterion for improper integrals
- If $f,g$ are integrable on $[a,b]$ then so are $\lvert f\rvert$, $f^{2}$, $fg$, $\max(f,g)$ and $\min(f,g)$, and $\bigl\lvert\int_a^b f\bigr\rvert \le \int_a^b\lvert f\rvert$
- Limits preserve non-strict inequalities
- Sequences of reals: bounded, eventually, frequently, tails, subsequences
- Every complete ordered field is Archimedean
- For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$
- Improper convergence is independent of finite truncations and split points
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 115 results over 27 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- William F. Trench, Introduction to Real Analysis, Theorem 3.4.9 (standard reference, not scraped)