# Initialized fractional calculus

(Redirected from Initialization of differintegrals)

In mathematical analysis, initialization of the differintegrals is a topic in fractional calculus.

## Composition rule of differintegral

A certain counterintuitive property of the differintegral operator should be pointed out, namely the composition law. Although

${\displaystyle \mathbb {D} ^{q}\mathbb {D} ^{-q}=\mathbb {I} }$

wherein Dq is the left inverse of Dq, the converse is not necessarily true:

${\displaystyle \mathbb {D} ^{-q}\mathbb {D} ^{q}\neq \mathbb {I} }$

### Example

It is instructive to consider elementary integer-order calculus to see what's happening. First, integrate then differentiate, using the example function 3x2 + 1:

${\displaystyle {\frac {d}{dx}}\left[\int (3x^{2}+1)dx\right]={\frac {d}{dx}}[x^{3}+x+c]=3x^{2}+1\,,}$

on exchanging the order of composition:

${\displaystyle \int \left[{\frac {d}{dx}}(3x^{2}+1)\right]=\int 6x\,dx=3x^{2}+c\,,}$

in which the constant of integration is c. Even if it wasn't obvious, the initialization terms ƒ'(0) = c, ƒ''(0) = d, etc. could be used. If we neglected those initialization terms, the last equation would show the composition of integration then differentiation (and vice versa) would not hold.

## Description of initialization

This is the problem that with the differintegral. If the differintegral is initialized properly, then the hoped-for composition law holds. The problem is that in differentiation, we lose information, as we lost the c in the first equation.

In fractional calculus, however, since the operator has been fractionalized and is thus continuous, an entire complementary function is needed, not just a constant or set of constants. We call this complementary function ${\displaystyle \Psi }$.

${\displaystyle \mathbb {D} _{t}^{q}f(t)={\frac {1}{\Gamma (n-q)}}{\frac {d^{n}}{dt^{n}}}\int _{0}^{t}(t-\tau )^{n-q-1}f(\tau )\,d\tau +\Psi (x)}$

Working with a properly initialized differintegral is the subject of initialized fractional calculus.