How to Fourier transform a tempered distribution?

Question

I was wondering how I should proceed to Fourier transform a tempered distribution like $$v=e^{-x^{2}}(\mathbb{pv} \frac{1}{x})$$ And more, how I can prove that this is a tempered distribution? What exactly means $\mathbb{pv}$? I know that is a tempered distribution, so by multiplying an object for such term I should have a tempered distribution. Thank you very much :).

Answer 1

You cannot define a functional as $$\left( \frac 1 t, \phi \right) \overset ? = \int_{-\infty}^{\infty} \frac {\phi(t)} t dt,$$ because the action is not defined for every test function. But this is a valid functional: $$\left( \frac 1 t, \phi \right) = \operatorname{p.\!v.} \int_{-\infty}^{\infty} \frac {\phi(t)} t dt,$$ where $\operatorname{p.\!v.}$ means that the integral is understood in the sense of the principal value. There are a number of notations for this functional.

To compute the Fourier transform, note that $v(x)$ is an odd function, so the transform is also an odd function, thus the $\cos$ component is zero, and the $\sin$ component can be computed as an ordinary function: $$\mathcal F[v] = \int_{-\infty}^{\infty}\frac 1 x e^{-x^2} i \sin{w x} \,dx = i \pi \operatorname{erf}\left( \frac w 2 \right).$$