- #1
nonequilibrium
- 1,412
- 2
Hello.
I understand that in the form of \int_{\mathbb R} f(x) \exp{2 \pi i tx} \mathrm d x the function f: \mathbb R \to \mathbb C: x \to \frac{1}{x} doesn't have a Fourier transform (because the function is not integrable).
But in my analysis course, there is a theorem that states that in L^2(\mathbb R) there is a unique extension to the Fourier transform for all square-integrable functions (so also the above-defined f). It does, however, not give an explicit form. So I was wondering, what would this generalized Fourier transform of f look like? Is it known explicitly?
Thank you!
EDIT: apologies! I was too fast in typing, of course 1/x is not square-integrable around the origin, so just cut out the [-1,1] interval and replace it with zero. My main question is how the function behaves for large x.
I understand that in the form of \int_{\mathbb R} f(x) \exp{2 \pi i tx} \mathrm d x the function f: \mathbb R \to \mathbb C: x \to \frac{1}{x} doesn't have a Fourier transform (because the function is not integrable).
But in my analysis course, there is a theorem that states that in L^2(\mathbb R) there is a unique extension to the Fourier transform for all square-integrable functions (so also the above-defined f). It does, however, not give an explicit form. So I was wondering, what would this generalized Fourier transform of f look like? Is it known explicitly?
Thank you!
EDIT: apologies! I was too fast in typing, of course 1/x is not square-integrable around the origin, so just cut out the [-1,1] interval and replace it with zero. My main question is how the function behaves for large x.