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I cannot understand the first and second equalities of the following.

$$\frac{C}{t^{n/2}} \int_{\mathbb{R}^n - B(x^0, \delta)} e^{-|y-x^0|^2 /{16t}}~dy = \frac{C}{t^{n/2}} \int_{\delta}^{\infty} e^{-r^2 /{16t}}~r^{n-1}~dr \rightarrow 0~~\text{as}~~t \rightarrow 0^+.$$

Obviously, the substitution is $y-x^0=r$ but I quite don't get the rest. Any help is much appreciated.

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    Actually, the substitution used is to pass to [spherical coordinates](https://en.wikipedia.org/wiki/N-sphere#Spherical_coordinates). – 2017-01-11
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    Right. But I don't understand the $r^{n-1}$ in the integrand. Of course, the limits are apparent being the complement of the open ball. – 2017-01-11
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    $r^{n-1}$ comes from Jacobian of the coordinate transformation. See, for example, [this](http://math.stackexchange.com/a/395114/250955). – 2017-01-11
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    Thank you. So for the $\rightarrow 0$ part, does it follow from the dominating term, the $e$ ? – 2017-01-11
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    If you use substitution $s=r/(4\sqrt{t})$, you get a factor of $t^{n/2}$ in front of integral that cancels with $t^{-n/2}$, so I don't think it actually tends to zero. – 2017-01-11

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The change of variable $r=4\sqrt{t}\,s$ (suggested by Zoran) transforms the last integral into $$ 4^n\,C\int_{\delta/(4\sqrt{t})}^\infty e^{–s^2}\,s^{n-1}\,ds. $$ Since $\int_0^\infty e^{–s^2}\,s^{n-1}\,ds$ is convergent, $\int_z^\infty e^{–s^2}\,s^{n-1}\,ds$ converges to $0$ as $z\to\infty$.

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    Thank you for the explanation. – 2017-01-12