Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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Linearity of convergent improper integrals

Statement

If the improper integrals of ff and gg converge over the same one-ended interval and r,sRr,s\in\mathbb R, then (rf+sg)=rf+sg.\int(rf+sg)=r\int f+s\int g. The same formula holds for mixed improper integrals when every singular-end piece of both integrals converges separately.

Facts & Assumptions

Proof

technique · direct
1.1

On every compact truncation, [L1] gives (rf+sg)=rf+sg\int(rf+sg)=r\int f+s\int g. Let the two truncation integrals tend to AA and BB. Given ε>0\varepsilon>0, [L2] makes their respective errors smaller than ε/(2(1+r))\varepsilon/(2(1+|r|)) and ε/(2(1+s))\varepsilon/(2(1+|s|)) sufficiently near the end. The triangle inequality then makes the error of the linear combination from rA+sBrA+sB smaller than ε\varepsilon. This proves the formula on every one-ended interval, including r=0r=0 or s=0s=0.

L1L2
2.1

For a mixed integral, apply step 1.1 to every separately convergent piece and then add the finitely many resulting identities as required by [L3]. No assertion is made when either side would contain an indeterminate difference of divergent quantities.

L3step 1.1

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 65 results over 16 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