Fractional integration

Define the integration operator I by

(I,f)(x) = int_a^x f(t), dt

so I f is an antiderivative of f.

Define the second antiderivative I2 by applying I to f twice:

(I_2, f)(x) = int_a^x int_a^t f(t_1), dt_1, dt

It turns out that

(I_2, f)(x) = int_a^x (x - t) , f(t), dt

To see this, notice that both expressions for I2 are equal when x = a, and they have the same derivative, so they must be equal everywhere.

Now define I3 by applying I to I2 and so forth defining In for all positive integers n. Then one can show that the nth antiderivative is given by

(I_n, f)(x) = frac{1}{Gamma(n)} int_a^x (x - t)^{n-1} , f(t), dt

The right side of this equation makes sense for non-integer values of n, and so we can take it as a definition of fractional integrals when n is not an integer.

Related post:

Fractional derivatives

4 thoughts on “Fractional integration

  1. In the Fourier domain, integration of f means divide the transform F by the frequency w. Thus one cam define the t-fractional integral as dividing by w to the power t.
    Are both definitions equivalent?

  2. It’s interesting to me that, although details are wrong, the
    expression is *kind of* right for non-positive integers n as well.. think Cauchy integral formula. Of course Gamma(n) needs to be replaced by frac{2pi i}{Gamma(1-n)}, having 2 endpoints is a little weird, and there is some contour selection issues… but the overall gist is right. That suggests to me that the expression could be tweaked a bit, I’m just not sure how at the moment.

  3. There are several approaches to fractional derivative. In my opinion that I’ve been stating in several papers, the most natural way of defining fractional derivative is through the limit of the fractional incremental ratio. Normally it is called Grunwald-Letnikov derivative, although it was also proposed (before) by Liouville. I made several generalizations. With it we can easily generalize the well-known properties of the transforms of the derivative, both for Laplace and Fourier transforms: multiplication by s^a or (jw)^a. The current Riemann-Liouville and Caputo derivatives can be deduced from the incremental ratio derivative. If anyone is interested, I can send copies of my papers.

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