CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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Henstock–Kurzweil integration by parts for differentiable factors

Statement

Let $a < b$ and let $F, G$ be differentiable on $[a, b]$ . Then $F^{'} G$ is HK integrable if and only if $F G^{'}$ is HK integrable, and whenever either condition holds,

$\int_{a}^{b} F^{'} G = F (b) G (b) - F (a) G (a) - \int_{a}^{b} F G^{'} .$

Facts & Assumptions

Given: Differentiable functions $F, G$ on $[a, b]$ .

[L1]

Every derivative is Henstock–Kurzweil integrable and its integral is the endpoint increment (Every derivative is Henstock–Kurzweil integrable and satisfies Newton–Leibniz).

[L2]

The product rule gives $(F G)^{'} = F^{'} G + F G^{'}$ (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 $g (c) \neq 0$ ).

[L3]

The Henstock–Kurzweil integral is linear (Linearity of the Henstock–Kurzweil integral).

Proof

technique · direct

1.1givenL1L2

By [L2] and [L1], $F^{'} G + F G^{'}$ is HK integrable and its integral equals $F (b) G (b) - F (a) G (a)$ .

2.1step 1.1L3algebra∎

If either summand is integrable, [L3] applied to its difference from the integrable sum in step 1.1 makes the other integrable; rearranging gives the formula, and the same argument in the other order proves the reverse implication.

Depends on

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Sources

Alessandro Fonda, The Kurzweil-Henstock Integral for Undergraduates, Ch. 1 (standard reference, not scraped)
Andrew Bruckner, Judith Bruckner and Brian Thomson, Real Analysis, Section 1.21 (standard reference, not scraped)