Consider the heat kernel for the hyperbolic plane $\mathbb{H}^2$ and the corresponding heat kernel:
$$k(x,y,t)=\frac{C}{t^{\frac{3}{2}}}\cdot e^{-t/4}\int_{d(x,y)}^{\infty}\frac{re^{-r^2/4t}}{\sqrt{\cosh(r)-\cosh d(x,y)}}dr$$
with appropriate constant $C>0$ and hyperbolic distance $d(x,y)$. Suppose we have $0<\delta <d(x,y).$
I am wondering why the integral on the right hand side is convergent? Is there a simple argument such that this is apparent?
Best wishes.
Here are some rough hints on how to analyze the convergence to get you started.
Since the integrand is continuous on $(d,\infty)$ it suffices to check the integral close to the singularity of the integrand at $x=d$ and in the tail that goes off to $x=\infty$.
For $r\gg 1$ we have $\cosh(r) \sim \frac{e^{r}}{2}$ and if $r\gg d$ then $\cosh(r) \gg \cosh(d)$ so in this limit the integrand becomes
$$f(r) \sim \frac{re^{-\frac{r^2}{4t}}}{\sqrt{e^{r}/2}} = \sqrt{2}re^{-\frac{(r+t)^2}{4t} + \frac{t}{4}}$$
so it reduces to showing that the integral $\int^\infty re^{-\frac{(r+t)^2}{4t}} {\rm d}r$ is convergent to show that the $\infty$-tail of the integral does not lead to any problems.
Close to $r=d$ we can Taylor expand $\cosh(r)$
$$\cosh(r) = \cosh(d) + \sinh(d)(r-d) + \mathcal{O}(r-d)^2$$
and the part of the integral which is close to the singularity becomes
$$\int_d^{d+\epsilon} \frac{re^{-\frac{r^2}{4t}}}{\sqrt{\cosh(r)-\cosh(d)}}{\rm d}r \simeq \int_d^{d+\epsilon} \frac{re^{-\frac{r^2}{4t}}}{\sqrt{\sinh(d)(r-d)}}{\rm d}r \leq \frac{d}{\sinh(d)} \int_d^{d+\epsilon}\frac{{\rm d}r}{\sqrt{r-d}} $$
and if you can show that the last integral is finite you are done.
This analysis can of course be done much more rigorously than what I did above. There are many good questions and answers on the convergence of improper integrals on this site that might be useful if you want to learn more about this.