Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
    
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
add comment

1 Answer

up vote 3 down vote accepted
+50

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.

share|improve this answer
    
good answer, +1! –  user01123581321345589144... 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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.