Inverse Fourier of $\omega^{-1+{\rm i}\alpha} u(\omega-1)$ Let $\alpha$ be an arbitrary real number and define
\begin{align}
\widehat{f}(\omega)=\left\{\begin{array}{ll}
\omega^{-1+{\rm i}\alpha}, & \omega>1,\\
0, & \textrm{otherwise}.
\end{array}
\right.
\end{align}
Are there any smoothness or boundedness results available for the inverse Fourier of $\widehat{f}$?
 A: Regarding smoothness: $f$ is smooth everywhere outside of 0, the singularity at zero can also be described. 
More precisely $f$ is the inverse Fourier transform of $(x_+)^{-1+i\alpha}-1_{[-1,1]}\cdot (x_+)^{-1+i\alpha}$, where $1_{[-1,1]}$ is the indicator function of the interval $[-1,1]$.
Notice that both expressions are well defined distributions. The second summand is a compactly supported distribution, hence its inverse Fourier transform is infinitely smooth. The inverse Fourier transform of the first summand is well known (see "Generalized functions" by Gelfand-Shilov). It has the form
$$|s|^{-i\alpha}(a+b\cdot sgn(s)),$$
where $a,b$ are non-zero constants which can be computed explicitly.
Thus $f(s)$ is a sum of the above expression and an infinitely smooth function.
A: Mathematica gives the following for the inverse Fourier transform:
$$
\frac{\left| s\right| ^{-1-i \alpha } \left(\left| s\right| ^{1+i \alpha } \left(\alpha 
   s \, _1F_2\left(\frac{i \alpha }{2}+\frac{1}{2};\frac{3}{2},\frac{i \alpha
   }{2}+\frac{3}{2};-\frac{s^2}{4}\right)+(1+i \alpha ) \, _1F_2\left(\frac{i \alpha
   }{2};\frac{1}{2},\frac{i \alpha }{2}+1;-\frac{s^2}{4}\right)\right)+\alpha  (\alpha
   -i) \left| s\right|  \cosh \left(\frac{\pi  \alpha }{2}\right) \Gamma (i \alpha
   )+\alpha  (\alpha -i) s \sinh \left(\frac{\pi  \alpha }{2}\right) \Gamma (i \alpha
   )\right)}{\sqrt{2 \pi } \alpha  (\alpha -i)}
$$
Presumably, at this point Abramovitz-Stegun can answer every possible question.
