What's the corresponding energy dispersion of Green function? I want write a "toy" Green function witch can describe the electrons in the band with a width of $±W$ with uniform density of states (DOS). The reference gives an explicit expression of imaginary-time Green function:
$$G(i\omega)=\frac{1}{2\pi}\ln \frac{i\omega+W}{i\omega+W}$$
with uniform DOS like this:

Thus, I am confused of the origin for above form of Green function?

In addition, I have made some attempts: from the start point of tight-binding model, (half-filling one dimension for simplicity), the Hamiltonian and Green function are:
$$H=-W\sum_{i,j}c_{i}^\dagger c_j+h.c.=-W \cos k c_k^\dagger c_k\\G(i\omega,k)=\frac{1}{i\omega+W\cos k}$$
to obtain the similar expression of initial form for Green function, I integral in term of momentum:
$$G(i\omega)=\int_{-\pi}^\pi G(i\omega,k)dk=\frac{-2i(-1)^{Floor[\frac{\pi-2Arg[i+\omega]+Arg[1+\omega^2]}{2\pi}]}}{\sqrt{1+\omega^2}} $$
the result is very tedious and the DOS is like this:
which is not similar to the initial form. Thus, I am also confused the explicit expression of band structure(energy dispersion), or model, corresponding to the initial form of Green function?
 A: To get this Green function you should take a wide-band limit. That is, a continuum limit where the energy of the electrons is $\epsilon_k = v_F k$ and then $k$ has limits $\pm W/v_F$. This is what you get when you linearize the spectrum about the Fermi energy. From the tight binding model you will get if you add a chemical potential $\mu$, and then linearize and take the continuum limit. You get something like $H = \sum_k v_F k c^{\dagger}_k c_k$, and $k=2\pi n/L$ and it is measured from $k_F$.
The single-particle Matsubara Green function is then $g_{\epsilon_k}(i\omega) = \left( i\omega-\epsilon_k\right)^{-1}$ and you sum over it to get the GF at a certain point [note that the factor of $1/L$ is added because we are looking at the correlation function of $\psi(x)$] $$G(i\omega) = \frac{1}{L} \sum_{k} g_{\epsilon_k}(i\omega) = \frac{1}{2\pi}\int_{-W/v_F}^{W/v_F}\! dk g_{\epsilon_k}(i\omega) = \frac{1}{2\pi v_F}\int_{-W}^{W}\frac{d\epsilon}{i\omega-\epsilon}$$
where we used $2\pi/L = dk$ in taking the continuum limit. This integral will result in what you wrote.
