Fractional calculus

The Fractional Calculus is the extension of the derivative term on non-integer orders. The term “fractional” is historically determined, the derivatives can generally be of a real or even complex order.

First of all: important functions and integral transformations

These functions and transformations usually each have their own article in Wikipedia. However, since they are of fundamental importance in the definition of the fractional integrals, they are briefly summarized here as definitions.

(Incomplete) gamma function

As a generalization of the factorial function , the gamma function is defined as follows:

${\displaystyle \Gamma (x)=\int _{0}^{\infty }t^{x-1}e^{-t}\mathrm {d} t}$

For integer arguments results . In the case of the incomplete gamma function, integration is not carried out to infinity, but only up to a certain value y: ${\displaystyle \Gamma (n+1)=n!}$

${\displaystyle \gamma (x,y)=\int _{0}^{y}t^{x-1}e^{-t}\mathrm {d} t}$.

Beta function

The beta function is defined as

${\displaystyle \mathrm {B} (x,y)=\int _{0}^{1}t^{x-1}(1-t)^{y-1}\,\mathrm {d} t}$,

whereby it can also be represented as a product of gamma functions

${\displaystyle \mathrm {B} (x,y)={\frac {\Gamma (x)\cdot \Gamma (y)}{\Gamma (x+y)}}}$.

Hypergeometric function

As an extension of the geometric series , the generalized hypergeometric function is defined as

${\displaystyle {}_{p}F_{q}(a_{1},\dots ,a_{p};b_{1},\dots ,b_{q};z)=\sum _{k=0}^{\infty }\prod _{i=1}^{p}{\frac {\Gamma (k+a_{i})}{\Gamma (a_{i})}}\prod _{j=1}^{q}{\frac {\Gamma (b_{j})}{\Gamma (k+b_{j})}}{\frac {z^{k}}{k!}}}$.

The special case is immediately apparent

${\displaystyle {}_{0}F_{0}(;;z)=\sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=e^{z}}$.

Fourier transform

For one defines ${\displaystyle f\in L^{1}(\mathbb {R} )}$

${\displaystyle {\tilde {f}}(k)=({\mathcal {F}}f)(k)=\int _{-\infty }^{\infty }e^{-2\pi ikx}f(x)\mathrm {d} x}$

as a Fourier transform , and

${\displaystyle f(x)=\left({\mathcal {F}}^{-1}{\tilde {f}}\right)(x)=\int _{-\infty }^{\infty }e^{2\pi ikx}{\tilde {f}}(k)\mathrm {d} k}$

as an inverse transformation.

Note that there are different ways of defining the Fourier transformation, which differ in which transformation you write the minus sign in the e-function or where the factor of 2π appears.

Translation: operator . ${\displaystyle (\tau _{a}f)(x)=f(x-a)\Rightarrow (\tau _{a}f)^{\sim }(k)=e^{-2\pi ika}{\tilde {f}}(k)\quad \land \quad \left(e^{2\pi iax}f\right)^{\sim }(k)=\left(\tau _{a}{\tilde {f}}\right)(k)={\tilde {f}}(k-a)}$

Yield: operator . ${\displaystyle \left(\delta _{\lambda }f\right)(x)=f\left({\frac {x}{\lambda }}\right)\quad \forall \lambda >0\qquad \Rightarrow \qquad \left(\delta _{\lambda }f\right)^{\sim }(k)=\lambda f(\lambda k)}$

Fold: . ${\displaystyle (f*g)(x)=\int _{-\infty }^{\infty }f(x-y)g(y)\mathrm {d} y\quad {\Biggl (}=\int _{-\infty }^{\infty }\left(\tau _{y}f\right)(x)g(y)\mathrm {d} y{\Biggr )}}$

From this follows the convolution theorem: For is ${\displaystyle f,g\in L^{1}(\mathbb {R} )}$

${\displaystyle \left({\mathcal {F}}(f*g)\right)(k)=({\mathcal {F}}f)(k)({\mathcal {F}}g)(k)}$.

The Fourier transform turns the convolution of two functions into the multiplication of their Fourier transforms.

