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?
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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 $\alpha$.
answered Apr 3, 2012 at 13:46
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3$\begingroup$ What happens with the $\textrm{sin}\pi (\lambda+1/2)$ for $\lambda=3/2$? $\endgroup$ Commented Apr 4, 2012 at 0:22
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2$\begingroup$ I thought nobody was going to ask this embarrassing questions. Here are two possible answers. You either let $\lambda\to\frac{3}{2}$ in the above formula or, better yet, consult the above reference where there is an alternate description involving the Bessel functions $N_\alpha$.. I did not include that formula since I would have had to explain what $N_\alpha$ is. $\endgroup$ Commented Apr 4, 2012 at 9:07
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1$\begingroup$ Liviu, thank you so much for your response. I believe $N_\alpha $ is the same as $Y_\alpha $, i.e. the Bessel function of the second kind, which is what I got by taking the limit. And thanks Robert for bringing up the discussion of the limiting cases. $\endgroup$– flavioCommented Apr 4, 2012 at 10:01
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1$\begingroup$ You are absolutely right. The label $N_\alpha$ is an old fashion one that survived mainly behind the Iron Curtain and it reflects the fact that $Y_\alpha=N_\alpha$ is sometime the Neumann function. $\endgroup$ Commented Apr 4, 2012 at 11:51