Iterated Integrals and Riemann-Liouville (Fractional) Derivatives

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.