Evaluate the integral $\int_{-\infty}^{+\infty}e^{-x^2}dx$

Let $A=\int_{-\infty}^{+\infty}e^{-x^2}dx$, and note also that $A=\int_{-\infty}^{+\infty}e^{-y^2}dy$.  Hence 
\[A^2=\int_{-\infty}^{+\infty}e^{-x^2}dx \times\int_{-\infty}^{+\infty}e^{-y^2}dy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy.\]
Thus using polar coordinates we have 
$$A^2=\int_{0}^{\infty}\int_0^{2\pi}e^{-r^2}rd\theta dr=2\pi\int_{0}^{\infty}e^{-r^2}r dr=\frac{2\pi}{-2}\int_{0}^{\infty}e^{-r^2}(-2r)$$ 
\[=-\pi e^{-r^2}\mid_0^{\infty}=-\pi(0-1)=\pi.\]
Hence 
\[A^2=\pi      \Rightarrow   A=\sqrt{\pi},\]
and therefore 
\[\int_{-\infty}^{+\infty}e^{-x^2}dx=\sqrt{\pi}.\]