# Fractional Order Derivative of the Exponential Function

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?

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.