# Iterated Integrals and Riemann-Liouville (Fractional) Derivatives

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...

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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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thanks. However, there is a constant multiplied to these iterated integrals. For $D^{-3}f(x)$ there is a $\frac{1}{2}$, and for $D^{-4}f(x)$ there is a $\frac{1}{2\cdot 3}$. – Mynameis Tiara Oct 11 '12 at 5:15
@MynameisTiara: I derived $f^{(-3)}(x)$, so you can see where the constants came from. – Mhenni Benghorbal Oct 11 '12 at 7:16
Thank you very much. :) – Mynameis Tiara Oct 11 '12 at 9:01
@MynameisTiara: You are welcome. – Mhenni Benghorbal Dec 7 '12 at 6:09
I think there's a typo in your derivation, e.g. $\int_0^t f^{(-3)}(t)$? – DanZimm Aug 31 '14 at 1:41