free Bose propagator at mass $\mu$. $G(z,\bar{z},\mu)$ can be written in terms of a modified Bessel function [@francesco] $$\label{boseG}
  G(x,\mu)=\int \frac{d^2p}{(2\pi)^2}\frac{e^{i x\cdot p}}{p^2+\mu ^2}=
\frac{K_0(|x|\mu)}{2\pi}.$$ Performing the differentiation we can then write as $$\begin{gathered}
\langle  T_{zz}(sz,s\bar{z})T_{zz}(0,0)\rangle = 
\frac{1}{2\cdot 48\, s^4 z^4}
\int_0^{\infty}d\mu^2\ \tilde{c}(\mu^2)\mu s|z|((\mu^3 s^3
  |z|^3  \\ + 24\mu s|z|)K_0(\mu s|z|)   
  +(8\mu^2 s^2 |z|^2+48)K_1(\mu s|z| )) .
\label{ftilde1}\end{gathered}$$ In the infra-red limit where $s\rightarrow \infty\ $ $K_0(\mu |z|s)$ and $K_1(\mu |z|s)$ have the asymptotic behaviour [@stegun] $e^{-\mu s | z |}$ and the only contribution to comes from the massless limit where $\mu
\rightarrow 0$, hence $$\label{infrar}
\langle  T_{zz}(sz,s\bar{z})T_{zz}(0,0)\rangle  \rightarrow 
\lim_{\epsilon \rightarrow 0}\, \frac{1}{2s^4z^4}
\int_0^\epsilon d\mu^2 \, \tilde{c}(\mu^2)=\frac{c_{IR}}{2s^4z^4}\ \ \ \mbox{for
  }s\rightarrow \infty,$$ so that $c_{IR}=\lim_{\epsilon \rightarrow 0}\int_0^\epsilon d\mu^2 \, \tilde{c}(\mu^2)$. This shows that $c_{UV}\ge c_{IR}$ as the spectral density is positive for a unitary theory, and this is another way of showing the $c$-theorem [@cappelli]. This representation of the central charge using the spectral representation also shows that the central charge measures the number of (massless) degrees of freedom of the CFT. We define the function[^5] $$\left.
\tilde{F}(s)=2z^4s^4\langle
T_{zz}(s z,s \bar{z})T_{zz}(0,0)\rangle\right|_{z=\bar{z}=1}$$ here $s$ and $\tilde{F}(s)$ are dimensionless and $s\in \mathbb{R}_+$. This function then satisfies $$\label{Fgraense}
\tilde{F}(s)\rightarrow \left\{ 
\begin{array}{ll}
c_{UV} 