From a general point of view there is not so much formal difference between the permittivity of a dielectric and conductance of a metal. For the latter, a crude classical model is that for a quasi-free gas of conduction electrons, moving in the external electric field and subject to friction from collisions. The equation of motion for such a conduction electron reads (in non-relativistic approximation, which is sufficient for any practical purposes)
[tex]m \dot{\vec{v}}=-m \gamma \vec{v}+q\vec{E}.[/tex]
Neglecting the spatial variation of [itex]\vec{E}[/itex] along the relevant distances of the electron's motion, we find by finding the retarded Green's function of the operator [itex]\partial_t+\gamma[/itex]
[tex]\vec{v}(t)=\int_{-\infty}^{\infty} \mathrm{d} t' \Theta(t-t') \exp[-\gamma(t-t')]\frac{q}{m} \vec{E}(t').[/tex]
The current is given by
[tex]\vec{j}(t)=n q \vec{v} = \frac{n q^2}{m}\mathrm{d} t' \Theta(t-t') \exp[-\gamma(t-t')]\frac{q}{m} \vec{E}(t'),[/tex]
where [itex]n[/itex] is the conduction-electron density, which we also consider as spatially homogeneous.
Writing the electric field as Fourier transform,
[tex]\vec{E}(t,\vec{x})=\int_{-\infty}^{\infty} \frac{\mathrm{d} \omega}{2 \pi} \exp(-\mathrm{i} \omega t) \tilde{\vec{E}}(\omega,\vec{x})[/tex]
and also
[tex]\vec{j}(t,\vec{x})=\int_{-\infty}^{\infty} \frac{\mathrm{d} \omega}{2 \pi} exp(-\mathrm{i} \omega t) \tilde{\vec{j}}(\omega,\vec{x}),[/tex]
we find in the frequency domain
[tex]\tilde{\vec{j}}(\omega,\vec{x})=\sigma(\omega) \tilde{\vec{E}}(\omega,\vec{x})[/tex]
with
[tex]\sigma(\omega)=\frac{n q^2}{m} \frac{1}{\gamma-\mathrm{i} \omega}.[/tex]
The only difference in the case of a non-conducting dieelectric is that here all the electrons are bound and in linear-response theory can be described as moving in a harmonic potential with frequency [itex]\omega_0[/itex] and some friction coefficient[itex]\gamma[/itex]. The only difference here is that we express the response to a (weak) electric field in terms of the polarization
[tex]\vec{P}=n q \vec{x}[/tex]
In Fourier space the Green's function gives the electric susceptibility,
[tex]\chi_e(\omega)=\frac{n q^2}{m} \frac{1}{\omega_0^2-\mathrm{i} \gamma \omega-\omega^2},[/tex]
leading to
[tex]\tilde{\vec{P}}(\omega,\vec{x})=\chi_e(\omega) \tilde{\vec{E}}(\omega,\vec{x}).[/tex]
In Heaviside-Lorentz units one thus has
[tex]\tilde{\vec{D}}=\tilde{\vec{E}}+\tilde{\vec{P}}=(1+\chi_e) \tilde{\vec{E}},[/tex]
and thus
[tex]\epsilon(\omega)=1+\chi_e(\omega).[/tex]
In a real material you have several eigen frequencies, corresponding to the different quantum-theoretical bound states of the electrons to their ions, and also in a conductor you usually have some bound electrons, so that [itex]\sigma[/itex] and [itex]\chi_e[/itex] are given by the sum of the corresponding terms.
That there is not so much difference in the two cases, because on a microscopic level the total current is given by the conduction current and the current due to the motion of the bound electrons. The latter obviously is given by
[tex]\vec{j}_{\text{bound}}=\partial_t \vec{P}[/tex]
or, in frequency space,
[tex]\tilde{\vec{j}}_{\text{bound}}=-\mathrm{i} \omega \tilde{\vec{P}}.[/tex]
Thus one finds the conductivity for [itex]\omega_0=0[/itex] for the dielectric case via this relation
[tex]\sigma=\left .-\mathrm{i} \chi_e \right|_{\omega_0=0}.[/tex]
