In the derivation of Riemann-Liouville derivatives, i got lost on the part when the pattern led to $D^{-2}f(x)=\int_0^xf(t)(x-t)dt$ $D^{-3}f(x)=\frac{1}{2}\int_0^xf(t)(x-t)^2dt$ $D^{-4}f(x)=\frac{1}{2\cdot 3}\int_0^xf(t)(x-t)^3dt$ $\vdots$ I was able to figure out everything except for the constants $\frac{1}{2}$, $\frac{1}{2\cdot 3}$. Where did they come from? Please please help me...
Iterated Integrals and Riemann-Liouville (Fractional) Derivatives
1 Answers
Related techniques: (I). Here is how you proceed,
$ f^{(-1)}(x) = \int_{0}^{x} f(t) dt$
$\implies f^{(-2)}(x) = \int_{0}^{x} \int_{0}^{t} f(\tau) d\tau dt = \int_{0}^{x} \int_{\tau}^{x} f(\tau) dt d\tau $
$ = \int_{0}^{x}f(\tau) \left( \int_{\tau}^{x} dt \right) d\tau = \int_{0}^{x}f(\tau) (x-\tau) d\tau = \int_{0}^{x}f(t) (x-t) dt \,.$
The whole idea is to change the order of integration. Let's derive $f^{(-3)}(x) $
$ f^{(-3)}(x) = \int_{0}^{x}f^{(-2)}(t)dt = \int_{0}^{x} \int_{0}^{t} (t-\tau)f(\tau) d\tau dt = \int_{0}^{x} f(\tau) \left(\int_{\tau}^{x}(t-\tau) dt\right) d\tau $
$ =\int_{0}^{x} f(\tau) \left[\frac{(t-\tau)^2}{2}\right]_{t=\tau}^{t=x} d\tau = \frac{1}{2}\int_{0}^{x}(x-\tau)^2f(\tau) d\tau = \frac{1}{2}\int_{0}^{x}(x-t)^2f(t) dt $
Now, you can see where the constants came from.
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0@$M$he$n$niBenghorbal right, that's what I meant :P at first it looked like you fixed only one of the typos, that's why I made the second comment. Apologies for any frustration – 2014-08-31