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While attempting to teach myself the fractional calculus, I encountered a tragically early roadblock. For non-power rule fractional derivatives, I am having a lot of trouble evaluating for a closed form.

Would someone mind walking me through the process for taking the half-derivative of $$f(x) = e^x$$

Really the most difficult part is evaluating

$$int_0^x frac{e^t}{sqrt{x-t}} dt$$

but a full hand-holding would be really helpful.

For the integral: Keep in mind that $x$ is a constant!

$$int_0^x frac{e^t}{sqrt{x-t}} dt$$

Use the substitution $u=x-t$, then $du=-dt$. This gives:

$$int_0^x -frac{e^{x-u}}{sqrt{u}} du$$

$$-e^xint_0^x frac{e^{-u}}{sqrt{u}} du$$

$$-e^xint_0^x u^{-1/2}e^{-u} du$$

$$-e^xgammaleft(frac{1}{2},xright)$$

Where $gamma$ is the incomplete lower gamma function.

This can also be written as
$$-e^x sqrt{x} E_{frac{1}{2}}(x)$$

using the exponential integral function. It has been proven there is no closed form of this function.

Let we assume that $x>0$. Since:
$$ e^x = sum_{ngeq 0}frac{x^n}{n!}tag{1} $$
and:
$$ D^{1/2} x^{m} = frac{x^{m-1/2},Gamma!left(m+1right)}{Gamma!left(m+frac{1}{2}right)}tag{2}$$
(look here, for instance) we have:
$$ D^{1/2} e^x = frac{1}{sqrt{x}}sum_{ngeq 0}frac{x^n}{Gamma!left(n+frac{1}{2}right)}=frac{1+e^xsqrt{pi x};text{Erf}(sqrt{x})}{sqrt{pi x}}tag{3}$$
where $text{Erf}$ is the usual error function.

$${1 over {Gamma(1/2)}} cdot {{d} over {dx}} int_0^x {{e^t} over {sqrt {x-t}}} dt$$
Where $Gamma(x)$ is the generalized factorial function. This equals
$${1 over {Gamma(1/2)}} cdot {{d} over {dx}} int_0^x {{e^t} over {sqrt {x-t}}} dt=e^x cdot operatorname{erf}(sqrt{x})$$
where $operatorname{erf}(u)$ is the error function. This is more of a definition than a technical thing, so you don't really need to prove the above per se. However the substitution $u=sqrt{x-t}$ and integration by parts brings the above into compliance with
$$operatorname{erf}(x)={2 over {sqrt{pi}}} cdot int_0^x e^{-t^2} dt$$

If this is really the integral you want, then notice you are integrating with respect to $t$ so treat $x$ as a constant. Then $$int frac{e^x}{x-t} dt = e^xintfrac{1}{x-t} dt$$ and the antiderivative of $frac{1}{x-t}$ with respect to $t$ is easily seen to be $-ln(x-t)+C$. Now evaluate the integral as you normally would with a definite integral.

This depends on a method and definition used, but all the other answers give unnatural expressions in my view.

The following definitions give a more natural answer and coincide with each other.

The first one is based on Newton series interpolation over consecutive integer derivatives:

$$f^{(s)}(x)=sum_{m=0}^{infty} binom {s}m sum_{k=0}^mbinom mk(-1)^{m-k}f^{(k)}(x)$$

the other one is based on Forier transform:

$$f^{(s)}(x)=frac{i}{2pi}int_{-infty}^{+infty} e^{- i omega x}{omega}^s int_{-infty}^{+infty}f(t)e^{iomega t}dt , domega$$

For $f(x)=e^x$ they both give the same result: the derivative of any order of this function is $e^x$. In other words, this function is invariant regarding differentiation and integration of any order.