Fractional Calculus - Heuristics

Heuristics

A fairly natural question to ask is whether there exists an operator, or half-derivative, such that

It turns out that there is such an operator, and indeed for any, there exists an operator such that

or to put it another way, the definition of can be extended to all real values of n.

Assuming a function that is defined where, form the definite integral from 0 to x. Call this

Repeating this process gives

and this can be extended arbitrarily.

The Cauchy formula for repeated integration, namely

leads in a straightforward way to a generalization for real n.

Using the Gamma function to remove the discrete nature of the factorial function gives us a natural candidate for fractional applications of the integral operator.

This is in fact a well-defined operator.

It is straightforward to show that the J operator satisfies

Proof
$\begin{align} (J^\alpha) (J^\beta f)(x) & = \frac{1}{\Gamma(\alpha)} \int_0^x (x-t)^{\alpha-1} (J^\beta f)(t) \; dt \\ & = \frac{1}{\Gamma(\alpha) \Gamma(\beta)} \int_0^x \int_0^t (x-t)^{\alpha-1} (t-s)^{\beta-1} f(s) \; ds \; dt \\ & = \frac{1}{\Gamma(\alpha) \Gamma(\beta)} \int_0^x f(s) \left( \int_s^x (x-t)^{\alpha-1} (t-s)^{\beta-1} \; dt \right) ds \end{align}$ where in the last step we exchanged the order of integration and pulled out the f(s) factor from the t integration. Changing variables to r defined by t=s+(x-s)r, $(J^\alpha) (J^\beta f)(x) = \frac{1}{\Gamma(\alpha) \Gamma(\beta)} \int_0^x (x-s)^{\alpha + \beta - 1} f(s) \left( \int_0^1 (1-r)^{\alpha-1} r^{\beta-1} \; dr \right) ds$ The inner integral is the beta function which satisfies the following property $\int_0^1 (1-r)^{\alpha-1} r^{\beta-1} \; dr = B(\alpha, \beta) = \dfrac{\Gamma(\alpha)\,\Gamma(\beta)}{\Gamma(\alpha+\beta)}$ Substituting back into our equation $(J^\alpha) (J^\beta f)(x) = \frac{1}{\Gamma(\alpha + \beta)} \int_0^x (x-s)^{\alpha + \beta - 1} f(s) \; ds = (J^{\alpha + \beta} f)(x)$ Interchanging α and β shows that the order in which the J operator is applied is irrelevant and completes the proof.

Proof

$\begin{align} (J^\alpha) (J^\beta f)(x) & = \frac{1}{\Gamma(\alpha)} \int_0^x (x-t)^{\alpha-1} (J^\beta f)(t) \; dt \\ & = \frac{1}{\Gamma(\alpha) \Gamma(\beta)} \int_0^x \int_0^t (x-t)^{\alpha-1} (t-s)^{\beta-1} f(s) \; ds \; dt \\ & = \frac{1}{\Gamma(\alpha) \Gamma(\beta)} \int_0^x f(s) \left( \int_s^x (x-t)^{\alpha-1} (t-s)^{\beta-1} \; dt \right) ds \end{align}$

where in the last step we exchanged the order of integration and pulled out the f(s) factor from the t integration. Changing variables to r defined by t=s+(x-s)r,

$(J^\alpha) (J^\beta f)(x) = \frac{1}{\Gamma(\alpha) \Gamma(\beta)} \int_0^x (x-s)^{\alpha + \beta - 1} f(s) \left( \int_0^1 (1-r)^{\alpha-1} r^{\beta-1} \; dr \right) ds$

The inner integral is the beta function which satisfies the following property

$\int_0^1 (1-r)^{\alpha-1} r^{\beta-1} \; dr = B(\alpha, \beta) = \dfrac{\Gamma(\alpha)\,\Gamma(\beta)}{\Gamma(\alpha+\beta)}$

Substituting back into our equation

$(J^\alpha) (J^\beta f)(x) = \frac{1}{\Gamma(\alpha + \beta)} \int_0^x (x-s)^{\alpha + \beta - 1} f(s) \; ds = (J^{\alpha + \beta} f)(x)$

Interchanging α and β shows that the order in which the J operator is applied is irrelevant and completes the proof.

This relationship is called the semigroup property of fractional differintegral operators. Unfortunately the comparable process for the derivative operator D is significantly more complex, but it can be shown that D is neither commutative nor additive in general.

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