Differintegral

Definitions via transforms

The definitions of fractional derivatives given by Liouville, Fourier, and Grunwald and Letnikov coincide. They can be represented via Laplace, Fourier transforms or via Newton series expansion.

Recall the continuous Fourier transform, here denoted ${\displaystyle \mathcal{F}}$: $${\displaystyle F(\omega )=\mathcal{F}\{f(t)\}=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(t)e^{-i\omega t}dt}$$

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, $${\displaystyle \frac{d^{n}f(t)}{d{t}^n}={\mathcal{F}}^{-1}\left\{(i\omega {)}^n\mathcal{F}[f(t)]\right\}}$$ which generalizes to $${\displaystyle {\mathbb{D}}^{q}f(t)={\mathcal{F}}^{-1}\left\{(i\omega {)}^q\mathcal{F}[f(t)]\right\}.}$$

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

Representation via Newton series is the Newton interpolation over consecutive integer orders:

$${\displaystyle {\mathbb{D}}^{q}f(t)=\sum _{m=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{q}{m})\sum _{k=0}^m(\genfrac{}{}{0ex}{}{m}{k})(-1{)}^{m-k}{f}^{(k)}(x).}$$

For fractional derivative definitions described in this section, the following identities hold:

${\displaystyle {\mathbb{D}}^q(t^n)=\frac{\mathrm{\Gamma }(n+1)}{\mathrm{\Gamma }(n+1-q)}{t}^{n-q}}$

${\displaystyle {\mathbb{D}}^q(\sin (t))=\sin \left(t+\frac{q\pi }{2}\right)}$

${\displaystyle {\mathbb{D}}^q(e^{at})=a^q{e}^{at}}$

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 $${\displaystyle {\mathbb{D}}_t^q(fg)=\sum _{j=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{q}{j}){\mathbb{D}}_t^j(f){\mathbb{D}}_t^{q-j}(g)}$$

In general, composition (or semigroup) rule is a desirable property, but is hard to achieve mathematically and hence is not always completely satisfied by each proposed operator; this forms part of the decision making process on which one to choose:

• ${\mathbb{D}}^a{\mathbb{D}}^{b}f={\mathbb{D}}^{a+b}f$ (ideally)
• ${\mathbb{D}}^a{\mathbb{D}}^{b}f\ne {\mathbb{D}}^{a+b}f$ (in practice)