nonequilibrium
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Hello.
I understand that in the form of [tex]\int_{\mathbb R} f(x) \exp{2 \pi i tx} \mathrm d x[/tex] the function [tex]f: \mathbb R \to \mathbb C: x \to \frac{1}{x}[/tex] 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 [tex]L^2(\mathbb R)[/tex] 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 [tex]\int_{\mathbb R} f(x) \exp{2 \pi i tx} \mathrm d x[/tex] the function [tex]f: \mathbb R \to \mathbb C: x \to \frac{1}{x}[/tex] 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 [tex]L^2(\mathbb R)[/tex] 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.