5
$\begingroup$

Question:

Suppose we are given the tempered distribution $\|x\|^{-\alpha}$. We want to compute the Fourier transform $\mathcal{F}[\|x\|^{-\alpha}](\xi)$.

  • What techniques are available for approaching this problem and/or others of a similar nature?
  • What are some references for learning about these techniques, or where I might see examples of such calculations?

Background:

I was recently reviewing Prof. Terry Tao's notes on distributions. In these notes, Prof. Tao presents a pretty miraculous procedure for computing the Fourier transform of the tempered distribution $\|x\|^{-\alpha}$. In particular, he presents the following:

Proceeding formally, we would like to evaluate the integral

$$ \mathcal{F}[\|x\|^{-\alpha}](\xi) = \int_{\mathbb{R}^d} \|x\|^{-\alpha} e^{-2\pi i \xi \cdot x} \mathrm{dx} $$

The integrand here is not absolutely integrable, so it would appear that our formula is not of much use. However, since we are treating $\|x\|^{-\alpha}$ as a tempered distribution, we know this Fourier transform does indeed exist. In order to transform the integral into something more manageable, we can write $\|x\|^{-\alpha}$ as an average of Gaussians, whose Fourier transforms we already know and love:

$$ \mathcal{F}[e^{-\pi t^2 \|x\|^2}](\xi) = t^d e^{-\pi \|\xi\|^2 /t^2}$$

To do this, we observe that $\|x\|^{-\alpha}$ is invariant under the scaling

$$ f(x) \mapsto t^{\alpha} f(tx)$$

Therefore, we average the Gaussian $e^{-\pi \|x\|^2}$ with respect to this scaling, then integrate against the multiplicative Haar measure $ \frac{\mathrm{dt}}{t}$. Doing this and making a change of variables, we get

$$ \int_0^\infty t^\alpha e^{-\pi t^2 \|x\|^2} \frac{\mathrm{dt}}{t} = \frac{1}{2} \pi^{-\alpha/2} \|x\|^{-\alpha} \Gamma\left( \frac{\alpha}{2} \right) $$

A formal Fourier transform of this equation gives

$$ \int_0^\infty t^\alpha t^{-d} e^{-\pi \|x\|^2/t^2} \frac{\mathrm{dt}}{t} = \frac{1}{2} \pi^{-\alpha/2} \mathcal{F}[\|x\|^{-\alpha}](\xi) \Gamma\left( \frac{\alpha}{2}\right)$$

another change of variables yields

$$ \int_0^\infty t^\alpha t^{-d} e^{-\pi \|x\|^2/t^2} \frac{\mathrm{dt}}{t} = \frac{1}{2} \pi^{-(d-\alpha)/2} \|\xi\|^{-(d-\alpha)} \Gamma\left( \frac{d - \alpha}{2} \right) $$

Combining these equations and solving for $\mathcal{F}[\|x\|^{-\alpha}](\xi)$ we can (formally) conclude

$$ \mathcal{F}[\|x\|^{-\alpha}](\xi) = \frac{\pi^{-(d - \alpha)/2} \Gamma\left(\frac{d - \alpha}{2}\right)}{\pi^{-\alpha/2}\Gamma\left( \frac{\alpha}{2}\right)} \|\xi\|^{-(d-\alpha)} $$

which can be verified rigorously by testing against an arbitrary Schwartz function. It is also mentioned that this solution can be continued meromorphically in $\alpha$.

As I mentioned above, this seems like a bit of a miracle to me. I am also curious about the precise meaning of "continuing the solution meromorphically in $\alpha$". I imagine the above is a particular example of some broader class of techniques for handling singular distributions. I would very much like to learn more about this so that I might be able to work similar miracles myself.

$\endgroup$
  • $\begingroup$ First, very nice question (+1)! Secondly i would guess that meromorphic extension means here that your first integral exists for $0< \alpha<1$ (at least in the sense of distributions) but the result can be continued to other values of $\alpha$ , meromorphic because the contiuned function now has simple poles at $\alpha=d$ $\endgroup$ – tired Sep 26 '15 at 12:31
  • $\begingroup$ Related: ($n$-dimensional) Inverse Fourier transform of $\frac{1}{\| {\omega} \|^{2\alpha}}$ $\endgroup$ – user147263 Sep 27 '15 at 5:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.