I'm trying to solve $$\int_{-\infty}^\infty \frac{Im(k(\omega'))}{\omega'^2 - \omega^2} d\omega'$$
Is this function a Dirac delta function? If so I'm not convinced, since a Dirac delta function is $0$ if $\omega' \neq \omega$ and $\infty$ if $\omega' = \omega$.
Furthermore, if the function above is a Dirac delta function, I'm wondering if the Dirac delta function is only this part $$\int_{-\infty}^\infty \frac{1}{\omega'^2 - \omega^2} d\omega'$$, thus to solve this integral I could use $$\int_{-\infty}^\infty f(\omega')\delta(\omega' - \omega) = f(\omega) = Im(k(\omega))$$
This is not a delta function. As is, the integral is undefined for real $\omega$, due to the singularities at $\omega' = \pm \omega$. For complex $\omega$, the integral converges, but is still not equal to a delta function. If $\omega$ has an infinitesimal imaginary part, then a delta function does appear, and its sign depends on the sign of the imaginary part of $\omega$ (look up the Sokhotski–Plemelj theorem).
Given the form of your integral, I assume that you are doing physics, and care about the Kramers-Kronig relations. In this case, $\omega$ is real, but the integral should be interpreted as a Cauchy principal value integral. In this case, the integral is related to the real part of $k(\omega)$. However, this only holds, since in physics, one can insist that (i) $k(\omega)$ is analytic in the upper-half plane of complex $\omega$ (this follows from causality), and (ii) $k(t)$ (the function in the time-domain) is real.
For details, please read over the Wikipedia page for the Kramers-Kronig relations (in particular, the section titled "Physical interpretation and alternate form"). If you have further questions, a better place to ask is probably physics stackexchange.