Further applies to ${\displaystyle f\in L^{1}(\mathbb {R} )}$

${\displaystyle \left({\frac {\partial }{\partial x_{j}}}f\right)^{\sim }\left({\vec {k}}\right)=2\pi ik_{j}{\tilde {f}}\left({\vec {k}}\right)}$.

Laplace transformation

Let be a locally integrable function , then the Laplace transformation is defined as ${\displaystyle f\in L_{loc}^{1}\left(\mathbb {R} ^{+}\right)}$

${\displaystyle {\tilde {f}}(u)=({\mathcal {L}}f)(u)=\int _{0}^{\infty }f(x)e^{-ux}\ \!\mathrm {d} x.}$

The Laplace convolution is defined similarly to the Fourier convolution and provides a similar relationship:

${\displaystyle (f\star g)(x)=\int _{0}^{\infty }f(x-y)g(y)\ \!\mathrm {d} y\Rightarrow {\mathcal {L}}(f\star g)(u)=({\mathcal {L}}f)(u)({\mathcal {L}}g)(u).}$

history

The mathematicians Gottfried Wilhelm Leibniz and Leonhard Euler were already concerned with the generalization of the term derivative. In a letter to Guillaume François Antoine, Marquis de L'Hospital, Leibniz describes the similarity between potencies and the product rule of derivatives:

${\displaystyle (x+y)^{n}=x^{n}y^{0}+nx^{n-1}y+\dots +x^{0}y^{n}}$

${\displaystyle \mathrm {d} ^{n}(xy)=\mathrm {d} ^{n}x\mathrm {d} ^{0}y+n\mathrm {d} ^{n-1}x\mathrm {d} y+\dots +\mathrm {d} ^{0}x\mathrm {d} ^{n}y}$

what is seemingly easy on

${\displaystyle \mathrm {d} ^{\alpha }(fg)=\mathrm {d} ^{\alpha }f\mathrm {d} ^{0}g+{\frac {\alpha }{1}}\mathrm {d} ^{\alpha -1}f\mathrm {d} g+{\frac {\alpha (\alpha -1)}{1\cdot 2}}\mathrm {d} ^{\alpha -2}f\mathrm {d} ^{2}g+{\frac {\alpha (\alpha -1)(\alpha -2)}{1\cdot 2\cdot 3}}\mathrm {d} ^{\alpha -3}f\mathrm {d} ^{3}g\dots }$

can be generalized (whereby one sets negative in the case of α-n ). However, problems arise with such naive use of symbolism. As an example, choose a function f such that ${\displaystyle \mathrm {d} ^{\alpha -n}=\int _{}^{n-\alpha }}$

${\displaystyle \mathrm {d} f=f\mathrm {d} x\land \mathrm {d} ^{2}f=f\mathrm {d} x^{2}}$

${\displaystyle \Rightarrow {\frac {\mathrm {d} ^{2}f}{\mathrm {d} x}}={\frac {\mathrm {d} \left({\frac {\mathrm {d} f}{\mathrm {d} x}}\right)}{\mathrm {d} x}}={\frac {\mathrm {d} f}{\mathrm {d} x}}=f\land \mathrm {d} x={\frac {\mathrm {d} f}{f}}}$

Note the mathematically incorrect "multiplying" with dx. With such a function, one thinks directly of the exponential function , which was not yet explicitly known as such at the time. Now where do you come across a contradiction when you look at it? To see that, just put : ${\displaystyle \mathrm {d} ^{\alpha }f=f\mathrm {d} x^{\alpha }}$ ${\displaystyle \alpha ={\frac {1}{2}}}$

${\displaystyle {\frac {\mathrm {d} ^{\frac {1}{2}}f}{\mathrm {d} x^{\frac {1}{2}}}}={\frac {\mathrm {d} ^{\frac {1}{2}}f}{\left({\frac {\mathrm {d} f}{f}}\right)^{\frac {1}{2}}}}={\sqrt {\frac {f}{\mathrm {d} f}}}\mathrm {d} ^{\frac {1}{2}}f\neq f.}$

Thus Leibniz's simple approach cannot be the appropriate solution to the problem.

