Taking the half-derivative of $e^x$ 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.
 A: 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:
$$\int_0^x -\frac{e^{x-u}}{\sqrt{u}} du$$
$$-e^x\int_0^x \frac{e^{-u}}{\sqrt{u}} du$$
$$-e^x\int_0^x u^{-1/2}e^{-u} du$$
$$-e^x\gamma\left(\frac{1}{2},x\right)$$
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. 
A: Let we assume that $x>0$. Since:
$$ e^x = \sum_{n\geq 0}\frac{x^n}{n!}\tag{1} $$
and:
$$ D^{1/2} x^{m} = \frac{x^{m-1/2}\,\Gamma\!\left(m+1\right)}{\Gamma\!\left(m+\frac{1}{2}\right)}\tag{2}$$
(look here, for instance) we have:
$$ D^{1/2} e^x = \frac{1}{\sqrt{x}}\sum_{n\geq 0}\frac{x^n}{\Gamma\!\left(n+\frac{1}{2}\right)}=\frac{1+e^x\sqrt{\pi x}\;\text{Erf}(\sqrt{x})}{\sqrt{\pi x}}\tag{3}$$
where $\text{Erf}$ is the usual error function.
A: $${1 \over {\Gamma(1/2)}} \cdot {{d} \over {dx}} \int_0^x {{e^t} \over {\sqrt {x-t}}} \ dt$$
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
$$\operatorname{erf}(x)={2 \over {\sqrt{\pi}}} \cdot \int_0^x e^{-t^2} \ dt$$
A: If this is really the integral you want, then notice you are integrating with respect to $t$ so treat $x$ as a constant. Then $$\int \frac{e^x}{x-t} dt = e^x\int\frac{1}{x-t} dt$$ 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.
A: 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:
$$f^{(s)}(x)=\sum_{m=0}^{\infty} \binom {s}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{(k)}(x)$$
the other one is based on Forier transform:
$$f^{(s)}(x)=\frac{i}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}{\omega}^s \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$
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.
