use hence $c_{IR}=\lim_{\rho
  \rightarrow \infty}I_{IR}(\rho)=\lim_{s\rightarrow \infty}F(s)$. From the asymptotic form of the modified Bessel functions $K_n(x)\sim e^{-x}$, together with , it follows that $c_{IR}=0$. Figure \[fig:boseint\]a shows the exact function $I_{UV}(\rho)$ computed by a numerical integration of using a NAG Fortran Library integration routine. The figure shows how the contribution from the cut vanishes in the limit of large $\rho$. The exact infra-red function $I_{IR}(\rho)$ is plotted in figure \[fig:boseint\]b, it is important to note that the minimum value of $I_{IR}(\rho)$ is very small, namely $-0.019$.

The free fermion
----------------

The free massive fermion has the action $$\label{actionfermion}
  S=\int d^2x\, \left( \bar{\psi}\p
  \bar{\psi}+\psi\bar{\p}\psi+i m\bar{\psi}\psi \right) .$$ A calculation analogous to the Bose case gives, for $k=4$, that $$\label{fermiF}
  F(s)=\frac{m^4s}{8}\left(K_1^2(ms^{1/4})\left(1+\frac{4}{m^2s^{1/2}}\right)-
  K_0^2(ms^{1/4})  \right) .$$ The same arguments as above yields $c_{UV}=\tfrac{1}{2}$, and the infra-red contribution $c_{IR}=\lim_{s \rightarrow \infty}F(s)=0$, again by using the asymptotic form of the Bessel functions. Again we can obtain the exact functions $I(\rho)$ by a numerical computation. We get the same behaviour of $I_{IR}(\rho)$ as in the bosonic case, but with the minimum value $-0.0048$, so this exact function does again not differ much from the infra-red central charge at its minimum value.

The unitary minimal models
--------------------------

The 