I've been working on a problem in physics in which I've encountered the following Fourier transform: $$ F(x) \equiv \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} d\omega\ \frac{\ln(\omega^2+i\epsilon)}{\sqrt{\omega^2 + i \epsilon}} e^{- i x \omega} $$
I am wondering if there is any way to attach a meaning to the above function in terms of distributions?
My attempt: Since $\omega^2 > 0$, this means that $\lim\limits_{\epsilon \to 0^{+}} \ln( \omega^2 + i \epsilon ) = \ln(\omega^2)$ and I am tempted to use the Sokhotski–Plemelj theorem in the following (probably wrong) way: $$ F(x) = \int_{-\infty}^{\infty} d\omega \frac{\ln(\omega^2)}{|\omega|} e^{- i x \omega} - i \pi \int_{-\infty}^{\infty} d\omega\ \delta(\omega)\ \ln(\omega^2) e^{- i x \omega} $$
where I've used the result of this post to replace $\delta(|\omega|) \mapsto \delta(\omega)$.
I know that the Fourier transform of $\frac{\log|\omega|}{|\omega|}$ is $\left( \log|x| - \gamma \right)^2 + C$ for some constant $C$ (I'm not sure what $C$ is though). However, the second term in the above is obviously non-sense since I get $\ln(0)$ after performing the integral.
I know that you can't multiply distributions, so I'm not sure how to interpret the $\ln(\omega^2 + i \epsilon)$. I also know that I can't so simply take the limit $\epsilon \to 0^{+}$ in each of the functions separately...where am I wrong?
Or is this function just complete non-sense?