is to obtain $F(\zeta)$ as given in (\[fdef\]) by integrating over $\vec{K}$. This is done numerically using Monte Carlo integration. The result is a function $F(\zeta)$ that is equal to $-\pi$ for $\zeta < -1$, where ${\cal R}_d$ is real and negative, equal to $0$ for $\zeta > 1$, where ${\cal R}_d$ is real and positive, and that is in between these two values for $\zeta \in [-1,1]$, where ${\cal R}_d$ is complex. It follows from the symmetry of ${\cal R}_d$ that $$F(-\zeta)=-\pi-F(\zeta).
\label{sym}$$ Figures \[f2\] and \[f3\] show $F(\zeta)$ for the square and cubic lattices.

Using $F(\zeta)$, $b_2^{\rm int}$ can be found from equations (\[b2e2\]) and (\[rhof\]) $$b_2^{\rm int}=-\frac{1}{\pi}\int_{-1}^{1} d\zeta \left(
\frac{\partial}{\partial \zeta} F(\zeta)\right) e^{-4d\beta
t(\zeta +1)}.
\label{mdb2}$$ Partially integrating, we find $$\begin{aligned}
b_2^{\rm int}&=&-e^{-4d\beta t} \left\{
e^{4d\beta t}+\frac{4d\beta t}{\pi} \int_{-1}^{1} d\zeta
F(\zeta) e^{-4d\beta t\zeta}\right\}\nonumber\\
&=&-e^{-4d\beta t} \left\{1-
\frac{8d\beta t}{\pi} \int_0^{1} d\zeta
F(\zeta) \sinh{4d\beta t\zeta}\right\}.\nonumber\\\end{aligned}$$ To find $b_2^{\rm int}$, the last equation can be numerically integrated for various values of $\beta t$.

As before, we compare this with the result for the ideal lattice gas (\[bl0\]) $$b_2^0=\frac{1}{2}b_1(2\beta)=\frac{1}{2}e^{-4d\beta t}
[I_0(4\beta t)]^d.
\label{mdb20}$$ The ratio $b_2^{\rm int}/b_2^0$ is given by $$b_2^{\rm int}/b_2^0=
\frac{-2}{[I_0(4\beta t)]^d}\left\{1-\frac{8d\beta t}{\pi}
\int_0^1d\zeta
F(\zeta) \sinh 4d\beta t\zeta\right\}.
\label{ratio}$$ 