Euler's approach

Euler considered integer derivatives of power functions z ^m . For these applies:

${\displaystyle {\frac {\mathrm {d} ^{n}z^{m}}{\mathrm {d} z^{n}}}={\frac {m!}{(m-n)!}}z^{m-n}.}$

He now tried to generalize this relationship to non-integer powers by replacing the faculty function with the gamma function he had found:

${\displaystyle {\frac {\mathrm {d} ^{\alpha }z^{\beta }}{\mathrm {d} z^{\alpha }}}={\frac {\Gamma (\beta +1)}{\Gamma (\beta -\alpha +1)}}z^{\beta -\alpha }.}$

This way too leads to contradictions. Consider again the e-function e ^λx , which n-times differentiated results in λ ⁿ e ^λx ; thus generalized:

${\displaystyle {\frac {\mathrm {d} ^{\alpha }e^{\lambda x}}{\mathrm {d} x^{\alpha }}}=\lambda ^{\alpha }e^{\lambda x}.}$

On the other hand, however, the exponential function is only an infinite power series, namely . ${\displaystyle e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}}$

Thus one has two possibilities to calculate the α-derivative of e ^x :

1. Directly: ${\displaystyle {\frac {\mathrm {d} ^{\alpha }e^{x}}{\mathrm {d} x^{\alpha }}}=e^{x}}$
2. Indirectly via the power series representation: ${\displaystyle {\frac {\mathrm {d} ^{\alpha }}{\mathrm {d} x^{\alpha }}}\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=\sum _{n=0}^{\infty }{\frac {\mathrm {d} ^{\alpha }}{\mathrm {d} x^{\alpha }}}{\frac {x^{n}}{n!}}=\sum _{n=0}^{\infty }{\frac {x^{n-\alpha }}{\Gamma (n-\alpha +1)}}\neq \sum _{n=0}^{\infty }{\frac {x^{n}}{\Gamma (n+1)}}=e^{x}}$

This contradiction can be explained with the example α = -1, if negative exponents of the differential operators are interpreted as integrals:

${\displaystyle {\frac {\mathrm {d} ^{-1}}{\mathrm {d} x^{-1}}}e^{x}=\int _{-\infty }^{x}e^{t}\mathrm {d} t=e^{x}}$

${\displaystyle {\frac {\mathrm {d} ^{-1}}{\mathrm {d} x^{-1}}}x^{\beta }=\int _{0}^{x}t^{\beta }\mathrm {d} t={\frac {x^{\beta +1}}{\beta +1}}}$

The different lower limits make it clear that with this approach you “have to know” from where to where you have to integrate in order to find the correct antiderivative. Thus, Euler's approach, although better in terms of its idea and execution, is not suitable for correctly generalizing the differential operator to real powers.

Definition of fractional integral operators

Iterative and fractional integrals

One possibility of defining fractional integrals without contradiction results from the formula

${\displaystyle \int _{a}^{x}\int _{a}^{y}f(z)\mathrm {d} z\mathrm {d} y=\int _{a}^{x}(x-y)f(y)\mathrm {d} y}$,

which converts the double integration over two variables with the same lower limit into a single integral. This formula can be extended to any number of integrals.

If one introduces the integral operator as follows ${\displaystyle I_{a+}}$

${\displaystyle (I_{a+}f)(x)=\int _{a}^{x}f(y)\mathrm {d} y=[F(y)]_{a}^{x}=F(x)-F(a)}$,

where F (x) is an antiderivative of f (x), then arbitrarily high powers of this operator can be reduced to a single integral thanks to the above formula of multiple integrals:

${\displaystyle (I_{a+}^{n}f)(x)=\int _{a}^{x}\int _{a}^{y_{1}}\int _{a}^{y_{2}}\dots \int _{a}^{y_{n-1}}f(y_{n})\mathrm {d} y_{n}\mathrm {d} y_{n-1}\dots \mathrm {d} y_{1}={\frac {1}{(n-1)!}}\int _{a}^{x}(x-y)^{n-1}f(y)\mathrm {d} y}$.

