to the mass of the field and in the context of AdS$_4$ it is $\Delta$ that provides the simplest representation of this parameter. Crucially, the Plancherel measure $\mu_s(\lambda)$ for the integration over the continuous eigenvalues $\lambda$ is different for bosons[^3] $$\mu_ s(\lambda) = (s+{1\over 2}) {\lambda^2 +(s +{1\over2})^2 \over 4\pi^2}  \lambda \tanh(\pi \lambda)~,$$ and for fermions $$\mu_ s(\lambda) = (s+{1\over 2}) {\lambda^2 +(s +{1\over2})^2 \over 4\pi^2}  \lambda \coth(\pi \lambda)~.$$ The distinction between bosons and fermions is inconsequential in the UV region where $\lambda\to\infty$ since then both $\tanh(\pi \lambda)\to 1$ and $\coth(\pi \lambda)\to 1$. It is instructive to evaluate the heat kernel (\[eq:KAmaster\]) such that this common feature is manifest. For bosons we write $\tanh(\pi \lambda) = 1- \frac{2}{e^{2\pi\lambda}+1}$ and then find $$\begin{aligned}
\label{eq:boseheatkernel}
K^{(s,\nu)}_{\rm boson}(t) &= \frac{s+{1\over 2}}{4 \pi^2\ell^4_A}e^{-{\frac{t\nu^2}{\ell_A^2}}} \left[ \int_0^\infty e^{-\frac{t\lambda^2}{\ell_A^2}}(\lambda^2 +(s+\frac{1}{2})^2)\lambda d\lambda
- 2 \int_0^\infty e^{-\frac{t\lambda^2}{\ell_A^2}} \frac{\lambda^2 +(s+\frac{1}{2})^2}{e^{2\pi \lambda}+1}\lambda d\lambda\right] \nonumber \\ 
& =  {s+\frac{1}{2}\over 8\pi^2\ell^4_A} e^{-\frac{t\nu^2}{\ell^2_A}}\left( {\ell^4_A\over t^2} + {\ell^2_A\over t}(s+{1\over 2})^2\right)
- {s+\frac{1}{2}\over 8\pi^2\ell^4_A}\left( {7\over 480} + {(s+{1\over 2})^2\over 12}\right) \nonumber \\
& = \frac{s+{1\over 2}}{8\pi^2\ell^4_A}\left[ {\ell_A^4\over t^2} + {\ell_A^2\over t}\left((s+{1\over 2})^2 - \nu^2\right)\right] + 
\frac{s+\frac{1}{2}}{16\pi^2\ell^4_A}\bigg[ \nu^4 - (s+\frac{1}{2})^2(2\nu^2+\frac{1}{6})-\frac{7}{240}\bigg]~.\end{aligned}$$ The first integral contains the UV terms that are common to bosons and fermions and is elementary 