\right) .
\nonumber$$ Knowing $F(s)$ exactly the central charges follow from directly, but let us use the relations , in the limit of large $\rho$ where the cut vanishes. For $c_{UV}$ we substitute $s'=ms^{1/4},\
\rho'=\frac{\rho}{m^4}$ in the contour $C_2$. From (\[align1\]) we then get that (for $r_{IR}=\tfrac{1}{m^4}$) $$\label{cUVbose}
  I_{UV}(\rho)=\frac{1}{4\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta\, 
         e^{4i\theta}e^{\rho'e^{2i\theta}}\left(4e^{-2i\theta}
K_1^2(e^{i\theta})+K_0^2(e^{i\theta})+ 
4e^{-i\theta}K_0(e^{i\theta})K_1(e^{i\theta}) \right) .$$ This is not expressible in terms of elementary functions, but it can be calculated using the analytical properties of the Bessel functions. The integration contour can be collapsed into a contour running along the imaginary axis together with an infinitesimal semi-circle around the origin. Setting $ix=e^{i\theta}$ we then get the 3 contributions $$\frac{1}{4\pi i}\left(\int_{-1}^{-\epsilon}\!\!\!dx\!+\!\int_{\epsilon}^{1}\!\!\!dx\!+\!\int_{C_\epsilon}\!\!\!dx\right)
e^{-\rho x^2}\left(-4xK_1(ix)^2+x^3K_0(ix)^2-4ix^2 K_0(ix)K_1(ix) \right).
\nonumber$$ For $\rho\rightarrow \infty$ only the contribution from the infinitesimal semi-circle $C_\epsilon$ survive and here we can insert the asymptotic form for $K_1(z)\sim z^{-1}$ and $K_0(z)\sim -\log z$. In this limit only the first term with $K^2_1$ will contribute as is seen from the asymptotic form. Taking into account that we only integrate over half a circle we then get the well known result $c_{UV}=1$, which is an exact result as the contribution from the cut vanishes in this limit[^6].

To calculate the infra-red central charge we 