Nederlands Wetenschappelijk Onderzoek’ (NWO).

The integral (\[2dr\]) can be expressed in terms of elliptic integrals as follows [@Mor; @Mont]. ${\cal R}_2$ is equal to $$\begin{aligned}
{\cal R}_2(\zeta,\vec{K})&=&\frac{1}{16t\pi}\int_{A^2>B^2} dk_y
\, \frac{\mathop{\rm sgn} (A)}{\sqrt{A^2-B^2}}\nonumber\\
&&-\frac{i}{16t\pi}\int
_{A^2<B^2} dk_y \frac{1}{\sqrt{B^2-A^2}},
\label{twoint}\end{aligned}$$ where $k_y\in [0,\pi]$. On making the substitution $p=\cos k_y$, and writing $$A^2-B^2=\frac{\cos^2 (K_y/2)}{4}(q_1-p)(q_2-p),
\label{ab}$$ where $$q_1=\frac{-2\zeta + \cos (K_x/2)}{\cos (K_y/2)}, ~~~~~
q_2=\frac{-2\zeta - \cos (K_x/2)}{\cos (K_y/2)},
\label{qs}$$ we end up with integrals of the type $$\int dp \frac{1}{\sqrt{|w|}},
\label{wint}$$ where $w=(1-p)(1+p)(q_1-p)(q_2-p)$ and $p\in [-1,1]$. By defining $$q_-=\min (q_1,q_2),~~~~~q_+=\max (q_1,q_2),
\label{+-}$$ we see that values of $p \in [q_-,q_+]$ give $A^2 < B^2$, and thus contribute to the imaginary part of ${\cal R}_2$. Values of $p$ outside this interval give $A^2 > B^2$ and thus contribute to the real part, for $p<q_-$ with $\mathop{\rm sgn} (A)=-\mathop{\rm sgn}
(\cos(K_y/2))$, and for $p>q_+$ with $\mathop{\rm sgn} (A)=\mathop{\rm sgn}
(\cos(K_y/2))$. The result is a sum of integrals of the type (\[wint\]) between limits that are consecutive zeroes of $w$. These integrals can all be expressed in terms of the complete elliptic integral of the first kind, $K(x)$ [@G+R], $$K(x)=\int_0^1 dp \frac{1}{\sqrt{(1-p^2)(1-x^2p^2)}}
{}~~~~~~(x\in [0,1]).
\label{ka}$$ There are several cases to be considered:

$q_-,q_+ < -1$ (1a) or $q_-,q_+ 