the contribution from the cut vanishes and we get $$\label{cirgranse}
  c_{IR}\approx \lim_{\rho \rightarrow \infty}I_n(\rho) =\lim_{s\rightarrow 
\infty}F_n(s)$$ where the last equality follows setting $s'=s/ \rho$ in and then taking the limit $\rho \rightarrow \infty$ in the integrand valid for all $r_{UV}>0$. Moving all scale dependence into the running coupling $\bar{g}(s)$ we can write $F_n(s)=\Phi_n(\bar{g}(s))$. Let $g^*_{IR}$ denote the first non trivial zero of the perturbative $\beta-$function, i.e.$\bar{g}(s)\rightarrow g^*_{IR}$ for $s\rightarrow \infty$. Then equation becomes $$\label{graense1}
  \lim_{s\rightarrow \infty}F_n(s)=\lim_{s\rightarrow
  \infty}\Phi_n(\bar{g}(s))=\Phi_n(g^*_{IR})=c^*_{IR}$$ which is the perturbative estimate of $c_{IR}$ we want to improve.

$F_n(s)$ is the ultra-violet perturbative approximation to $F(s)$ and the integration range in $I(\rho)$ is compact so $\lim_{\rho \rightarrow 0}I_{n}(\rho)=\lim_{\rho \rightarrow
  0}I(\rho)=c_{UV}$. $I_n(\rho)$ therefore provides a good approximation to $I(\rho)$, for small enough $\rho$, since the power series expansion of $I(\rho)$ is controlled by the small $s$ expansion of $F(s)$ for which perturbation theory applies. This is illustrated in below for the minimal models. For larger values of $\rho$ higher order terms in the expansion of $I_n(\rho)$ become important and the coefficients of the expansion of $I_n(\rho)$ and $F_n(\rho)$ part company. If $c_{IR}<c^*_{IR}$ and if the region where $I_n(\rho)$ is a good approximation to $I(\rho)$ is large enough, then $I_n(\rho)$ will have 