$k$ of $\tilde{F}(s)$ by inserting a series expansion for $\tilde{F}(s)$.

The contour integral in is independent of the contour $\tilde{C}'$, i.e. in the choice of $\tilde{\epsilon}$ for $\tilde{\epsilon}\in
(0,\pi-2\epsilon')$. This shows that equation holds for all $k$ with $F(s) \equiv \tilde{F}(s^{1/k})$ and $\alpha>\pi/k$, because we can always find an $\tilde{\epsilon}>0$ so that $\alpha>(\pi+\tilde{\epsilon})/k$, which again implies that $\tilde{C}'\subset
S(\alpha)$ and $|\arg (\tilde{\rho}/s)^k| > \pi/2$ which again makes the contour integral well defined.

We have then shown that: [*the infra-red limit of a physical quantity, $F_{IR}$, is the limiting value of the Borel transform of $\tilde{F}(s)$, and changing the way in which the scale parameter is introduced amounts to changing the order of the Borel transform. The order $k$ has to satisfy the bound $\alpha>\pi/k$ where $\alpha$ determines the analytic sector of the physical quantity $\tilde{F}(s)$*]{}.

One way of calculating the contour integral in is to insert a series expansion of $\tilde{F}(s)$ around the origin, but because $\tilde{F}(s)$ is only analytic in a sector this series can only be an asymptotic expansion of $\tilde{F}(s)$ [^1]. An analytic function has a unique asymptotic expansion, but an asymptotic series might be the asymptotic expansion of several analytic functions. However in the case where 