In contrast to the formulas at the beginning, this integral operator can be generalized relatively easily from whole numbers n to real (or complex) numbers α by replacing n by α and the factorial by the gamma function and requires that : ${\displaystyle -\infty \leq \alpha <x<\infty \land f\in L_{loc}^{1}(a,\infty )}$

${\displaystyle (I_{a+}^{\alpha }f)(x)={\frac {1}{\Gamma (\alpha )}}\int _{a}^{x}(x-y)^{\alpha -1}f(y)\mathrm {d} y}$

This is called the right-sided fractional Riemann-Liouville integral . Similarly, through

${\displaystyle (I_{b-}^{\alpha }f)(x)={\frac {1}{\Gamma (\alpha )}}\int _{x}^{b}(y-x)^{\alpha -1}f(y)\mathrm {d} y}$

the left-hand equivalent can be defined.

Reduction of fractional integrals to convolution

If the distribution is defined , the fractional integral can be reduced to a Laplace convolution: ${\displaystyle x_{a+}^{\alpha }={\begin{cases}x^{\alpha }&\forall x>a\\0&\forall x\leq a\end{cases}}}$

${\displaystyle {\mathcal {L}}\left(I_{a+}^{\alpha }f\right)(u)=({\mathcal {L}}f)(u)\left({\mathcal {L}}{\frac {x_{a+}^{\alpha }}{\Gamma (\alpha )}}\right)(u)}$,

there ${\displaystyle {\mathcal {L}}\left(x_{a+}^{\alpha }\right)(u)=\Gamma (\alpha -1)u^{-\alpha -1}}$

Fractional Weyl integrals

If one approaches infinity in the above equations or in terms of absolute values, one obtains the so-called Weyl integrals and the corresponding partial integral operators ${\displaystyle a}$ ${\displaystyle b}$

${\displaystyle {\begin{aligned}\left(I_{+}^{\alpha }f\right)(x)&={\frac {1}{\Gamma (\alpha )}}\int _{-\infty }^{x}(x-y)^{\alpha -1}f(y)\mathrm {d} y\\\left(I_{-}^{\alpha }f\right)(x)&={\frac {1}{\Gamma (\alpha )}}\int _{x}^{\infty }(y-x)^{\alpha -1}f(y)\mathrm {d} y\end{aligned}}}$

for and with the definition set . This condition is fulfilled , for example, for with . ${\displaystyle -\infty <x<\infty }$ ${\displaystyle \{f{\mathcal {|}}I_{\pm }^{\alpha }f<\infty \}}$ ${\displaystyle f\in L^{p}(\mathbb {R} )}$ ${\displaystyle 1\leq p\leq {\frac {1}{\alpha }}}$

Fractional Weyl integrals and folds

Fractional Weyl integrals can also be traced back to convolution. However, these are Fourier convolutions, since Weyl integrals have an infinite lower or upper limit.

${\displaystyle \left(I_{+}^{\alpha }f\right)(x)={\frac {1}{\Gamma (\alpha )}}\int _{-\infty }^{x}(x-y)^{\alpha -1}f(y)\mathrm {d} y}$

what can be converted into ${\displaystyle x-y=t}$

${\displaystyle {\frac {1}{\Gamma (\alpha )}}\int _{0}^{\infty }t^{\alpha -1}f(x-t)\mathrm {d} t={\frac {1}{\Gamma (\alpha )}}\int _{-\infty }^{\infty }t_{+}^{\alpha -1}f(y)\mathrm {d} y}$ With ${\displaystyle t_{+}^{\alpha }={\begin{cases}t^{\alpha }&\forall t>0\\0&\forall t\leq 0\end{cases}}}$

${\displaystyle \Rightarrow \left(I_{+}^{\alpha }f\right)(x)=\left(f*{\frac {x_{+}^{\alpha -1}}{\Gamma (\alpha )}}\right)(x)}$

Hence the Fourier transform for ${\displaystyle 0<\alpha <1}$

${\displaystyle \left(I_{+}^{\alpha }f\right)^{\sim }(k)=\left({\mathcal {F}}I_{+}^{\alpha }f\right)(k)=\left({\mathcal {F}}{\frac {x_{+}^{\alpha -1}}{\Gamma (\alpha )}}\right)(k)({\mathcal {F}}f)(k)=(2\pi ik)^{-\alpha }{\tilde {f}}(k)}$

One can see that the fractional Riemann-Liouville integral operator is diagonalized by the Laplace convolution, the fractional Weyl integral operator by the Fourier convolution.

