Inverse Fourier of $\omega^{-1+{\rm i}\alpha} u(\omega-1)$

Question

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 $\widehat{f}$?

Answer 1

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.

Answer 2

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.