origin. It follows that this contour integral is independent of the choice of $\tilde{x}$ and $\epsilon$ as long as $\epsilon'<\epsilon<\tfrac{\pi}{2}$. The upper limit ensures that the rays stay in the negative half-plane (where $e^{\rho/s}$ is a damping factor), and the lower limit that $F(s)$ is analytic on the contour. Using this contour the ultra-violet limit $r_{UV}\rightarrow 0$ can be taken explicitly by extending the rays to the origin.

Using that $F(s)=\tilde{F}(s^a)$ for $a=\tfrac{\alpha}{2(\pi-\epsilon')}$ then amounts to $$\label{newcontour}
  I_{IR}(\rho)=\frac{1}{2\pi i}\int_{\tilde{C}}\frac{ds}{s}e^{\rho/s}F(s)=
\frac{1}{2\pi i}\int_{\tilde{C}}\frac{ds}{s}e^{\rho/s}\tilde{F}(s^a)=
\frac{k}{2\pi i}\int_{\tilde{C}'}\frac{ds}{s}e^{(\frac{\tilde{\rho}}{s})^k}
\tilde{F}(s),$$ where $k=\tfrac{1}{a}$, $\tilde{\rho}=\rho^{1/k}$ and $\tilde{C}'$ is the contour where the rays satisfy $|\arg(s)|=|(\pi-\epsilon'')/k|$ for any $\epsilon'' \in
(\epsilon',\tfrac{\pi}{2})$. We will write $\epsilon''=(\pi-\tilde{\epsilon})/2$ for any $\tilde{\epsilon}\in (0,\pi-2\epsilon')$, then $|\arg((\tfrac{\tilde{\rho}}{s})^k)|>\pi/2$ and the integrand in is again damped on the rays. Also, on the rays is $|\arg
(s)|=(\pi+\tilde{\epsilon})/2k=\tfrac{\alpha}{4}\tfrac{\pi+\tilde{\epsilon}}{\pi-\epsilon'}<\alpha/2$ so that $\tilde{F}(s)$ is analytic on the contour, $\tilde{C}'\subset
S(\alpha)$, and the contour integral is therefore well defined.

Equation then shows that $I_{IR}(\tilde{\rho})$ is the Borel transform of $\tilde{F}(s)$ of order $k$ [@balser]. The Borel transform of order $k$ of a formal power series $h(z)=\sum_n
h_nz^n$ is given by ${\cal B}_k (h)(\rho)=\sum_n \tfrac{h_n\rho^n}{\Gamma
  (1+n/k)}$, using the integral representation $\tfrac{1}{\Gamma
  (1+n)}=\tfrac{1}{2\pi i}\int_{\tilde{C}'}\tfrac{ds}{s}e^{1/s}s^n$ it then follows that is the Borel transform of order 