Examples

${\displaystyle {\underline {f(x)=x^{\beta }}}}$:

${\displaystyle \left(I_{0+}^{\alpha }f\right)(x)={\frac {1}{\Gamma (\alpha )}}\int _{0}^{x}(x-y)^{\alpha -1}y^{\beta }\mathrm {d} y={\frac {x^{\alpha -1}}{\Gamma (\alpha )}}\int _{0}^{x}\left(1-{\frac {y}{x}}\right)^{\alpha -1}y^{\beta }\mathrm {d} y}$

Substitute ${\displaystyle z(y)=y/x\rightarrow dy=xdz}$

${\displaystyle \left(I_{0+}^{\alpha }f\right)(x)={\frac {x^{\alpha +\beta }}{\Gamma (\alpha )}}\int _{0}^{1}(1-z)^{\alpha -1}z^{\beta }\mathrm {d} z={\frac {\mathrm {B} (\beta +1,\alpha )x^{\alpha +\beta }}{\Gamma (\alpha )}}={\frac {\Gamma (\beta +1)}{\Gamma (\alpha +\beta +1)}}x^{\alpha +\beta }}$

In the special case it becomes α = 1

${\displaystyle \left(I_{0+}^{1}f\right)(x)={\frac {\Gamma (\beta +1)}{\Gamma (\beta +2)}}x^{1+\beta }={\frac {\Gamma (\beta +1)}{(\beta +1)\Gamma (\beta +1)}}x^{\beta +1}={\frac {1}{\beta +1}}x^{\beta +1}}$

${\displaystyle {\underline {f(x)=e^{ax}}}}$with the substitution : ${\displaystyle z=a(x-y)}$

${\displaystyle \left(I_{+}^{\alpha }f\right)(x)={\frac {1}{\Gamma (\alpha )}}\int _{-\infty }^{x}(x-y)^{\alpha -1}e^{ay}\mathrm {d} y=-{\frac {1}{\Gamma (\alpha )}}\int _{\infty }^{0}\left({\frac {z}{a}}\right)^{\alpha -1}e^{ax-z}{\frac {1}{a}}\mathrm {d} z={\frac {a^{-\alpha }e^{ax}}{\Gamma (\alpha )}}\int _{0}^{\infty }z^{\alpha -1}e^{-z}\mathrm {d} z=a^{-\alpha }e^{ax}}$

So you can see that with this integral operator, similar to Euler's approach, you “have to know” from where to where you have to integrate in order to get the actual antiderivative of a function, but this is explicitly included in the operator definition. Thus, the lower limit has to be chosen in such a way that in (see above, definition of I _{a +} ) the F (a) vanishes and one obtains F (x) (or the fractional equivalent to it). In this second example we have integrated from -∞ to x, knowing full well that e ^ax approaches 0 for a → -∞. Therefore we simply integrate this function again, but this time with the lower limit 0 and the substitution : ${\displaystyle F(x)-F(a)}$ ${\displaystyle e^{ax}}$ ${\displaystyle z=x-y}$

${\displaystyle \left(I_{0+}^{\alpha }e^{ax}\right)(x)={\frac {1}{\Gamma (\alpha )}}\int _{0}^{x}(x-y)^{\alpha -1}e^{ay}\mathrm {d} y={\frac {1}{\Gamma (\alpha )}}\int _{0}^{x}z^{\alpha -1}e^{a(x-z)}\mathrm {d} z={\frac {e^{ax}}{\Gamma (\alpha )}}\int _{0}^{x}z^{\alpha -1}e^{-z}\mathrm {d} z}$

If one now substitutes az with t, the result is:

