\zeta \leq 1)\\
0 & ~~( \zeta > 1).
\end{array}
\right.
\label{1df}\end{aligned}$$ Using $\zeta$ instead of $E$ in (\[b2e2\]) as integration variable we have $$b_2^{\rm int} = \frac{1}{N}  \int_{-1}^1 d\zeta \left[
\rho_2(\zeta) - \rho_2^0(\zeta)\right] e^{-4\beta t(\zeta +1)} .
\label{rhozeta}$$ Due to the differentiation with respect to $z$ in (\[rhof\]) only values of $\zeta$ between $-1$ and 1 contribute to the integral. From (\[1df\]) we see that for $|\zeta| \leq 1$ $$\rho_2(\zeta) - \rho_2^0(\zeta)= -\frac{N}{\pi} \frac{\partial}
{\partial
 \zeta} F(\zeta)=
-\frac{N}{\pi} \frac{1}{\sqrt{1-\zeta^2}}.
\label{rho1d}$$ Substituting this into (\[rhozeta\]) we find $$b_2^{\rm int}=-e^{-4\beta t} I_0(4\beta t).
\label{1dbint}$$

We now compare this with the result for an ideal lattice gas, given by (\[bl0\]) $$b_2^0=\frac{1}{2}b_1(2\beta)=\frac{1}{2}e^{-4 \beta t}
I_0(4\beta t)=
-\frac{1}{2} b_2^{\rm int}.
\label{1db20}$$ So the hard-core interaction giving rise to $b_2^{\rm int}$ changes the value of $b_2$ from $b_2^0$ into its opposite, or in other words, the Bose value is turned into the Fermi value. This is exactly what has to be expected from the well known fact that a hard-core Bose gas in one dimension is equivalent to an ideal Fermi gas (for all virial coefficients).

The picture shown in figure \[fifig\] for the phase $\Phi$ has some general validity, in the sense that for 