DOS of Van Hove singularity in 2D square lattice tight binding model

For the simplest example, 2D square lattice tight binding model gives the energy band as $$\varepsilon_k=-2t(\cos k_x+\cos k_y) \, .$$ We know that $\mathbf{k}=(0,\pi)$ and related momentum points are saddle points which give $|\nabla_k \varepsilon_k|=0$ and thus some kind of singularity in density of state (DOS) since $$\rho(\varepsilon)\propto \int_{\varepsilon=\text{const}} \frac{d S}{|\nabla_k \varepsilon_k|} \, .$$ How can I get the $\ln$ divergence for DOS near $\varepsilon=0$? Should I only care about those singularity points and omit the integral from normal parts and do Taylor expansion near those saddle points?

Moreover, why are points like $(\pi/2,-\pi/2)$ or $(2\pi/3,-\pi/3)$ not called Van Hove points when those points are also lie in the $\varepsilon=0$ line and give $|\nabla_k \varepsilon_k|=0$?

• ε=0 at (π/2,−π/2) (2π/3,−π/3), while $|\nabla_kε_k|$ is not equal 0. Note that $\nabla_k$ is an vector. – Zili Von Feb 2 '18 at 8:29