${\displaystyle ={\frac {e^{ax}}{\Gamma (\alpha )a^{\alpha -1}}}\int _{0}^{ax}t^{\alpha -1}e^{-t}{\frac {1}{a}}\mathrm {d} t={\frac {\gamma (\alpha ,ax)}{\Gamma (\alpha )}}a^{-\alpha }e^{ax}}$

${\displaystyle {\underline {f(x)=(x+c)^{\beta }}}}$:

${\displaystyle \left(I_{0+}^{\alpha }f\right)(x)={\frac {1}{\Gamma (\alpha )}}\int _{0}^{x}(x-y)^{\alpha -1}(y+c)^{\beta }\mathrm {d} y={\frac {c^{\beta }x^{\alpha -1}}{\Gamma (\alpha )}}\int _{0}^{x}\left(1-{\frac {y}{x}}\right)^{\alpha -1}\left(1+{\frac {y}{c}}\right)^{\beta }\mathrm {d} y}$

Substitution of z with y / x lists

${\displaystyle \left(I_{0+}^{\alpha }f\right)(x)={\frac {c^{\beta }x^{\alpha }}{\Gamma (\alpha )}}\int _{0}^{1}(1-z)^{\alpha -1}\left(1+{\frac {xz}{c}}\right)^{\beta }\mathrm {d} z}$

Compare this with . One can see that one simply has to set a = 1, b = -β, c = α + 1 and z = -x / c to get the above integral. So is ${\displaystyle \int _{0}^{1}y^{a-1}(1-y)^{c-a-1}(1-zy)^{-b}\mathrm {d} y={\frac {\Gamma (c-a)\Gamma (a)}{\Gamma (c)}}{}_{2}F_{1}(a,b;c;z)}$

${\displaystyle \left(I_{0+}^{\alpha }f\right)(x)={\frac {x^{\alpha }c^{\beta }}{\Gamma (\alpha +1)}}{\frac {\Gamma (\alpha +1-1)\Gamma (1)}{\Gamma (\alpha +1)}}{}_{2}F_{1}\left(1,-\beta ;\alpha +1;-{\frac {x}{c}}\right)={\frac {c^{\beta }}{\Gamma (\alpha +1)}}{}_{2}F_{1}\left(1,-\beta ;\alpha +1;-{\frac {x}{c}}\right)x^{\alpha }}$

Integration of hypergeometric functions

Since many other functions can be represented with hypergeometric functions, it is advisable to represent a formula for their integration here.

${\displaystyle f(x)={}_{p}F_{q}(a_{1},\dots ,a_{p};b_{1},\dots ,b_{q};x)}$

${\displaystyle (I_{0+}^{\alpha }f)(x)={\frac {x^{\alpha }}{\Gamma (\alpha +1)}}{}_{p+1}F_{q+1}(a_{1},\dots ,a_{p},1;b_{1},\dots ,b_{q},1+\alpha ;x)}$

Duality of the operators + and -

The following applies in general

${\displaystyle \int _{a}^{b}f(x)\left(I_{a+}^{\alpha }g\right)(x)\mathrm {d} x=\int _{a}^{b}\left(I_{b-}^{\alpha }f\right)(x)g(x)\mathrm {d} x}$

the two Riemann-Liouville and the two Weyl integrals are therefore each dual to one another. In integrals, for example, the fractional integral can be shifted from one function to the one that is easier to integrate.

Definition of general fract. Integral operators I ^α and I ^{⋆ α}

First generalization of fractional integration

Through the approach

${\displaystyle \left(I^{\alpha }f\right)(x)={\frac {1}{C(\alpha )}}\int _{-\infty }^{\infty }\vert x-y\vert ^{\alpha -1}f(y)\mathrm {d} y}$

we shall try to define a more general integral operator. The amount bars instead of simple round brackets already indicate that a kind of spherical symmetry is assumed for this. C (α) should be determined in such a way that the additivity of the order (I ^α I ^β = I ^{α + β} ) still applies. One can already assume that this operator is simply a linear combination of the already known Weyl integral operators, which one can also prove:

