$|s|\rightarrow
\infty$ from $S$, hence $\mbox{Im}[F(re^{-i(\pi-\epsilon)})]\rightarrow 0$ for $r\rightarrow \infty$. If $\mbox{Im}[F]$ falls off like $r^{-\delta}$ for some $\delta>0$ then in the limit $r_{IR}\rightarrow \infty$, $\exists\ k_1,k_2,k_3>0$: $$\label{cutirlim}
  \left|cut_{IR}\right|<k_1\int_1^\infty dr \frac{\left(\sin
  (k_2/r)+k_3r^{-\delta}\right)}{r}<\infty \mbox{   for all  } \rho \in
  \mathbb{R}_+ .$$ For a general $F(s)$ where the fall off might be slower we keep a finite $r_{IR}$ then $|cut_{IR}|$ is again finite for all $\rho$ and a finite $r_{IR}$ introduces a $O(\tfrac{1}{r_{IR}})$ term in which is negligible for $r_{IR}$ large. As $\epsilon\ll 1$ we can replace $\tfrac{1}{2(\pi-\epsilon)i}$ with $\tfrac{1}{2\pi i}$ given a term $O(\epsilon)$ on the right hand side in which is again negligible for small $\epsilon$. We can then define $$\label{firrho}
  I_{IR}(\rho)=\lim_{r_{UV}\rightarrow 0}\ 
\frac{1}{2\pi i}\int_{C_0} ds \frac{e^{\rho/s}}{s}F(s)$$ so that $F_{IR}=\lim_{\rho \rightarrow \infty}I_{IR}(\rho)$ and we have succeeded in writing the infra-red limit as the limiting value of a contour integral in the ultra-violet region.

We will now use Cauchy’s theorem again to rewrite $I_{IR}(\rho)$ and this will show that $I_{IR}(\rho)$ is the Borel transform of $\tilde{F}(s)$. Instead of integrating over $C_0$ we will integrate over $\tilde{C}$ given as the path from the origin along the ray $\arg (s)=-\pi+\epsilon$ and then anti-clockwise along $|s|=\tilde{x}>0$ until $\arg (s)=\pi-\epsilon$ and then back to the 