Density of states (DOS) integral when surface is not closed

According to the density of states (DOS) formula $$\rho(\varepsilon)\propto \int_{\varepsilon=\text{const}}\frac{dS}{|\nabla_k \varepsilon_k|}.$$

Since there is an integral on the constant energy surface in the DOS formula, what if the constant energy surface is not closed and thus an infinite one? For example, in 2D $$\varepsilon_k=v_xk_x+v_yk_y \, .$$ The isoenergy surface will be plane extended to infinite and thus ill-defined in the integral $$\int dS$$. The DOS integral is divergent and gives no finite result.

Some one may argue that this kind of Hamiltonian must come from effective theory which is only valid for low energy and thus has a valid region beyond which the dispersion relation doesn't work. Well, if I only care about the divergent behavior and don't want to go further to the 'real' dispersion which is hard to manipulate (perhaps no analytical expression at all). Can I get the divergent information only from the linear dispersion? Should I give $$k$$ region a cut off and then calculate?

• What exactly are you trying to calculate? What do you mean be "divergent information"? Questions like "should I impose a cut off" normally depend a lot on what is reasonable in the context and on exactly what you want to know. – By Symmetry Aug 2 at 11:34

If you're talking about a solid, the integral is bounded to the first Brillouin zone. This gives you a finite number of states. Otherwise, you have for each energy an infinite set of possible $k_z$ so the DOS diverges.