${\displaystyle \left(I^{\alpha }f\right)(x)={\frac {1}{C(\alpha )}}\left[\int _{-\infty }^{x}(x-y)^{\alpha -1}f(y)\mathrm {d} y+\int _{x}^{\infty }(y-x)^{\alpha -1}f(y)\mathrm {d} y\right]={\frac {\Gamma (\alpha )}{C(\alpha )}}\left[\left(I_{+}^{\alpha }f)(x)+(I_{-}^{\alpha }f\right)(x)\right]}$

So is

${\displaystyle I^{\alpha }={\frac {\Gamma (\alpha )}{C(\alpha )}}\left(I_{+}^{\alpha }+I_{-}^{\alpha }\right)}$

which can also be generalized to higher dimensions:

${\displaystyle \left(I^{\alpha }f\right)\left({\vec {x}}\right)={\frac {1}{C(d,\alpha )}}\int _{\mathbb {R} ^{d}}\vert {\vec {x}}-{\vec {y}}\vert ^{\alpha -d}f\left({\vec {y}}\right)\mathrm {d} {\vec {y}}}$

Now the question is how C (d, α) can be determined. If one wants to make the choice in such a way that the following applies, then after a detailed study of the Fourier transform for C (d, α): ${\displaystyle \left(I^{\alpha }f\right)^{\sim }\left({\vec {k}}\right)=\left|2\pi {\vec {k}}\right|^{-\alpha }{\tilde {f}}\left({\vec {k}}\right)}$

${\displaystyle C(d,\alpha )={\frac {2^{\alpha }\pi ^{\frac {d}{2}}\Gamma \left({\frac {\alpha }{2}}\right)}{\Gamma \left({\frac {d-\alpha }{2}}\right)}}}$.

Using the formulas and, this results in the one-dimensional case: ${\displaystyle \Gamma (2z)={\frac {2^{2z-1}}{\sqrt {\pi }}}\Gamma (z)\Gamma \left(z+{\frac {1}{2}}\right)}$ ${\displaystyle \Gamma \left({\frac {1-z}{2}}\right)\Gamma \left({\frac {1+z}{2}}\right)={\frac {\pi }{\cos \left(z{\frac {\pi }{2}}\right)}}}$

${\displaystyle \left(I^{\alpha }f\right)(x)={\frac {1}{2\Gamma (\alpha )\cos \left(\alpha {\frac {\pi }{2}}\right)}}\int _{-\infty }^{\infty }\vert x-y\vert ^{\alpha -1}f(y)\mathrm {d} y.}$

This is called the fractional Riesz-Feller integral .

Further generalization of fractional integration

The formula allows us to conclude that there are other such general integral operators through ${\displaystyle I^{\alpha }={\frac {\Gamma (\alpha )}{C(\alpha )}}\left(I_{+}^{\alpha }+I_{-}^{\alpha }\right)}$

${\displaystyle I^{\alpha }={\frac {\Gamma (\alpha )}{C(\alpha )}}\left(c_{+}I_{+}^{\alpha }+c_{-}I_{-}^{\alpha }\right)}$

can define what makes the Riesz-Feller integral the special case c ₊ = c _- = 1. E.g. for c ₊ = 1 and c _- = -1

${\displaystyle \left({I^{\star }}^{\alpha }f\right)(x)={\frac {\Gamma (\alpha )}{C(\alpha )}}\left(\left(I_{+}^{\alpha }f\right)(x)-\left(I_{-}^{\alpha }f\right)(x)\right)={\frac {1}{2\Gamma (\alpha )\sin \left(\alpha {\frac {\pi }{2}}\right)}}\int _{-\infty }^{\infty }\operatorname {sign} (x-y)\vert x-y\vert ^{\alpha -1}f(y)\mathrm {d} y}$

These two operators are linked by the Hilbert transformation :

${\displaystyle {I^{\star }}^{\alpha }={\mathcal {H}}I^{\alpha }\land I^{\alpha }={\mathcal {H}}{I^{\star }}^{\alpha }}$

Feller has for integrals of form

${\displaystyle \left(I_{\delta }^{\alpha }f\right)(x)={\frac {1}{\Gamma (\alpha )\sin(\alpha \pi )}}\int _{-\infty }^{\infty }\sin \left(\alpha \left({\frac {\pi }{2}}+\delta \operatorname {sign} (y-x)\right)\right)\vert x-y\vert ^{\alpha -1}f(y)\mathrm {d} y}$

proved that the additivity of order holds. These integrals can also be represented as a linear combination of the above form, you just have to

${\displaystyle c_{\pm }={\frac {\sin \left(\alpha {\frac {\pi }{2}}\mp \delta \alpha \right)}{\sin(\alpha \pi )}}}$

choose.

