& \mbox{for }s\rightarrow 0_{+}, \\
c_{IR} & \mbox{for }s\rightarrow \infty. 
\end{array} \right.$$

Analyticity of $\tilde{F}(s)$
-----------------------------

We will show that $\tilde{F}(s)$ is an analytic continuation of $\tilde{F}(x),\ x\in \mathbb{R}_+$, for $s\in S=S(\pi-\epsilon')$ (with $\epsilon'\ll 1$); to show this write $\tilde{F}(s)=\int_M d\nu f(s,\nu)$. For $\tilde{F}$ to be holomorphic in $S$ then $f(s,\nu)$ must be holomorphic in $S$ for all $\nu \in M$, and both $f$ and $\frac{df}{ds}$ must be integrable over the set $M$. Using (\[ftilde1\]) above we can write $\tilde{F}(s)$ as $$\label{F2}
\tilde{F}(s)=\frac{1}{48}\int_0^\infty\! \! \! d\mu^2 \, \tilde{c}(\mu^2)
 \mu s \left( ({\mu}^{3}{s}^{3}\!+\!24\mu 
s) K_0(\mu s)\!+\!(8 \mu^2 s^{2}\!
+\!48) K_1(\mu s)\right ).$$ $K_{\nu}(z)$ is holomorphic in $S$ so $f(s,\nu)$ is clearly holomorphic in $S$. The modified Bessel functions also satisfy that $|K_\nu(z)|$ is bounded for $|z|\geq \epsilon$ for any $\epsilon\in \mathbb{R}_+$ and $|\arg z|<\frac{\pi}{2}$, which is the case for $z=\mu s$ when $s\in S$. For large values of $|s|$ integrability is ensured by $\int_0^{\infty}d\mu^2\
\tilde{c}(\mu^2)=c_{UV}<\infty$ and the asymptotic behaviour $K_\nu(z)=
\sqrt{\frac{\pi}{2z}}e^{-z}(1+O(z^{-1}))$. Around the origin $K_0(z) z\rightarrow 0$ for $z\rightarrow 0$ and $K_1(z) z\rightarrow 1$ for $z\rightarrow 0$, hence $\tilde{F}(s)$ is integrable. $\frac{d \tilde{F}(s)}{ds}$ is shown to be integrable in a similar way and the analyticity is shown.

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