Fractional Order Derivative of the Exponential Function

Question

I'm new to these concepts. Just curious from an engineering point of view. I'm using this definition of the fractional order derivative:

$D^{\alpha}f(x)=\frac{1}{\Gamma(1-\alpha)}\frac{d}{dx}\int_0^x \frac{f(t)}{(x-t)^\alpha}dt$

I know that $D^0 e^x=e^x$ and that $D^1 e^x=e^x$ so my very little understanding makes me believe that any $0<\alpha<1$ should give me an exponential... if fractional-order operators provide smooth operators that are somewhat in between. Yet the fractional-order derivative seams to yield a different function, with a vertical asymptote in $x=0$.

I implemented the fractional order operator with the following Mathematica code:

    FractionalD[nu_, f_, t_, opts___] := 
Module[{x},g=(f /. t -> x);
  Integrate[(t - x)^(-nu - 1)g , {x, 0, t}, opts]/Gamma[-nu]]

FractionalD[mu_?Positive, f_, t_, opts___] :=
  Module[{m = Ceiling[mu]},
    D[FractionalD[-(m - mu), f, t, opts], {t, m}]
  ]

Below you may find in purple $e^x$, $D^\alpha e^x$ in blue with $\alpha=2/90$, in orange with $\alpha = 2/9$, in green with $\alpha = 8/9$ and in red with $\alpha = 89/90$.

It appears that with $\alpha ->1$ the fractional-order derivative finds it hard to get rid of that asymptote and restore its appearance as a first-order derivative.

Is my understanding of the meaning of a fractional order derivative wrong?

Answer 1

Indeed your definition of fractional derivative/integral does not fix the exponential function, but has the effect $D^\alpha x^\beta=\frac{\Gamma(\beta+1)}{\Gamma(-\alpha+\beta+1)}x^{\beta-\alpha}$. In particular, fractional derivative of the constant function is $[D^\alpha 1](x)=x^{-\alpha}/\Gamma(1-\alpha)$, with property the factor $1/\Gamma(1-\alpha)\to 0$ in the limit $\alpha\uparrow 1$ to recover the pointwise $D1=0$, which is what you observe: the fractional integral/derivative of the exponential function is $$ [D^{\nu}\exp](x)=\sum_{k=0}^\infty\frac{x^{k-\nu}}{\Gamma(1+k-\nu)} $$ for all $\nu\notin\{1,2,3,\dots\}$ and otherwise take the limiting value over other $\nu$s (equivalently remove the singular part).

Remark: There are other versions of fractional calculus that try to interpolate $D\exp(ax)=a\exp(ax)$, e.g., Weyl integral.