Consider the fourier transform of $e^{it|\xi|^{2\alpha}}$ ($\alpha>0$)in $\mathbb{R}^n$,let $K_{\alpha}=\mathcal{F}(e^{it|\xi|^{\alpha}})$,so $K$ is a tempered distribution.Now I want to know if there is a explicit expression of $K$,for the simpliest case,namely $\alpha=1$,it's well known that $$ K_1=(4\pi it)^{-\frac{n}{2}}e^{-\frac{|x|^2}{4it}} $$ Another special case is $\alpha=\frac{1}{2}$,since we know that $\mathcal{F}e^{-t|\xi|}=C_{n}\frac{t}{(t^{2}+|\xi|^2)^{\frac{n+1}{2}}}$,where $t>0$,let $t=-it$,so at least formally, $$ K_{\frac{1}{2}}=C_{n}\frac{-it}{(|\xi|^2-t^{2})^{\frac{n+1}{2}} }$$

My question is how about general $\alpha$ ?,so far I have known that when $0<\alpha<\frac{1}{2}$,$\alpha=\frac{1}{2}$,$\alpha>\frac{1}{2}$,the singularity of $K$ lies at $0$,$t=|x|$,$\infty$ respectively.