= Solution
Let $J=\int_0^\infty e^{-x^2}\,dx$. It is positive and finite: on $(0,1)$ the integrand is at most one, and on $[1,\infty)$ it is at most $e^{-x}$. Since the planar integrand is nonnegative and measurable, <Tonelli's theorem> gives
$$
J^2=\int_{(0,\infty)^2}e^{-(x^2+y^2)}\,dx\,dy.
$$
The <change of variables formula> in <polar coordinates>, with $x=r\cos\theta$, $y=r\sin\theta$ and absolute <Jacobian determinant> $r$, yields
$$
J^2=\int_0^{\pi/2}\int_0^\infty e^{-r^2}r\,dr\,d\theta
=\frac\pi2\cdot\frac12=\frac\pi4.
$$
One can apply the formula first on $\delta<r<R$, $0<\theta<\pi/2$, where the coordinate map is a <diffeomorphism>, and then exhaust the quadrant using the <monotone convergence theorem>. Taking the positive square root proves the half-line <Gaussian integral>:
$$
\boxed{J=\frac{\sqrt\pi}{2}.}
$$
Here $dx$ is understood in the one-dimensional integral; it is missing from the printed question.
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