TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11

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Linearity of convergent improper integrals

Statement

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

Facts & Assumptions

Given: Convergent improper integrals of $f$ and $g$ on the indicated domain, and scalars $r, s$ .

[L1]

Proper Riemann integration is linear (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ).

[L2]

Infinite and one-sided endpoint limits use the usual epsilon definitions (Limits at $+ \infty$ and $- \infty$ , and infinite limits at a point, The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (- \infty, c)$ and $A \cap (c, \infty)$ ).

[L3]

Mixed convergence is defined separately on every singular piece (Improper integrals with several singular ends).

Proof

technique · direct

1.1L1L2

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

2.1L3step 1.1∎

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.

Depends on

Dependency tree · 2 levels

Sources

William F. Trench, Introduction to Real Analysis, Section 3.4