I'm reading the proof of solution to the heat equation (homogeneous problem, Theorem 1, pg. 48) and the last step is left to the reader, to verify that the integral $\frac{1}{t^{\frac{n}{2}}}\int_{\delta}^{+\infty} e^{-\frac{-r^2}{16t}}r^{n-1}dr \to 0$ as $t\to0^+$.
If I substitute $-\frac{-r^2}{16t}$ with $u$ I don't see where from there.
On
Let $r^2/16t=s$ or $r=4\sqrt{st}$. Then$$ \frac{1}{t^{n/2}}\int_{\delta}^{\infty}e^{-r^2/16t}r^{n-1}dr = \frac{1}{t^{n/2}}\int_{\delta^2/16t}^{\infty}e^{-s}(4\sqrt{st})^{n-1}2\sqrt{t}\frac{ds}{\sqrt{s}} \\ = \frac{(4\sqrt{t})^{n-1}2\sqrt{t}}{t^{n/2}}\int_{\delta^2/16t}^{\infty}e^{-s}s^{-1+n/2}ds \\ = 4^{n+1}\int_{\delta^2/16t}^{\infty}e^{-s}s^{-1+n/2}ds. $$
A change of variable $u= \frac{r}{\sqrt{t}}$ reduces the integral to $$\int_{\delta /\sqrt{t}}^{\infty} e^{-u^{2}/16}u^{n-1}\, du$$ Now if you know the dominated convergence theorem, then you are done. If not, then we simply borrow a factor from the exponential. That is for every $u\in [\frac{\delta}{t},\infty)$ we have that
$$e^{-u^{2}/16}=e^{-u^{2}/32}\, e^{-u^{2}/32} \leq e^{-\delta^{2}/32t} \, e^{-u^{2}/32}$$
This implies that your integral is bounded from above by a constant times $e^{-\delta^{2}/32t}$ which tends to zero, as $t\rightarrow 0 +$