# Differintegral

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

${\displaystyle \mathbb {D} ^{q}f}$

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.
{\displaystyle {\begin{aligned}{}_{a}\mathbb {D} _{t}^{q}f(t)&={\frac {d^{q}f(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{aligned}}}
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.
{\displaystyle {\begin{aligned}{}_{a}\mathbb {D} _{t}^{q}f(t)&={\frac {d^{q}f(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{aligned}}}
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 ${\displaystyle {\mathcal {F}}}$ :

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

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

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

So,

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

which generalizes to

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

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

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

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

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

## Basic formal properties

Linearity rules

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

Zero rule

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

Product rule

${\displaystyle \mathbb {D} _{t}^{q}(fg)=\sum _{j=0}^{\infty }{q \choose j}\mathbb {D} _{t}^{j}(f)\mathbb {D} _{t}^{q-j}(g)}$

In general, composition (or semigroup) rule is not satisfied[1]:

${\displaystyle \mathbb {D} ^{a}\mathbb {D} ^{b}f\neq \mathbb {D} ^{a+b}f}$

## A selection of basic formulæ

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