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Note: I first posted question on math.stackexchange and I got one reply, which was a bit helpful (I'm still trying to understand it fully), but did not explore the two solution cases that I mentioned. Besides I would appreciate additional insight on this question, which I hope its not too trivial for mathoverflow. Thanks.


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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  • $\begingroup$ @Olumide: in this case you should bump your Math.SE question. In anycase, Zarrax already gave you the answer to your question. Why not just follow it up there? $\endgroup$ Commented Jul 2, 2011 at 11:57
  • $\begingroup$ Will do. I just assumed I'd find more professional mathematicians, and thus new perspectives here on MO. $\endgroup$
    – Olumide
    Commented Jul 2, 2011 at 12:33

1 Answer 1

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Let's start with a simple change of notation. Let me consider first on $\mathbb R^n_x$ the function $f_{\beta}(x)=\Vert x\Vert^{\beta-n}$ for $0<\beta< n$, which is locally integrable and homogeneous with degree $\beta-n$. Its Fourier transform is also a radial distribution, i.e. such that $$ \forall A\in O(n),\quad \hat{f_\beta}=\hat{f_\beta}\circ A, $$ which is homogeneous with degree $-\beta+n-n=-\beta.$ A direct computation shows that $$ \hat{f_\beta}(\xi)=\Vert \xi\Vert^{-\beta} \frac{\pi^{-\beta/2}\Gamma(\beta/2)}{\pi^{-(n-\beta)/2}\Gamma((n-\beta)/2)}. $$ There are various extensions of this formula to other values of $\beta$, but one has to pay attention that for $\beta\le 0$, the ``function" $f_\beta$ is not locally integrable. It is then necessary to define an homogeneous distribution with degree $\beta-n$ which coincides with $\Vert x\Vert^{\beta-n}$ on $\mathbb R^n\backslash\{0\}$, which is possible uniquely if $\beta$ is not a nonpositive integer.

Bazin.

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