A book on probability theory I am reading asserts the following: for $x, y \in \mathbb{R}^n, t \geq 0$, consider the function $h_t(x, y) = \frac{1}{t^{n/2}}e^{-\frac{r^2}{4t}}$, where $r = |x - y|$. Then the following integral is finite: $$ H(x, y) = \int_0^{r^2} h_t(x, y)dt.$$
It is not clear to me how they are calculating it. Furthermore, it seems from the discussion (I am not completely sure here) that the final answer is something like (constant)$r^{\frac{n - 2}{2}}$. I would really appreciate a little help with this. Thanks in advance!
Set $u=\frac{1}{t}$, then (assuming $r>0$) $$ \int_0^{r^2}t^{-\frac{n}{2}}e^{-\frac{r^2}{4t}}\;dt=\int_{r^{-2}}^{\infty}u^{\frac{n}{2}-2}e^{-\frac{r^2u}{4}}\;du $$ which is finite because of the exponential decay at infinity. Then setting $v=\frac{r^2}{4}u$, we have $$ \int_{r^{-2}}^{\infty}u^{\frac{n}{2}-2}e^{-\frac{r^2u}{4}}\;du=\frac{4^{\frac{n}{2}-1}}{r^{n-2}}\int_{\frac{1}{4}}^{\infty}v^{\frac{n}{2}-2}e^{-v}\;dv=\frac{4^{\frac{n}{2}-1}}{r^{n-2}}\Gamma\Big(\frac{n}{2}-1,\frac{1}{4}\Big)$$ where $\Gamma(s,x)$ is the (upper) incomplete Gamma function.