= Solution
Let $I$ be the positive <Gaussian integral>. Its <improper integral> converges because $e^{-x^2}\le e^{-|x|}$ for $|x|\ge1$. Multiplying two copies and using the <polar coordinates> <area element> gives
$$
I^2=\int_{\mathbb R^2}e^{-(x^2+y^2)}\,dx\,dy
=\int_0^{2\pi}\int_0^\infty e^{-r^2}r\,dr\,d\theta
=2\pi\left[-\frac12e^{-r^2}\right]_0^\infty=\pi.
$$
Nonnegativity justifies the product <double integral> and the change of integration order. Since $I>0$,
$$
\boxed{\int_{-\infty}^\infty e^{-x^2}\,dx=\sqrt\pi.}
$$
Back to article page