Examples of fractional integral equations in physics

Tautochron problem

Problem in two dimensions: a point of mass falls under the influence of gravity along a fixed but unknown path y = h (x) from height y ₀ to height y ₁ ; the time it takes for this is given as = time of the fall from fixed y ₀ to variable y ₁ . The question now is: can h (x) be determined from knowledge of the fall times alone? ${\displaystyle T_{y_{0}}\left(y_{1}\right)}$ ${\displaystyle T_{y_{0}}(y)\;\forall \;y\in \left[y_{1},y_{0}\right]}$

We set v (y) equal to the amount of the current speed, then for the duration of the fall from P ₀ to P ₁ : with s (y) equal to the value covered as a function of the altitude. If y = h (x) is invertible, then x = h ⁻¹ (y) = Φ (y) and the arc length differential with the arc length ${\displaystyle T_{y_{0}}(y_{1})=\int _{P_{0}\left(x_{0},y_{0}\right)}^{P_{1}\left(x_{1},y_{1}\right)}{\frac {\mathrm {d} s(y)}{v(y)}}}$ ${\displaystyle \mathrm {d} s={\sqrt {1+\Phi '(y)^{2}}}\mathrm {d} y}$ ${\displaystyle s=\int _{y_{0}}^{y_{1}}{\sqrt {1+\Phi '^{2}}}\mathrm {d} y}$

It follows from the law of energy . Inserting into the equation for T gives ${\displaystyle mg(y_{0}-y)={\frac {m}{2}}v(y)^{2}}$ ${\displaystyle v(y)={\sqrt {2g\left(y_{0}-y\right)}}}$

${\displaystyle T_{y_{0}}\left(y_{1}\right)=\int _{y_{0}}^{y_{1}}{\sqrt {\frac {1+\Phi '(y)^{2}}{2g(y_{0}-y)}}}\mathrm {d} y}$

If you now define (and consider that it is), then this results ${\displaystyle f(y)={\sqrt {\frac {\pi }{2g}}}{\sqrt {1+\Phi '(y)^{2}}}}$ ${\displaystyle {\sqrt {\pi }}=\Gamma \left({\frac {1}{2}}\right)}$

${\displaystyle T_{y_{0}}\left(y_{1}\right)={\frac {1}{\Gamma \left({\frac {1}{2}}\right)}}\int _{y_{0}}^{y_{1}}{\frac {f(y)}{\left(y-y_{0}\right)^{\frac {1}{2}}}}\mathrm {d} y=\left(I_{y_{0}+}^{\frac {1}{2}}f\right)\left(y_{1}\right)}$

literature

• Richard Herrmann: Fractional calculus. An introduction for physicists. BoD, Norderstedt 2014, ISBN 978-3-7357-4109-7 .
• Richard Herrmann: Fractional Calculus - An Introduction for Physicists. World Scientific, Singapore 2014, ISBN 978-981-4551-07-6 .
• F. Mainardi: Fractional Calculus and Waves in Linear Viscoelasticity: An Introduction to Mathematical Models. Imperial College Press, 2010, ISBN 978-1-84816-329-4 .
• VE Tarasov: Fractional Dynamics: Applications of Fractional Calculus to Dynamics of Particles, Fields and Media. Springer, 2010, ISBN 978-3-642-14002-0 .
• VV Uchaikin: Fractional Derivatives for Physicists and Engineers. Springer, Higher Education Press, 2012, ISBN 978-3-642-33910-3 .

Web links

Mathematical journals

• Fractional Calculus and Applied Analysis
• Fractional Differential Equations (FDE)
• Journal of Fractional Calculus and Applications (JFCA)