Can anyone clarify why this is? The question is to prove: $$I=\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}$$
Let $$I_r=\int_{-r}^{r}e^{-x^2}dx \implies I_{r^2}=\int_{-r}^{r}e^{-x^2}dx\int_{-r}^{r}e^{-y^2}dy=\iint_{[-r,r]^2} e^{-x^2-y^2}dxdy$$$x^2+y^2\leq r^2=D_r$
$$K_r=\iint_{D_r}{e^{-x^2-y^2}}dxdy$$
$$0\leq I_r-K_r=\int_{Q_r \setminus D_r}e^{-x^2-y^2}dxdy\leq e^{-r^2}\mu(Q_r)=e^{-r^2}4r^2\rightarrow0; r\rightarrow \infty$$. The rest is clear to me just this last inequality: $\leq e^{-r^2}\mu(Q_r)$ $;\mu(A)$-Measure of A. I'll type out the rest of the proof if need be, for educational purposes..
 A: I'm not sure if you are getting at the usual proof of this, which is
$$
\begin{align}
\left(\int_{-R}^R e^{-x^2}\,\mathrm{d}x\right)^2
&=\int_{-R}^R e^{-x^2}\,\mathrm{d}x\int_{-R}^R e^{-y^2}\,\mathrm{d}y\\
&=\int_{-R}^R\int_{-R}^R e^{-x^2-y^2}\,\mathrm{d}x\,\mathrm{d}y
\end{align}
$$
and by sandwiching the square between two circles,
$$
\int_0^{2\pi}\int_0^Re^{-r^2}r\,\mathrm{d}r\,\mathrm{d}\theta\le\int_{-R}^R\int_{-R}^R e^{-x^2-y^2}\,\mathrm{d}x\,\mathrm{d}y\le\int_0^{2\pi}\int_0^{\sqrt2R}e^{-r^2}r\,\mathrm{d}r\,\mathrm{d}\theta
$$
which combined with
$$
\int_0^{2\pi}\int_0^Re^{-r^2}r\,\mathrm{d}r\,\mathrm{d}\theta=\pi\left(1-e^{-R^2}\right)
$$
and the Squeeze Theorem yields
$$
\left(\int_{-\infty}^\infty e^{-x^2}\,\mathrm{d}x\right)^2=\pi
$$
A: I would go for some integral mean value theorem:
$$
\int\limits_{Q_r \setminus D_r} e^{-x^2 -y^2} \,dx \,dy = e^{-(X^2+Y^2)} \int\limits_{Q_r \setminus D_r} \,dx \,dy = e^{-(X^2+Y^2)} \mu (Q_r \setminus D_r) 
\le e^{-(X^2 + Y^2)} \mu(Q_r)
$$
where $(X,Y) \in Q_r \setminus D_r$ ("square with circular hole") and the minimum of $X^2+Y^2$ is at $\partial D_r$ with $r^2$, so 
$$
e^{-(X^2 + Y^2)} \le e^{-r^2}
$$ 
for all $(X,Y) \in Q_r \setminus D_r$.
This leads then to 
$$
e^{-(X^2 + Y^2)} \mu(Q_r) \le e^{-r^2} \mu(Q_r)
$$
