ultra-violet and infra-red central charges in the limits $z \rightarrow 0$ and $z \rightarrow \infty$ respectively. In [@cappelli] $\langle T_{\mu \nu}(x)T_{\rho \sigma}(0)\rangle$ was written using the Källen-Lehmann spectral representation: $$\label{correllator}
\langle T_{\mu \nu}(x)T_{\rho \sigma}(0)\rangle = 
\frac{\pi}{3\cdot 16}\int_0^\infty  d\mu^2 \, \tilde{c}(\mu^2)
\int \frac{d^2 p}{(2\pi)^2} e^{ipx}\, \frac{(g_{\mu \nu}p^2-p_\mu
  p_\nu)(g_{\rho \sigma}p^2 - p_\rho p_\sigma)}{p^2+\mu^2} ,
\nonumber$$ and it follows that $$\label{spectral}
\langle T_{zz}(z,\bar{z})T_{zz}(0,0)\rangle = 
\frac{\pi}{3\cdot 16}\int_0^\infty  d\mu^2 \, \tilde{c}(\mu^2)
\int \frac{d^2 p}{(2\pi)^2} \frac{e^{\frac{i}{2}(p\bar{z}+\bar{p}z)}}
{p\bar{p}+\mu^2} \, \bar{p}^4,$$ where we use the usual complex variables[^3] $z,\bar{z}$, and $\tilde{c}(\mu^2)d\mu^2$ is the spectral density which represents the density in degrees of freedom of the quantum field theory at the mass $\mu$. If we scale $z,\bar{z}$ by a positive real dimensionless parameter $s$ ($s>0$) we get $$\label{Fscaled}
\langle T_{zz}(s z,s\bar{z})T_{zz}(0,0)\rangle=
\frac{\pi}{3\cdot 16}\int_0^\infty d\mu^2 \, \tilde{c}(\mu^2)
\int \frac{d^2 q}{(2\pi)^2} \frac{e^{\frac{i}{2}(q\bar{z}+\bar{q}z)}}
{q\bar{q}+s^2\mu^2} \, \frac{\bar{q}^4}{s^4}.$$ In the ultra-violet limit where $s\rightarrow 0$ then $\frac{\bar{q}^4}{q\bar{q}+s^2\mu^2}\rightarrow \frac{\bar{q}^4}{q\bar{q}}$ and becomes[^4] $$\label{ultrav}
\langle  T_{zz}(sz,s\bar{z})T_{zz}(0,0)\rangle \rightarrow 
  \frac{1}{2s^4z^4}\int_0^\infty d\mu^2 \, \tilde{c}(\mu^2)=
  \frac{c_{UV}}{2s^4z^4} \ \ \mbox{for} \ s \rightarrow 0.$$ The ultra-violet central charge is therefore $c_{UV}=\int_0^\infty d\mu \, \tilde{c}(\mu)$. To calculate the infra-red limit we first note that $$\label{frie}
 \int \frac{d^2 q}{(2\pi)^2} \frac{e^{\frac{i}{2}(q\bar{z}+\bar{q}z)}}
{q\bar{q}+s^2\mu^2} \, \bar{q}^4 = 2^4 \left(\frac{\p}{\p z}\right)^4 
G(z,\bar{z},s\mu),$$ where $G(z,\bar{z},\mu)$ is equal to the 