\nonumber$$ we get that $$F_{UV}=\frac{-1}{2(\pi-\epsilon) i}\left(\int_{C_2}ds \frac{e^{\rho s}}{s}F(s)+
\int_{C_1}ds \frac{e^{\rho s}}{s}F(s)  \right).
\label{nycuv}$$ We denote the integral along the contour $C_1$ (close to the cut) by $$cut(\rho)=\frac{1}{2(\pi-\epsilon) i}\int_{C_1}ds \frac{e^{\rho/ s}}{s}F(s).$$ The integrand in this integral is damped by the factor $e^{-\rho/ |s|}$, and we will show that $\lim_{\rho \rightarrow \infty}cut(\rho)=0$ by showing that it is bounded by a finite integral for all $\rho$ allowing us to take the limit $\rho \rightarrow\infty$ in the integrand. We substitute $s=re^{\pm i(\pi-\epsilon)}$ for points on $C_1$ in the upper and lower half-plane, the integral becomes $$\label{label1}
\begin{split}
  cut(\rho)=&\ \frac{-1}{2(\pi-\epsilon) i}\left(\int_{r_{IR}}^{r_{UV}}dr \frac{e^{\rho 
      e^{-i(\pi
        -\epsilon)}/r}}{r}F(re^{i(\pi-\epsilon)}) +\! \int_{r_{UV}}^{r_{IR}}dr\frac{e^{\rho 
      e^{i(\pi -\epsilon)}/r}}{r}F(re^{-i(\pi-\epsilon)}) \right)\\
=&\
\frac{-1}{(\pi-\epsilon)}\int_{r_{UV}}^{r_{IR}}dr \frac{e^{\rho \cos (\pi
    -\epsilon)/r}}{r}\mbox{Im} [ e^{i\rho \sin (\pi-\epsilon)/r}F(re^{-i(\pi-\epsilon)}) ]
      \\
=&\ \frac{-1}{(\pi-\epsilon)}\int_{r_{UV}}^{r_{IR}}dr\frac{e^{\rho \cos (\pi
    -\epsilon)/r}}{r}\left(\cos (\rho \sin
      (\pi-\epsilon)/r)\mbox{Im}[F(re^{-i(\pi-\epsilon)})] \right. \\ 
& \left. \ \ \ \ + \ \ \mbox{Re}[F(re^{-i(\pi-\epsilon)})]
\sin (\rho \sin (\pi-\epsilon))\right).
\end{split}
\nonumber$$ We divide the $r$ interval into $( r_{UV},1)$ and $( 1,r_{IR})$ and write $cut(\rho)=cut_{UV}+cut_{IR}$. The function $F(s)$ is finite in $S$ hence there exists a constant $q>0$ so that in the limit $r_{UV}\rightarrow 0$ $$\label{iir1}
  |cut_{UV}|< q \int_0^1 dr \frac{e^{-\rho /r}}{r}\rightarrow 0 \ \ 
\mbox{    for } \ \rho \rightarrow \infty.$$ We know that $F(s)\rightarrow F_{IR}\in \mathbb{R}$ for 