Solution
= Solution
Let
$$
I=\int_0^\infty e^{-ax^2}\,dx.
$$
Then
$$
I^2=\int_0^\infty\int_0^\infty
e^{-a(x^2+y^2)}\,dx\,dy.
$$
Using <polar coordinates> on the first quadrant,
$$
I^2
=\int_0^{\pi/2}\int_0^\infty e^{-ar^2}r\,dr\,d\theta
=\frac{\pi}{4a}.
$$
Since $I>0$, the <Gaussian integral> is
$$
\boxed{
\int_0^\infty e^{-ax^2}\,dx
=\sqrt{\frac{\pi}{4a}}}.
$$