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Let $h $ be Riemann integrable on $[c,d]$.Let $a\in [c,d]$ and $g(x)=\int_a^x h(t)dt $. Suppose $f $ is Riemann integrable on $g([c,d])$ How do you show the integral $\int_c^df(g(x))h(x)dx$ exists and is equal to $\int_{g(c)}^{g(d)}f(x)dx $ ?

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Can you post what you have tried for this? This type of problem is reasonably standard and, as far as I can tell, should be provable directly from the definition of a Riemann-Stieltjes integral (the former quantity being $\int_c^d f(g(x))dg(x)$). As a hint, the fact that $h(t)$ is properly Riemann integrable on $[c,d]$ shows you that $g(x)$ is a function of bounded variation. –  Chris Janjigian Feb 22 '12 at 2:17
    
Simply prove that $g'(x) = h(x)$, which has probably been proved in your book or class, and then this is the change of variables theorem. To prove the change of variables theorem, you could for instance use the chain rule to differentiate $F(g(x))$, where $F$ is a primitive of $f$. –  Pedro Feb 22 '12 at 3:37
    
@Pedro $h(x)$ is not given to be continuous, so that is only true $a.e.$. –  Chris Janjigian Feb 22 '12 at 4:30
    
thanks @Chris , I've forgoten something about iemann-Stieltjes integral. –  Leitingok Feb 22 '12 at 10:04
    
@Chris, this problem is not as easy as your hint. I find a paper . jstor.org/pss/3614764 –  Leitingok Feb 23 '12 at 3:12

1 Answer 1

With a substitution. Let $y=g(x)$. Then $dy=g'(x) dx= h(x) dx$ where we have used the fundamental theorem of calculus. Thus

$\int_c^df(g(x))h(x)dx =\int_{g(c)}^{g(d)}f(y)dy$

and you are done.

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sorry,$g$ need not be differentiable. –  Leitingok Feb 20 '12 at 7:09
    
Please read Rudin's Real & Complex Chapter 8. g will have a derivative a.e. in the classical sense. This is rather tricky to prove. –  chango Feb 20 '12 at 8:06
    
But I need a elementary proof. And the theorem you give does not seems slove my problem. –  Leitingok Feb 20 '12 at 8:44
    
It clearly DOES solve your problem! Look at the proof first and then decide. There might not be a more elementary proof...what leads you to think there is one? –  chango Feb 20 '12 at 11:42
    
Nik, you mean if y=g'(x) a.e, then $\int_{c}^{d}f(g(x))g'(x)dx = \int_{g(c)}^{g(d)} f(x)dx $ ? Generally, that is incorect. –  Leitingok Feb 21 '12 at 1:33

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