Fourier transform of $x \in \mathbb{R} \mapsto (1+x^2)^{-\alpha}$

Question

I hope that the answer is not somewhere in the forum. I searched without success. I would like to compute the fourier transform of $f:x \in \mathbb{R} \mapsto (1+x^2)^{-\alpha}$ where $\alpha > 1/2$. Or, if there is no closed form, i would like to obtain approximation of $\mathcal{F}[f](k)$ when $k \to 0$. I know how to deal with the case $\alpha = 1$ using complex integration and inverse fourier formula. But I struggle to generalize this method. Do you have any hint or suggestion of approach? Thanks,

Answer 1

These are called Bessel potentials https://en.wikipedia.org/wiki/Bessel_potential. You can already find several formulas and the asymptotic behavior on the Wikipedia page. You can find some other properties in Stein Singular integrals and differentiability properties of functions (1970), Chapter V, Section 3. In particular, a formula I like to get properties of these functions, is the following integral representation $$ B_s(x) = \frac{\omega_s}{2} \int_0^\infty t^{\frac{d-s}{2}-1}\, e^{-\pi/t}\, e^{-\pi |x|^2 t}\,\mathrm d t $$ where $B_s(x) = \mathcal F((1+|x|^2)^{-s/2})$ and $\omega_s = \frac{2\,\pi^{s/2}}{\Gamma(s/2)}$. To prove it, you can proceed as I did here in the case of $s = d+1$.

At $x=0$, it is convergent if $s>d$, and then $B_s(0) = \frac{\omega_s}{2} \int_0^\infty t^{\frac{d-s}{2}-1}\, e^{-\pi/t}\,\mathrm d t = \frac{\omega_s}{\omega_{s-d}}$.

If $s<d$, it behaves in the same way as $\mathcal F(|x|^{-s}) = \frac{\omega_s}{\omega_{d-s}\,|x|^{d-s}}$.