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LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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Improper convergence is independent of finite truncations and split points

Statement

Changing a finite lower endpoint of ∫a∞f\int_a^\infty f, a finite upper endpoint of ∫−∞bf\int_{-\infty}^b f, or a compact truncation beside a finite singular endpoint neither creates nor destroys improper convergence. The values change by the corresponding oriented proper integral.

Consequently, convergence and value in the definitions of an interior-singularity integral and a whole-line integral are independent of the chosen finite split point.

Facts & Assumptions

Given: Local Riemann integrability on every compact interval away from the stated singular ends.

[L1]

Proper integrals are additive over adjacent intervals (For a<c<ba<c<b: ff is integrable on [a,b][a,b] if and only if it is integrable on [a,c][a,c] and on [c,b][c,b], and then ∫abf=∫acf+∫cbf\int_a^b f = \int_a^c f + \int_c^b f; with the oriented form for arbitrary a,b,ca,b,c).

[L2]

At an infinite or finite one-sided singular end, adding or subtracting a fixed finite constant preserves convergence and translates the limit by that constant; this follows directly from the same epsilon estimate in the defining limits (Improper integrals over unbounded intervals, Improper integrals at a finite singular endpoint).

[L3]

Improper integrals require separate finite limits at each singular end (Improper integrals over unbounded intervals, Improper integrals at a finite singular endpoint, Improper integrals with several singular ends).

Proof

technique · direct
1.1

If a<a′a<a', then for every R>a′R>a', [L1, L2] ∫aRf=∫aa′f+∫a′Rf.\int_a^R f=\int_a^{a'}f+\int_{a'}^R f. The first term is fixed and finite. Subtracting it changes the absolute error from a proposed translated limit by exactly the same amount, so [L2] shows that either truncation limit exists exactly when the other does and that their values differ by ∫aa′f\int_a^{a'}f. The other three one-ended orientations follow by the same identity with endpoints reversed.

L1L2
2.1

Let s<ts<t be two split points on the whole line. Step 1.1 transfers the finite proper integral ∫stf\int_s^t f from the right tail to the left tail, so the two sums agree. The same calculation around an interior singularity changes only the nonsingular finite portion. Because [L3] continues to require both pieces separately, no cancellation of divergent pieces is introduced.

step 1.1L1L3∎

Depends on

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Sources