# Differintegral

(Redirected from Fractional integration)
Fractional integration redirects here. Not to be confused with Autoregressive fractionally integrated moving average

In fractional calculus, an area of applied mathematics, the differintegral is a combined differentiation/integration operator. Applied to a function ƒ, the q-differintegral of f, here denoted by

$\mathbb{D}^qf$

is the fractional derivative (if q > 0) or fractional integral (if q < 0). If q = 0, then the q-th differintegral of a function is the function itself. In the context of fractional integration and differentiation, there are several legitimate definitions of the differintegral.

## Standard definitions

The three most common forms are:

This is the simplest and easiest to use, and consequently it is the most often used. It is a generalization of the Cauchy formula for repeated integration to arbitrary order.
\begin{align} {}_a\mathbb{D}^q_tf(t) & = \frac{d^qf(t)}{d(t-a)^q} \\ & =\frac{1}{\Gamma(n-q)} \frac{d^n}{dt^n} \int_{a}^t (t-\tau)^{n-q-1}f(\tau)d\tau \end{align}
The Grunwald–Letnikov differintegral is a direct generalization of the definition of a derivative. It is more difficult to use than the Riemann–Liouville differintegral, but can sometimes be used to solve problems that the Riemann–Liouville cannot.
\begin{align} {}_a\mathbb{D}^q_tf(t) & = \frac{d^qf(t)}{d(t-a)^q} \\ & =\lim_{N \to \infty}\left[\frac{t-a}{N}\right]^{-q}\sum_{j=0}^{N-1}(-1)^j{q \choose j}f\left(t-j\left[\frac{t-a}{N}\right]\right) \end{align}
This is formally similar to the Riemann–Liouville differintegral, but applies to periodic functions, with integral zero over a period.

## Definitions via transforms

Recall the continuous Fourier transform, here denoted $\mathcal{F}$ :

$F(\omega) = \mathcal{F}\{f(t)\} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(t) e^{- i\omega t}\,dt$

Using the continuous Fourier transform, in Fourier space, differentiation transforms into a multiplication:

$\mathcal{F}\left[\frac{df(t)}{dt}\right] = i \omega \mathcal{F}[f(t)]$

So,

$\frac{d^nf(t)}{dt^n} = \mathcal{F}^{-1}\left\{(i \omega)^n\mathcal{F}[f(t)]\right\}$

which generalizes to

$\mathbb{D}^qf(t)=\mathcal{F}^{-1}\left\{(i \omega)^q\mathcal{F}[f(t)]\right\}.$

Under the Laplace transform, here denoted by $\mathcal{L}$, differentiation transforms into a multiplication

$\mathcal{L}\left[\frac{df(t)}{dt}\right] = s\mathcal{L}[f(t)].$

Generalizing to arbitrary order and solving for Dqf(t), one obtains

$\mathbb{D}^qf(t)=\mathcal{L}^{-1}\left\{s^q\mathcal{L}[f(t)]\right\}.$

## Basic formal properties

Linearity rules

$\mathbb{D}^q(f+g)=\mathbb{D}^q(f)+\mathbb{D}^q(g)$
$\mathbb{D}^q(af)=a\mathbb{D}^q(f)$

Zero rule

$\mathbb{D}^0 f=f \,$

Product rule

$\mathbb{D}^q_t(fg)=\sum_{j=0}^{\infty} {q \choose j}\mathbb{D}^j_t(f)\mathbb{D}^{q-j}_t(g)$

In general, composition (or semigroup) rule

$\mathbb{D}^a\mathbb{D}^{b}f = \mathbb{D}^{a+b}f$

is not satisfied.[1]

## Some basic formulae

$\mathbb{D}^q(t^n)=\frac{\Gamma(n+1)}{\Gamma(n+1-q)}t^{n-q}$
$\mathbb{D}^q(\sin(t))=\sin \left( t+\frac{q\pi}{2} \right)$
$\mathbb{D}^q(e^{at})=a^q e^{at}$