How do you take the fractional derivative $J^\frac{1}{2}f(x)$ where $f(x)=w\sin(x)$, and where $w$ is a constant? What is $y$ in $$J^\frac{1}{2}f(x)=y$$
$$f(x)=w\sin(x)$$ 
where $w$ is a constant?
 A: Allow $\frac{d^n}{dx^n}e^{cx}=c^ne^{cx}$.  This should be fairly obvious and holds by induction.
Now let $c=i$ and we can get Euler's formula to do all the work.
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
$$\frac{d^n}{dx^n}\sin(x)=\frac{d^n}{dx^n}\frac{e^{ix}-e^{-ix}}{2i}$$
$$=\frac{i^ne^{ix}-(-i)^ne^{-ix}}{2i}$$
$$=\frac{e^{i(x+\frac{n\pi}2)}-e^{-i(x+\frac{n\pi}2)}}{2i}$$
$$=\sin(x+\frac{n\pi}2)$$
$$\frac{d^n}{dx^n}w\sin(x)=w\frac{d^n}{dx^n}\sin(x)=w\sin(x+\frac{n\pi}2)$$
For $n=1/2$, you get $w\sin(x+\frac\pi4)$.
A: Hint : 
The fractional derivative of the sinusoidal functions :

Copy from page 46, Eqs. (14-4) and (14-5) in the technical report "Use of ractional derivatives to express the properties of energy storage in electrical networks"(1982), no longer available but cited on page 13, ref.[9] in : https://fr.scribd.com/doc/71923015/The-Phasance-Concept
The special functions  $S$  and $C$  are the "Generalized Fresnel Integrals", from : M.Abramowitz and I.A.Stegun, Handbook of Mathematical Functions, p.262, Dover Publ. N-Y. 9th print (1970).
In the case of the present question concerning fractional integration, $\nu=-\frac{1}{2}$  and  $\omega=1$ 
$$\frac{d^{-1/2}}{dt^{-1/2}}\sin(t)=\sin(t-\frac{\pi}{4}) -\frac{1}{\sqrt{\pi}}\left( \sin(t)C(t,\frac{1}{2}) -\cos(t)S(t,\frac{1}{2})\right)$$ 
because $\Gamma(\frac{1}{2})=\sqrt{\pi}$
Note that, for large  $t$  the generalized Fresnel integrals tend to  $0$  and in steady state  $\frac{d^{-1/2}}{dt^{-1/2}}\sin(t)\simeq\sin(t-\frac{\pi}{4})$
In the case of partial integration of order  $\frac{1}{2}$  that is to say   $\nu=-\frac{1}{2}$  a more direct approach is :
$$\frac{1}{\Gamma(\frac{1}{2})}  \int_0^t \frac{\sin(\omega \tau)} {(t-\tau)^{\frac{1}{2}} }d\tau = \sqrt{\frac{2}{\omega}}\left( \sin(\omega t)C\left( \sqrt{\frac{2\omega t}{\pi}} \right) - \cos(\omega t)S\left( \sqrt{\frac{2\omega t}{\pi}} \right) \right)$$
This is the same as above, thanks to the relationships between the Fresnel integrals and the generalized Fresnel integrals :
$C(x,\frac{1}{2})=\sqrt{2\pi}\left( \frac{1}{2}-C\left( \sqrt{\frac{2x}{\pi}} \right)   \right)$  and   $S(x,\frac{1}{2})=\sqrt{2\pi}\left( \frac{1}{2}-S\left( \sqrt{\frac{2x}{\pi}} \right) \right)$  
In the field of electrotechnic, the particular case  $\nu=\frac{1}{2}$  is related to the so-called Warbourg impedance and the related fractional derivative is :
$$\frac{d^{1/2}}{dt^{1/2}}\sin(\omega t)=\sqrt{\omega} \left(\sin(\omega t+\frac{\pi}{4}) +\frac{1}{2\sqrt{\pi}}\left( \sin(t)C(\omega t,-\frac{1}{2}) -\cos(t)S(\omega t,-\frac{1}{2})\right)\right)$$
In steady state :  $\frac{d^{1/2}}{dt^{1/2}}\sin(\omega t)\simeq \sqrt{\omega}\sin(\omega t+\frac{\pi}{4})$ 
