for all $t$. The second integral is special to bosons. It is finite for small $t$ so we evaluate it at $t=0$, omitting higher powers in $t$. It is evident from this structure that only the first integral contributes to the terms that are divergent in the UV limit $t\to 0$.

We next compare with the fermion heat kernel where we write $\coth(\pi \lambda) = 1 + \frac{2}{e^{2\pi\lambda}-1}$ and find $$\begin{aligned}
\label{eq:fermiheatkernel}
K^{(s,\nu)}_{\rm fermion}(t) &= \frac{s+{1\over 2}}{4 \pi^2\ell^4_A}e^{-{\frac{t\nu^2}{\ell_A^2}}} \left[ \int_0^\infty e^{-\frac{t\lambda^2}{\ell_A^2}}(\lambda^2 +(s+\frac{1}{2})^2)\lambda d\lambda
+ 2 \int_0^\infty e^{-\frac{t\lambda^2}{\ell_A^2}} \frac{\lambda^2 +(s+\frac{1}{2})^2}{e^{2\pi \lambda}-1}\lambda d\lambda\right]  \nonumber \\
& = \frac{s+{1\over 2}}{8\pi^2\ell^4_A}\left[ {\ell_A^4\over t^2} + {\ell_A^2\over t}\left((s+{1\over 2})^2 - \nu^2\right)\right] + 
\frac{s+\frac{1}{2}}{16\pi^2\ell^4_A}\bigg[ \nu^4 - (s+\frac{1}{2})^2(2\nu^2-\frac{1}{3})+\frac{1}{30}\bigg]~.\end{aligned}$$ Since the first integral is the same in the boson and fermion heat kernels (\[eq:boseheatkernel\], \[eq:fermiheatkernel\]) these expressions have the same divergences in the UV limit $t\to 0$. It is for the same reason that they have the same dependence on conformal dimension appearing through $\nu^2 = (\Delta - \frac{3}{2})^2$. However, the two cases are of course different due to the second integral and this is reflected in the terms that are constant and independent of $\nu$.

We are particularly interested in massless particles since those are the ones that 