analytic in $S(\alpha)$. Choosing another positive power $f(s^\gamma x)|_{x=1}$, $\gamma>0$, gives the same ultra-violet and infra-red limits, but different intermediate behaviour. Let us now introduce the scale parameter so that the opening of the analytic sector is $2(\pi-\epsilon')$ with $\epsilon'\ll 1$. $F(s)=\tilde{F}(s^a)$ is analytic in $S=S(2(\pi-\epsilon'))$ provided that $a=\tfrac{\alpha}{2(\pi-\epsilon')}$. We will now express the infra-red limit as an integral entirely in the ultra-violet region using analyticity.

Using Cauchy’s theorem the analyticity of $F(s)$ implies that the contour integral $$\label{nycontourint2}
  \frac{1}{2(\pi-\epsilon) i}\int_C ds \, \frac{e^{\rho /s}}{s}F(s)=
\frac{1}{2(\pi-\epsilon) i}\left(\int_{C_0}ds+\int_{C_1}ds+\int_{C_2}ds\right)\frac{e^{\rho /s}}{s}F(s)=0,
\nonumber$$ vanishes in the sector $S$, where $\rho \in \mathbb{R}_+$ and the contour $C$ is given in figure 1; $\epsilon>\epsilon'$ so that $C\subset S$. In the limit where $r_{IR}\rightarrow \infty$, the contribution from the contour $C_{2}$ (see figure 1) becomes $$\label{nyC22}
\lim_{r_{IR} \rightarrow \infty}
\frac{1}{2(\pi-\epsilon) }\int_{\pi-\epsilon}^{-\pi+\epsilon}d\theta \, e^{\rho e^{-i\theta}/r_{IR}}
F(r_{IR}e^{i\theta})= -\lim_{|s|\rightarrow \infty}
F(s) = -F_{IR}.$$ The angular integral and the limit $r_{IR}\rightarrow \infty$ can be interchanged as the integrand in is bounded by a constant in the limit $r_{IR}\rightarrow \infty$. We then get the integral representation of $F_{IR}$ $$F_{IR}=\frac{1}{2(\pi-\epsilon) i}\left( \int_{C_0}ds \frac{e^{\rho /s}}{s}F(s)+
\int_{C_1}ds \frac{e^{\rho /s}}{s}F(s)  \right) .
\label{nyCir}$$ Analogously by considering $$\label{nycontourint}
  \frac{1}{2(\pi-\epsilon) i}\int_C ds \, \frac{e^{\rho
      s}}{s}F(s)=\frac{1}{2(\pi-\epsilon)  i}
\left(\int_{C_0}ds+\int_{C_1}ds+\int_{C_2}ds\right)\frac{e^{\rho s}}{s}F(s)=0,
