Is Permittivity Equivalent to EM Conductivity?

[Dickfore]

So in this case we get division by zero?

Provided that \hat{\sigma}(\omega) \neq o(\omega), \ \omega \rightarrow 0. This is true for conductors, but not for insulators (dielectrics).

EDIT:
Remember, we are working wIth the Fourier transforms of these quantities in the frequency domain. This simply tells us that there ought to be a pole of the dielectric response function at \omega = 0. Going back to time domain, we get:
<br /> \epsilon&#039;&#039;(t -t&#039;) = \frac{\sigma}{\epsilon_0} \, \theta(t - t&#039;)<br />
which gives the following relation between the polarization and the electric field:
<br /> \mathbf{P}(t) = \sigma \, \int_{-\infty}^{t}{\mathbf{E}(t&#039;) \, dt&#039;}<br />
or, the current density due to bound charges is:
<br /> \mathbf{J}(t) = \dot{\mathbf{P}}(t) = \sigma \, \mathbf{E}(t)<br />
But, this is just Ohm's law in differential form, provided the "bound" charges are the free ones, as is the case for conductors.