Fourier transform of an inverse power of one plus x squared ($\mathcal{F}[(1+x^2{)}^{-\alpha /2}]=h_{\alpha }$)

For $\alpha >0$, put $a=\alpha /2$. The regular distribution of $(1+x^2{)}^{-a}$ is a tempered distribution, since the function is bounded. Its Fourier transform is the integrable functionwith arbitrary finite value at zero when the displayed integral diverges there. To prove the identity, the gamma function gives $(1+x^2{)}^{-a}=\Gamma (a{)}^{-1}{\int }_0^{\infty }{t}^{a-1}{e}^{-t}{e}^{-t{x}^2}dt$. Pair this with a Schwartz function, use the Fourier transform of a Gaussian, and interchange integrals by Fubini's theorem. The absolute bound for the transformed pairing follows from Tonelli's theorem:The positive kernel and dominated convergence give positivity and continuity away from zero. It has a continuous representative at zero exactly when $\alpha >1$, in which case $h_{\alpha }(0)=\sqrt{\pi }\Gamma ((\alpha -1)/2)/\Gamma (\alpha /2)$. For $\alpha \le 1$, Fatou's lemma gives $h_{\alpha }(\xi )\to \infty$ as $\xi \to 0$, so no change on a set of zero Lebesgue measure can give continuity.