# Differintegral

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

$\mathbb {D} ^{q}f$ 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 four 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. Here, $n=\lceil q\rceil$ .
{\begin{aligned}{}_{a}^{RL}\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.
{\begin{aligned}{}_{a}^{GL}\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.

In opposite to the Riemann-Liouville differintegral, Caputo derivative of a constant $f(t)$ is equal to zero. Moreover, a form of the Laplace transform allows to simply evaluate the initial conditions by computing finite, integer-order derivatives at point $a$ .
{\begin{aligned}{}_{a}^{C}\mathbb {D} _{t}^{q}f(t)&={\frac {d^{q}f(t)}{d(t-a)^{q}}}\\&={\frac {1}{\Gamma (n-q)}}\int _{a}^{t}{\frac {f^{(n)}(\tau )}{(t-\tau )^{q-n+1}}}d\tau \end{aligned}} ## 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,

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

$\mathbb {D} ^{q}f(t)={\mathcal {F}}^{-1}\left\{(i\omega )^{q}{\mathcal {F}}[f(t)]\right\}.$ Under the bilateral Laplace transform, here denoted by ${\mathcal {L}}$ and defined as ${\mathcal {L}}[f(t)]=\int _{-\infty }^{\infty }e^{-st}f(t)\,dt$ , 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

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

$\mathbb {D} _{t}^{q}(fg)=\sum _{j=0}^{\infty }{q \choose j}\mathbb {D} _{t}^{j}(f)\mathbb {D} _{t}^{q-j}(g)$ In general, composition (or semigroup) rule is not satisfied:

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