Half derivative of $e_{\alpha} = \frac{x^{\alpha - 1}}{\Gamma(\alpha)}$ Let $e_{\alpha} = \frac{x^{\alpha - 1}}{\Gamma(\alpha)}$. We assume that $\alpha > \frac{1}{2}$. I know that the half derivative of a function can be defined as $$ D_{\frac{1}{2}}f(x) = \frac{d}{dx} \int_0^x \frac{f(y)}{\sqrt{x - y}}dy$$
Substituting the desired function into this definition gives
$$ D_{\frac{1}{2}}e_{\alpha} = \frac{1}{\Gamma(\alpha)}\frac{d}{dx} \int_0^x \frac{y^{\alpha - 1}}{\sqrt{x - y}}dy$$
To evaluate this, I know I can treat $x$ as a constant, so I was thinking of using a substitution like $k = x - y$, or something similar, but I end up with a huge mess in the integrand. 
This is supposed to be a stepping stone to computing the half derivative of $Sin(x)$.
 A: Hint: Substitute $u=y/x$ and express the result in terms of Beta function
$$\int_0^x \frac{y^{\alpha-1}}{\sqrt{x - y}}dy=\int_0^1 \frac{(ux)^{\alpha-1}}{\sqrt{x - ux}}xdu=x^{\alpha-1/2}\int_0^1 u^{\alpha-1}(1 - u)^{-1/2}du=x^{\alpha-1/2}B(\alpha,1/2)$$
Furthermore
$$B(\alpha,1/2)=\frac{\Gamma(\alpha)\Gamma(1/2)}{\Gamma(\alpha+1/2)}=\frac{\sqrt\pi\ \Gamma(\alpha)}{\Gamma(\alpha+1/2)}$$
So your half derivative is:
$$\frac{\sqrt\pi}{\Gamma(\alpha-\frac{1}{2})}x^{\alpha-\frac{3}{2}}$$
A: Through the Laplace transform:
$$ D^{1/2} f(x) = \mathcal{L}^{-1}\left(\sqrt{s}\cdot(\mathcal{L}f)(s)\right)(x)\tag{1} $$
so, if $f(x)=\frac{x^{a-1}}{\Gamma(a)}$, from $(\mathcal{L}f)(s) = s^{-a}$ we get:
$$ D^{1/2} f(x) = \mathcal{L}^{-1}\left(s^{1/2-a}\right)(x)=\frac{x^{a-3/2}}{\Gamma(a-1/2)}\tag{2} $$
and by assuming linearity:
$$\begin{eqnarray*}D^{1/2} \sin(x) &=& D^{1/2}\sum_{n\geq 1}\frac{(-1)^{n+1} x^{2n-1}}{\Gamma(2n)}=\sum_{n\geq 1}\frac{(-1)^{n+1}x^{2n-3/2}}{\Gamma(2n-1/2)}\\&=&2\sqrt{\frac{x}{\pi}}\;\phantom{}_1 F_2\left(1;\frac{3}{4},\frac{5}{4};-\frac{x^2}{4}\right).\tag{3} \end{eqnarray*}$$
That can be re-written as:
$$\begin{eqnarray*} D^{1/2}\sin(x) &=& \frac{4\sqrt{x}}{\pi}\sum_{n\geq 0}\frac{(-1)^n x^{2n}}{(2n)!}\int_{0}^{\pi/2}\cos^{4n+1}(\theta)\,d\theta \\ &=&\frac{4\sqrt{x}}{\pi}\int_{0}^{\pi/2}\cos(\theta)\cos(x\cos^2\theta)\,d\theta\\&=&\frac{4\sqrt{x}}{\pi}\int_{0}^{+\infty}\cos\left(\frac{x}{1+t^2}\right)\frac{dt}{(1+t^2)^{3/2}}\\&=&\frac{4\sqrt{x}}{\pi}\int_{0}^{1}\frac{t \cos(xt^2)}{\sqrt{1-t^2}}\,dt\\&=&\frac{2\sqrt{x}}{\pi}\int_{0}^{1}\frac{\cos(xt)}{\sqrt{1-t}}\,dt\tag{4}\end{eqnarray*}$$
or as:

$$ D^{1/2}\sin(x) = \sqrt{\frac{8}{\pi}}\left[\sin(x)\,S\!\left(\sqrt{\frac{2x}{\pi}}\right)+\cos(x)\,C\!\left(\sqrt{\frac{2x}{\pi}}\right)\right]\tag{5} $$

where $C$ and $S$ are Fresnel integrals.
