Editing Fractional calculus (section)

==Computing the fractional integral==
Let <math>f(x)</math> be a function defined for <math>x>0</math>. Form the definite integral from 0 to <math>x</math>. Call this
<math display="block">( J f ) ( x ) = \int_0^x f(t) \, dt \,.</math>

Repeating this process gives
<math display="block">\begin{align}
\left( J^2 f \right) (x) &= \int_0^x (Jf)(t) \,dt \\
&= \int_0^x \left(\int_0^t f(s) \,ds \right) dt \,,
\end{align}</math>

and this can be extended arbitrarily.

The [[Cauchy formula for repeated integration]], namely
<math display="block">\left(J^n f\right) ( x ) = \frac{1}{ (n-1) ! } \int_0^x \left(x-t\right)^{n-1} f(t) \, dt \,,</math>
leads in a straightforward way to a generalization for real {{mvar|n}}: using the [[gamma function]] to remove the discrete nature of the factorial function gives us a natural candidate for applications of the fractional integral operator as
<math display="block">\left(J^\alpha f\right) ( x ) = \frac{1}{ \Gamma ( \alpha ) } \int_0^x \left(x-t\right)^{\alpha-1} f(t) \, dt \,.</math>

This is in fact a well-defined operator.

It is straightforward to show that the {{mvar|J}} operator satisfies
<math display="block">\begin{align}
\left(J^\alpha\right) \left(J^\beta f\right)(x) &= \left(J^\beta\right) \left(J^\alpha f\right)(x) \\
   &= \left(J^{\alpha+\beta} f\right)(x) \\
   &= \frac{1}{ \Gamma ( \alpha + \beta) } \int_0^x \left(x-t\right)^{\alpha+\beta-1} f(t) \, dt \,.
\end{align}</math>

{{Collapse top|title=Proof of this identity}}
<math display="block">
\begin{align}
\left(J^\alpha\right) \left(J^\beta f\right)(x) & = \frac{1}{\Gamma(\alpha)} \int_0^x (x-t)^{\alpha-1} \left(J^\beta f\right)(t) \, dt \\
& = \frac{1}{\Gamma(\alpha) \Gamma(\beta)} \int_0^x \int_0^t \left(x-t\right)^{\alpha-1} \left(t-s\right)^{\beta-1} f(s) \, ds \, dt \\
& = \frac{1}{\Gamma(\alpha) \Gamma(\beta)} \int_0^x f(s) \left( \int_s^x \left(x-t\right)^{\alpha-1} \left(t-s\right)^{\beta-1} \, dt \right) \, ds
\end{align}
</math>

where in the last step we exchanged the order of integration and pulled out the {{math|''f''(''s'')}} factor from the {{mvar|t}} integration.

Changing variables to {{mvar|r}} defined by {{math|1=''t'' = ''s'' + (''x'' − ''s'')''r''}},
<math display="block">\left(J^\alpha\right) \left(J^\beta f\right)(x) = \frac{1}{\Gamma(\alpha) \Gamma(\beta)} \int_0^x \left(x-s\right)^{\alpha + \beta - 1} f(s) \left( \int_0^1 \left(1-r\right)^{\alpha-1} r^{\beta-1} \, dr \right)\, ds</math>

The inner integral is the [[beta function]] which satisfies the following property:
<math display="block">\int_0^1 \left(1-r\right)^{\alpha-1} r^{\beta-1} \, dr = B(\alpha, \beta) = \frac{\Gamma(\alpha)\,\Gamma(\beta)}{\Gamma(\alpha+\beta)}</math>

Substituting back into the equation:
<math display="block">\begin{align}
\left(J^\alpha\right) \left(J^\beta f\right)(x) &= \frac{1}{\Gamma(\alpha + \beta)} \int_0^x \left(x-s\right)^{\alpha + \beta - 1} f(s) \, ds \\
   &= \left(J^{\alpha + \beta} f\right)(x)
\end{align}</math>

Interchanging {{mvar|α}} and {{mvar|β}} shows that the order in which the {{mvar|J}} operator is applied is irrelevant and completes the proof.
{{Collapse bottom}}

This relationship is called the semigroup property of fractional [[differintegral]] operators. 

===Riemann–Liouville fractional integral===
The classical form of fractional calculus is given by the [[Riemann–Liouville integral]], which is essentially what has been described above. The theory of fractional integration for [[periodic function]]s (therefore including the "boundary condition" of repeating after a period) is given by the [[Weyl integral]]. It is defined on [[Fourier series]], and requires the constant Fourier coefficient to vanish (thus, it applies to functions on the [[unit circle]] whose integrals evaluate to zero). The Riemann–Liouville integral exists in two forms, upper and lower. Considering the interval {{closed-closed|''a'',''b''}}, the integrals are defined as
<math display="block">\begin{align}
\sideset{_a}{_t^{-\alpha}}D f(t) &= \sideset{_a}{_t^\alpha}I f(t) \\
   &=\frac{1}{\Gamma(\alpha)}\int_a^t \left(t-\tau\right)^{\alpha-1} f(\tau) \, d\tau \\
\sideset{_t}{_b^{-\alpha}}D f(t) &= \sideset{_t}{_b^\alpha}I f(t) \\
  &=\frac{1}{\Gamma(\alpha)}\int_t^b \left(\tau-t\right)^{\alpha-1} f(\tau) \, d\tau
\end{align}</math>

Where the former is valid for {{math|''t'' > ''a''}} and the latter is valid for {{math|''t'' < ''b''}}.<ref>{{cite book |last=Hermann |first=Richard |date=2014 |title=Fractional Calculus: An Introduction for Physicists |edition=2nd |location=New Jersey |publisher=World Scientific Publishing |page=46 |isbn=978-981-4551-07-6 |doi=10.1142/8934 |bibcode=2014fcip.book.....H}}</ref> 

It has been suggested<ref name=Mainardi/> that the integral on the positive real axis (i.e. <math>a = 0</math>) would be more appropriately named the Abel–Riemann integral, on the basis of history of discovery and use, and in the same vein the integral over the entire real line be named Liouville–Weyl integral.

By contrast the [[Grünwald–Letnikov derivative]] starts with the derivative instead of the integral.

===Hadamard fractional integral===
The ''Hadamard fractional integral'' was introduced by [[Jacques Hadamard]]<ref>{{cite journal |last=Hadamard |first=J. |date=1892 |title=Essai sur l'étude des fonctions données par leur développement de Taylor |url=http://sites.mathdoc.fr/JMPA/PDF/JMPA_1892_4_8_A4_0.pdf |journal=Journal de Mathématiques Pures et Appliquées |volume=4 |issue=8 |pages=101–186}}</ref> and is given by the following formula,
<math display="block">\sideset{_a}{_t^{-\alpha}}{\mathbf{D}} f(t) = \frac{1}{\Gamma(\alpha)} \int_a^t \left(\log\frac{t}{\tau} \right)^{\alpha -1} f(\tau)\frac{d\tau}{\tau}, \qquad t > a\,.</math>

===Atangana–Baleanu fractional integral (AB fractional integral)===
The Atangana–Baleanu fractional integral of a continuous function is defined as:
<math display="block">\sideset{_{\hphantom{A}a}^\operatorname{AB}}{_t^\alpha}I f(t)=\frac{1-\alpha}{\operatorname{AB}(\alpha)}f(t)+\frac{\alpha}{\operatorname{AB}(\alpha)\Gamma(\alpha)}\int_a^t \left(t-\tau\right)^{\alpha-1} f(\tau) \, d\tau </math>