What is the Fourier transform of 1/x?

[nonequilibrium]

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.

 

[mathman]

1/x is not square integrable. (Behavior as x -> 0 is problem).

 

[nonequilibrium]

the "EDIT" was there well before you posted though

 

[zetafunction]

1/x is not square integrable. (Behavior as x -> 0 is problem).

it exist distributionally as $$\frac{\pi}{2}(H(u)-H(-u))$$

remember the integral of $$\frac{sin(ax)}{x}$$ and you will get the answer