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The exercise state as follows: Let $f:R \rightarrow R$ be an absolutely continuous function over every compact in $R$. I'm asked to prove that: $\displaystyle \frac{d}{dy}\int_a^b f(x+y)dx=\int_a^b f'(x+y)dx $
for every $y \in R$.

I thought that if $f$ is absolutely continuous over the compacts subset of $R$ so is also the translation $f(x+y)$. I don't think I need to prove this. Do you think I should?

I tried to apply the fundamental theorem but I didn't get anything useful: $ f(b)=f(a)+\int_a^b f'(x+y)dx$
I also considered the derivative respect $x$ or $y$ the function $f'(x+y)$ is the same and it's equal to the derivative $f'(x)$. Is it right?
But any of these idea seems to be helpful, I've also tried to apply the Lagrange theorem to the derivative and the function over the set $f(b)-f(a)$ but still no results.
I hadn't tried with results on derivating under integral because the information on f and on its derivative seems definitely too weak in my opinion.So I don't know what to do now. Any idea?

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2 Answers 2

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Hint: Given any integrable function $g$, $$\int_a^bg(x+y)dx=\int_{a+y}^{b+y}g(x)dx.$$

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I tried to develop your hint in the answer above. –  Laura Dec 4 '12 at 19:41
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$$\int_a^bg(x+y)dx=\int_{a+y}^{b+y}g(x)dx=G(b+y)-G(a+y).$$
Where $$G(x)=\int_a^x g(t)dt$$ Since g is absolutely continuous we got the last part.
Now if we calculate the derivative: $$\frac{d}{dy}\int_a^bg(x+y)dx=\frac{d}{dy}(G(b+y)-G(a+y))=g(b+y)-g(a+y)=\int_{a+y}^{b+y}g'(x)dx=\int_a^b g'(x+y)dx$$
Is it right? I think that everypart is correct since the function is absolutely continuous and we can use the fundamental theorem and the translation don't change the derivative.

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It is correct except for some typo: in the last two integrals, $g$ should be replaced by $g'$. –  23rd Dec 4 '12 at 19:45
    
Fixed, thanks for the hint rewrite all my calculus quitting the $x+y$ let me work better.So simple and so useful! –  Laura Dec 4 '12 at 19:56
    
You are welcome. –  23rd Dec 4 '12 at 20:30
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