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I was wondering what are the weakest hypothesis for applying integration by parts to calculate $$\int_a^b fg dm,$$ where $m$ denotes the Lebegue measure on $\mathbb R$.

Is it enough that $f$ be differentiable on $(a,b)$, $f' \in L^1(a,b)$ and $g \in L^1(a,b)$ to write

$$ \int_a^b fg dm = \left[f(x)\int_a^x g(t) dm(t)\right]_a^b - \int_a^b f'(x) \left(\int_a^x g(t) dm(t) \right) dm(x)? $$

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the primitive of $g$ must also be integrable, no? – suissidle Mar 28 '13 at 23:54
What is the primitive of g? – Cantor Mar 29 '13 at 11:17
$\int_a^x g(t) dm(t)$ – suissidle Mar 29 '13 at 15:42
up vote 3 down vote accepted

Proposition: Assume that $f$ is absolutely continuous on $[a,b]$ and $g\in L^1([a,b])$. For $x\in[a,b]$, denote $G(x)=\int_a^xg~dm$. Then $$\int_a^b fg~dm=f(b)G(b)-f(a)G(a)-\int_a^bf'G~dm.\tag{1}$$

Proof: Since both $f$ and $G$ are absolutely continuous on $[a,b]$, $fG$ is also absolutely continuous on $[a,b]$, and
$$(fG)'(x)=f'(x)G(x)+f(x)g(x)\quad a.e. x\in[a,b].\tag{2}$$ Integrating $(2)$ over $[a,b]$, $(1)$ follows. $\quad\square$

Remark: If $f$ is differentiable on $[a,b]$ and $f'\in L^1([a,b])$, then $f$ is absolutely continuous on $[a,b]$. See, for example, Theorem 7.21 in Real and Complex Analysis(Third Edition) by Walter Rudin.

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good answer, +1! – user67133 Apr 12 '13 at 15:23
@user67133: Thank you! – 23rd Apr 12 '13 at 15:25
Can you provide the proof of $G$ being absolutely continuous? Also what assumptions do you need to integrate the right side of equation (2)? – Cantor Apr 13 '13 at 8:29
@Cantor: See the same wikipedia page as given in my answer. There is a list of equivalent conditions of absolute continuity. In particular, (3) implies that $G$ is absolute continuous. Moreover, $G'=g$ almost everywhere. Since $f'$(resp. $g$)$\in L^1$ and $G$(resp. $f$) is bounded, $f'G\in L^1$(resp. $fg\in L^1$), so RHS in $(2)$ is integrable. – 23rd Apr 13 '13 at 9:18

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