TITLE: Fourier transform
QUESTION [8 upvotes]: Does anyone know what the Fourier transform (in the sense of distributions) of 
$$
f(x) = (x^2 - 1)^{1/2}x, \quad |x|\ge 1, 
$$
and $f(x) = 0$ otherwise, is?

REPLY [17 votes]: First of all observe that
$$ f(x)=\frac{1}{3}\frac{d}{dx} (x^2-1)^{\frac{3}{2}}_+, $$
where  for any real number $t$ we set $t_+=\max(t,0)$.   Thus it suffices  to compute the Fourier  transform of $(x^2-1)^{\frac{3}{2}}_+$.
In Section 2.5 Chapter 2  of   the book by Gelfand and Shilov, Generalized Functions, vol.1, Academic Press 1964,  the authors  compute the Fourier transform of $(ax^2+bx+c)^\lambda_+$. Your example corresponds to Case (3) discussed there.  More precisely  the Fourier transform of $(x^2-1)^\lambda_+$ is the function
$$\Gamma(\lambda+1)\sqrt{\pi}\left|\frac{\xi}{2}\right|^{-\lambda-\frac{1}{2}}\frac{\cos\pi(\lambda+\frac{1}{2}) J_{-\lambda-\frac{1}{2}}(|\xi|)-J_{\lambda+\frac{1}{2}}(|\xi|)}{\sin \pi(\lambda+\frac{1}{2})}, $$
where $J_\alpha$ denotes the Bessel function  of order $\alpha$.