= Fourier transform of an inverse power of one plus x squared
{c}
{title2=$\mathcal F[(1+x^2)^{-\alpha/2}]=h_\alpha$}
= Gaussian mixture formula for an inverse-power Fourier transform
{c}
{synonym}
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 function>
$$
h_\alpha(\xi)=\frac{\sqrt\pi}{\Gamma(a)}\int_0^\infty t^{a-3/2}e^{-t-\xi^2/(4t)}\,dt\qquad(\xi\ne0),
$$
with arbitrary finite value at zero when the displayed <integral> diverges there. To prove the identity, the <gamma function> gives $(1+x^2)^{-a}=\Gamma(a)^{-1}\int_0^\infty t^{a-1}e^{-t}e^{-tx^2}\,dt$. 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>:
$$
\|h_\alpha\|_1=\frac{\sqrt\pi}{\Gamma(a)}\int_0^\infty t^{a-3/2}e^{-t}\,2\sqrt{\pi t}\,dt=2\pi.
$$
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 $\alpha\leq1$, <Fatou's lemma> gives $h_\alpha(\xi)\to\infty$ as $\xi\to0$, so no change on a set of zero <Lebesgue measure> can give continuity.
Back to article page
