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I'd like to find the $n$-dimensional inverse Fourier transform of $\frac{1}{\| \mathbf{\omega} \|^{2\alpha}}$ i.e. $$ \int_{\mathbb{R}^n} \frac{1}{ \| \mathbf{\omega} \|^{2\alpha}} e^{2 \pi i \mathbf{\omega}\cdot \mathbf{x} } d \mathbf{\omega} $$ where $\mathbf{x} = ( x_0 , x_1 , \cdots , x_n )$ is a spatial parameter in $\mathbb{R}^n$, $\mathbf{\omega} = ( \omega_0 , \omega_1 , \cdots , \omega_n )$, and $$ \| \omega\| = \omega_0^2 + \omega_1^2 + \cdots + \omega_n^2 $$ All I've been able to come up with in the one-dimensional case is that the integral $$ \int_{-\infty}^{+\infty} \frac{1}{ \| \omega \|^{2\alpha}} e^{2 \pi i \omega x } d \mathbf{\omega} $$ diverges because the lower power terms $\omega^p$ terms, for which $p < 2\alpha$, in expansion of the exponential $$ e^{2 \pi i \omega x } = \sum_{p = 0}^{\infty} \frac{(2 \pi i \omega x)^p}{p!} $$ do not prevent $\frac{1}{\| \omega \|^{2\alpha}}$ from blowing up at the origin.

I know that one possible way of regularizing this integral is to include a test function and consider the limit of the resulting integral, but I don't quite know how to do so. I've tried reading Gelfand and Shilov's Gneralized Functions vol 1 and while I understand bits of it on the whole its a bit heavy for me.


Based on the papers that I've read I know that there are two cases (the latter of which appears to me more general) and two solutions in each.

  • Case 1: 2$\alpha$ is an odd/even integer
  • Case 2: 2$\alpha$ is integer or otherwise

I'd appreciate help, if possible, coming up with both solutions.

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Are you sure you want $\|\omega \| = \sum \omega_i^2$? Are do you want $\|\omega\|^2 = \sum \omega+i^2$? –  Willie Wong Jul 2 '11 at 12:26
    
Quite a bit of this is also discussed at a more accessible/practical level (compared to Gelfand and Shilov) in M. Taylor's Partial Differential Equations, volume 1, Section 3.8. (In my copy pages 239 - 245). –  Willie Wong Jul 2 '11 at 12:51

1 Answer 1

I won't include all details since you don't seem to want that...

Up to minus signs and constants your problem is the same as trying to find the Fourier transform of $f(\omega) = \frac{1}{\|\mathbf{\omega}\|^{2\alpha}}$. This distribution is radial and homogeneous of degree $-2\alpha$. You can use the scaling properties of the Fourier transform to show that this means that its Fourier transform is radial and homogeneous of degree $-n + 2\alpha$. Then you can show that this only happens when this Fourier transform is of the form $c_\alpha \frac{1}{\|\mathbf{\xi}\|^{n - 2\alpha}}$.

So it remains to determine $c_\alpha$. For this you can use Plancherel's theorem in conjunction with the fact that $e^{-\pi |\omega|^2}$ is its own Fourier transform, so that you have $$\int_{R^n} \frac{1}{\|\mathbf{\omega}\|^{2\alpha}}e^{-\pi|\omega|^2}\,d\omega = c_{\alpha} \int_{R^n} \frac{1}{\|\mathbf{\xi}\|^{n - 2\alpha}}e^{-\pi|\xi|^2}\,d\xi$$ You can turn both integrals into one dimensional integrals using polar coordinates and then solve for $c_{\alpha}$. The result will be a ratio of gamma functions.

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+1. Where did you learn this trick? I have seen it for the first time in Stein's "Singular integrals and differentiability properties of functions". –  Jonas Teuwen Jun 29 '11 at 17:52
    
From one of Stein's books originally, so it was probably that. More recently I saw it in Duoandikoetxea's Fourier analysis book, but most likely he got it from Stein too. –  Zarrax Jun 29 '11 at 19:58
    
Somewhere it should be said that $2\alpha < n$. Else $f(\omega)$ is not locally integrable, and is not the representation of a distribution. To extend for $2\alpha \geq n$, see en.wikipedia.org/wiki/Homogeneous_distribution –  Willie Wong Jul 2 '11 at 12:04
    
And if you are too lazy to compute the constants, you can read it off from item 311 here en.wikipedia.org/wiki/Fourier_transform#Distributions –  Willie Wong Jul 2 '11 at 12:22
    
@Zarrax in which of Stein's books did you find this trick? I've got his book "Singular Integrals and Differentiability Properties of Functions" with me but I can't find it. –  Olumide Oct 10 '11 at 1:47

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