vector $G$ enters into the problem. In figure \[bzfig\] we have divided the original Brillouin zone $-\pi < k_1 < \pi\;,\; -\pi < k_2 < \pi$ into four domains, I, II, III, and IV. Domains I and IV, as well as II and III, refer to the same states, as they are obtained from each other through interchanging $k_1
\leftrightarrow k_2$. So it suffices to take one of each, say I and III. Now III is equivalent to III$'$, which follows from III by shifting $k_1$ over $2\pi$. The combined domain I and III$'$ is given in $K,
k$ space by $0 < K < 2\pi\;,\; 0 < k < \pi$.

Using this parameter space one can write (\[r2\]) explicitly as (dropping the primes) $${\cal R}(z,K)=\frac{1}{2\pi} \int_0^{\pi} dk\,
\frac{1}{z-4t(1-\cos\frac{K}{2}
\cos k)} .
\label{1dr}$$ For convenience we put $z=4t(\zeta+1+i\epsilon)$, and have $${\cal R}(\zeta,K)=\frac{1}{8 \pi t}\int_0^{\pi} dk \,\frac{1}
{\zeta
+i\epsilon+\cos
\frac{K}{2} \cos k}.
\label{1dr2}$$ According to equations (\[rhof\]) and (\[fdef\]) we are interested in real values of $\zeta$. So, letting $\epsilon \downarrow 0$, the integral has a real part given by a principal value integral, and an imaginary part given by an integral over a delta function $$\begin{aligned}
{\cal R}(\zeta,K)&=& \frac{1}{8 \pi 