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I'm having difficulty justifying the change of limits in the derivation of the Riemann-Liouville derivative at xuru.org. What I don't undestand is how $\int_0^{t_2}$ becomes $\int_{t_1}^x$ in the following statement,

$\int_0^x \int_0^{t_2} f(t_1) dt_1 dt_2 = \int_0^x \int_{t_1}^x f(t_1) dt_2 dt_1$

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There is a typo in your second integral, where one of the lower limits should be $t_1$ instead of $t_2$. Actually, I now speculate that that misreading may be the entire source of your question, so I am deleting my answer. Please clarify if there is something else. –  Jonas Meyer Oct 16 '10 at 1:43
    
I've fixed the typo, but I still don't understand the reason for the change in the integral. Would you kindly consider reposting your answer. Thanks. –  Olumide Oct 16 '10 at 2:21
    
Thanks. I just undeleted my answer. –  Jonas Meyer Oct 16 '10 at 2:26
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2 Answers

Notice that $0\leq t_2\leq x$ and $0\leq t_1\leq t_2$ is equivalent to $0\leq t_1\leq x$ and $t_1\leq t_2\leq x$. You could shorten the description to $0\leq t_1\leq t_2\leq x$. Whether you believe me or not, I recommend that you draw a picture.

The point is that both integrals are over the same triangular region.

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Let $A$ be the first set of integration and let $B$ be the second set of integration, and then you have to prove $A=B$. It is just a rephrase of Jonas answer. –  AD. Oct 16 '10 at 4:59
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I liked the geometriYou can use integration by parts, following the well known formula:

\begin{align*} \int_a^b f(x) \frac{dg(x)}{dx} dx = [f(x)g(x)]_a^b - \int_a^b \frac{df(x)}{dx} g(x) dx \end{align*}

setting $g(x)=x$ and $f(x)=\int_a^x f(\xi) d\xi$ you have your result :)

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