I find two main sources on how to compute the half-derivative of $e^x$. Both make sense to me, but they give different answers.
Firstly, people argue, that $$\begin{align} \frac{\mathrm{d}}{\mathrm{d} x} e^{k x} &= k e^{k x} \\[4pt] \frac{\mathrm{d}^2}{\mathrm{d} x^2} e^{k x} &= k^2 e^{k x} \\[4pt] \frac{\mathrm{d}^n}{\mathrm{d} x^n} e^{k x} &= k^n e^{k x} \end{align}$$
Therefore, it seems very reasonable, that $$\frac{\mathrm{d}^{1/2}}{\mathrm{d} x^{1/2}} e^{k x} = \sqrt{k} e^{k x}$$
But this is not what the usual formula gives $$ \frac{\mathrm{d}^{1/2}}{\mathrm{d} x^{1/2}} e^{k x} = \frac{1}{\Gamma (1/2)} \frac{\mathrm{d}}{\mathrm{d} x} \int \limits_0^x \mathrm{d} t \frac{e^{k t}}{\sqrt{x-t}} = \frac{1}{\Gamma (1/2)} \frac{\mathrm{d}}{\mathrm{d} x} e^{k x} \int \limits_0^x \mathrm{d} u \frac{e^{- k u}}{\sqrt{u}} = \\ = \frac{1}{\Gamma (1/2)} \frac{\mathrm{d}}{\mathrm{d} x} \frac{e^{k x}}{\sqrt{k}} \int \limits_0^{\sqrt{k x}} \mathrm{d} s \, e^{-s^2} $$
We can already see that this is not equal to $\sqrt{k} e^{k x}$.
So who is right?
Why do we use the latter formula in almost all cases but somehow we settle for the simpler formula when it comes to the exponential?
Note that both satisfy that if we apply the half-derivative twice, we get the usual first derivative. (In the first case, it's simply because $\sqrt{k} \sqrt{k} = k$; in the second case, there's a proof on Wiki using the properties of beta and gamma functions - plus I verified this numerically even though I couldn't express the integrals in a closed-form.)
I also have hard time accepting this second, complicated, formula, mainly because any integer derivative of the exponential gives exponential, but for the half-derivative we get this weird monstrosity. On the other hand, it should be consistent with the formulas for the half-derivative for all powers of $x$ when it's put together in the infinite series for $e^{k x}$.
Can anyone shed some light on this issue for me, please?
Both are right. Think of it like this: what is "the" square root of $i$? Even for $1$ there are two of them, but we forget because of the habit to take the positive root. For $i$ there is no habit available. Things get even more plural when we ask for square roots of $I$, the $n\times n$ identity matrix. There are $2^n$ square roots just among diagonal matrices, and if $A$ is a square root so is $SAS^{-1}$ for any invertible matrix $S$. So we should not expect that square root of $D=\frac{d}{d x}$ is a single something either.
A better analogy here is not derivative but antiderivative, which is, after all $D^{-1}$. There is no "the" antiderivative, there is always a family of them. Again, we tend to overlook that, because they only differ by constants of integration. And we could force uniqueness by declaring the definite integral $\int_0^xf(x)\,dx$ to be "the" antiderivative of $f(x)$. But if we do "the" antiderivative of $e^x$ will be not our favorite $e^x$, but $e^x-1$. To get $e^x$, we should choose $\int_{-\infty}^xf(x)\,dx$ to be "the" antiderivative instead, but then this choice won't work for functions that do not quickly vanish at $-\infty$. There just is no perfect choice. Just as antiderivatives need a reference value to become definitive ($x_0=0$ or $-\infty$ above) so do fractional derivatives more generally. The case of the derivative, and its integer powers, is an exception rather than a rule.
For half-derivatives $D^{\frac12}$ the integral is a bit more involved, in one of the forms
$$D^{\frac12}f(x)=\frac{1}{\Gamma\left(\frac{1}{2}\right)}\frac{d}{dx}\int \frac{f(u)}{(x-u)^\frac12}du,$$
I left out the limits of integration. As shown in Fractional Calculus on Mathpages, if we choose the limits from $0$ to $x$ the answer is $e^x\left(\operatorname{erf}\sqrt{x} + \frac{e^{-x}}{\sqrt{\pi x}}\right)$. On the other hand, with the already familiar choice from $-\infty$ to $x$ the answer is the desired $e^x$. Moreover, then $D^\nu(e^{kx})=k^\nu e^{kx}$ for any $k$ and $\nu$. This is easier to check by using the inverted formula:
$$D^{-\frac12}f(x)=\frac{1}{\Gamma\left(\frac{1}{2}\right)}\int_{-\infty}^x \frac{f(u)}{(x-u)^\frac12}du.$$
The difference in the answers is more visible now, because it is not by an additive constant of integration, but the principle is the same - we get a whole $1$-parameter family of half-derivatives.