Taking the half-derivative of $e^x$

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.

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.

$${1 \over {\Gamma(1/2)}} \cdot {{d} \over {dx}} \int_0^x {{e^t} \over {\sqrt {x-t}}} \ dt$$ 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 $$\operatorname{erf}(x)={2 \over {\sqrt{\pi}}} \cdot \int_0^x e^{-t^2} \ dt$$