What is the Fourier transform of the bump function $e^{-frac{1}{1-|x|^2}}$?

Mathematics Asked by Medo on January 5, 2022

e^{-frac{1}{1-|x|^2}}, & hbox{$|x|<1$;} \
0, & hbox{$|x|geq1$.}

This is a generic bump function (a smooth positive bounded function with compact support). It is a Schwartz function, so it has a Schwartz Fourier transform.

My question is about calculating its Fourier transform $hat{f}$:

Since $f$ is radial (i.e. rotationally invariant), then so is $hat{f}$. So $hat{f}(xi)=hat{f}(|xi|,0,…,0)$ for all $xi in mathbb{R}^{n}$. Therefore, denoting $x^prime=(x_2,x_3,…,x_n)$, we have

$$hat{f}(xi)=int_{|x|leq 1}e^{dot{imath}x_{1}|xi|}e^{-frac{1}{1-|x|^2}}dx=int_{|x|leq 1}e^{dot{imath}x_{1}|xi|}e^{-frac{1}{1-|x|^2}}dx\= int_{-1}^{1}e^{dot{imath}x_{1}|xi|}int_{|x^prime|leq sqrt{1-x_1^2}}e^{-frac{1}{1-x_1^2-|x^prime|^2}}dx^prime dx_1\= int_{-1}^{1}e^{dot{imath}x_{1}|xi|}int_{mathbb{S}^{n-2}}int_{0}^{sqrt{1-x_1^2}}e^{-frac{1}{1-x_1^2-rho^2}}rho^{n-2}drho domega_{n-2} dx_1\
=|mathbb{S}^{n-2}|int_{-1}^{1}e^{dot{imath}x_{1}|xi|}int_{0}^{sqrt{1-x_1^2}}e^{-frac{1}{1-x_1^2-rho^2}}rho^{n-2}drho dx_1$$

And I am stuck here!

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