Fourier transform of $|\omega|/\sqrt{\omega^2-a^2}$

Question

I am trying to find the Fourier transform of $|\omega|/\sqrt{\omega^2-a^2}$. The transform of $\mathrm{sgn}(\omega)/\sqrt{\omega^2-a^2}$ is found in terms of Bessel functions, with the result being $K_0(iat)-K_0(-iat)=-i\pi J_0(at)$. Naively, the transform of $|\omega|/\sqrt{\omega^2-a^2}$ would be $i\frac{d}{dt}\big(-i\pi J_0(at)\big)=-\pi a J_1(at)$. However, since the original transformand $|\omega|/\sqrt{\omega^2-a^2}$ is not square integrable, and asymptotically becomes a constant, $1$, this method likely misses additional contributions arising due to this feature.

While this is just one specific case, how does one generalise and obtain Fourier transforms of functions $f(\omega)$ which are not square-integrable, but the transform for $f(\omega)/\omega$ exists.

Answer 1

I tried to get the searched Fourier Transform using Mathematica by the following code:

H[a_, t_] := Abs[t]/(Sqrt[t^2 - a^2]);
FourierTransform[H[a, t], t, \[Omega]]

As a result we obtain the function:

$$ \sqrt{\frac{\pi }{2}} a \sqrt{-a^2} \left(-\sqrt{-\frac{1}{a^2}}\right) J_1(a \left| \omega \right| )-\sqrt{\frac{\pi }{2}} \sqrt{-a^2} \pmb{H}_{-1}(a \omega )+\sqrt{2 \pi } \delta (\omega ) $$

Here $J_1$ is Bessel function of the first kind and $H_{-1}$ represents the Struve function and $\delta$ is the Dirac delta function.

When choosing $a=i$, we are able to generate plots of this Fourier transform.

For $0\le\omega\le1000$:

For $0\le\omega\le100$:

For $30\le\omega\le30.5$: