What is $\epsilon_\infty$ in this equation and why can it be neglected in the IR? I'm reading this paper (warning, PDF) and they mention that the complex permittivity $\epsilon$ and complex conductivity $\sigma$ are related through the equation
$$\epsilon - \epsilon_\infty = (4\pi i \sigma)/\omega$$
Then they say that in the far-IR region, $\epsilon_\infty$ can be neglected, so it simplifies the equation.
What does $\epsilon_\infty$ represent physically?
From Jackson, he says that if some electrons in the material are "free", then the complex permittivity is
$$\epsilon(\omega) = \epsilon_b(\omega) + i\frac{Ne^2f_0}{m\omega(\gamma_0-i\omega)}$$, which is pretty similar in form to the first equation. In Jackson, $ \epsilon_b(\omega)$ is the contribution from the electrons that are dipoles rather than free.
So why can they be ignored in the FIR?
 A: For a metal, the permittivity can is typically described by the Drude model with a permittivity given by,
\begin{equation}
\epsilon = \epsilon' - i\epsilon'' = \epsilon_\infty - \frac{\omega_p^2}{\omega(\omega - i\gamma)} = \epsilon_\infty - \frac{\omega_p^2}{\omega^2 + \gamma^2} - i\frac{\gamma}{\omega}\frac{\omega_p^2}{\omega^2 + \gamma^2}
\end{equation}
where $\omega_p^2 = Nq^2 /\epsilon _0 m^*$ is the plasma frequency of the electron gas. This is equivalent to the form you have given. $\epsilon_\infty$ is the dielectric constant, and is a positive value related to the atomic polarizability. It is determined by the atomic properties and the crystal structure of your material, and is typically weakly dependent upon the frequency. For example, in Silicon we have $\epsilon_\infty \approx 3.6 $ at $\lambda = 1500\ nm$ and $\epsilon_\infty \approx 3.4 $ at $\lambda = 10.6\  \mu m$.
If the material has a very large electron concentration, typically on the order of $10^{22}$ in metals, then it turns out that the plasma frequency $\omega_p$ corresponds to a frequency either in the red portion of the visible spectrum, or in the near-infrared. If $\omega > \omega_p$, then we will have $\epsilon \approx \epsilon_\infty$. If $\omega < \omega_p$, then the Drude terms will be very large and negative, especially in the infrared. $\epsilon_\infty$ is typically around 10-20, but in the far infrared we will often have the real part of the permittivity $\epsilon'$ possessing negative values greater than $1000$. So in this frequency regime, it's usually safe to neglect the dielectric constant.
A: Because in the FIR, $\omega \rightarrow 0$ and therefore the $4\pi i \sigma/\omega$ term dominates the $\epsilon_\infty$ term. It can therefore be safely neglected. You are right that the $\epsilon_\infty$ represents the contribution to $\epsilon(\omega)$ of the bound (or dipole-like) electrons.
