Fourier transform $1/\sqrt{x}$ weighted by a gaussian noise.

Question

I tried computing the Fourier transform of $\frac1{\sqrt{\vert x\vert}}$ i.e. $$\int_{\mathbb{R}}e^{-i\omega x}\frac{1}{\sqrt{\vert x\vert}}dx.$$

The function in not in $L^{1}$ neither in $L^2.$ So the integral is a notation and rigorously it must be seen as the Fourier transform of tempered distribution. Ok and apparently it is a classical result.

However I get the following idea: Let $\varepsilon>0$ and denote $f_{\varepsilon}(x)=\frac{e^{-\varepsilon x^2}}{\sqrt{\vert x\vert}}.$

How can I compute the Fourier transform of $f_\varepsilon$?

Wolfram answer with the following $$\frac{\sqrt{\pi}\sqrt{\vert\omega\vert}e^{-\omega^2/8\varepsilon}}{2\sqrt{\varepsilon}}I_{-1/4}(\frac{\omega^2} {8\varepsilon})$$

where $I$ is the modified bessel function of the first Kind.

Answer 1

Instead of doing that trick, we will use a clever property of the Fourier Transform involving homogeneity.

$\mathbf{\text{Def.}}$ A function $f\in\mathcal{S}(\mathbb{R})$ is homogeneous of degree s if $\forall a \neq 0$

$$f(ax) = a^sf(x)$$

$\mathbf{\text{Lemma.}}$ Let $f$ be homogeneous degree s. Then $\hat{f}$ is homogeneous degree -1-s.

Proof: $$\hat{f}(a\xi) = \frac{1}{\sqrt{2\pi}}\int_\mathbb{R}f(x)e^{-ia\xi x}dx = \frac{1}{\sqrt{2\pi}}\int_\mathbb{R}f(\frac{u}{a})e^{-i\xi u}\frac{du}{a}$$ $$\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}\left(\frac{1}{a}\right)^sf(u)e^{-i\xi u}\frac{du}{a} = a^{-1-s}\hat{f}(\xi)$$

Even though we proved this in the Schwartz function case, the same property applies to tempered distributions, albeit with a lot more work.

Using the above lemma, since $f(x)=|x|^{-1/2}$ is homogeneous degree $-\frac{1}{2}$, $\hat{f}(\xi)$ must be homogeneous degree $-\frac{1}{2}$ The only possible homogeneous function with that degree would be $C|\xi|^{-1/2}$ for some constant C (no deltas because the integral around 0 will converge for all Schwartz test functions).

By the Fourier inversion formula, this same homogeneity property would have to hold in reverse, so $\mathcal{F}^{-1}\mathcal{F}(f) = C^2f \implies C = 1$ ($C=-1$ does not hold as a solution). Therefore we have that:

$$\hat{f}(\xi) = \frac{1}{\sqrt{|\xi|}}$$