Fourier-transformation

As

$$R_{q}(t)={\bf I}^{q+1}R_{-1}(t)={\bf I}^{q+1}\delta(t)$$

then  

$${\bf F}_{t}R_{q}(t)= ({1\over -i\omega})^{q+1}\cdot 1 =i\vert\omega \vert^{-q-1} \exp (i{q\pi \over 2}{\rm sgn} (\omega )).$$ (15)