= Solution
The integrand in the preceding expression is nonnegative. <Tonelli's theorem> permits squaring as a quadrant integral; then <polar coordinates> give
$$
\begin{aligned}
\Gamma(1/2)^2&=4\int_0^\infty\int_0^\infty e^{-(u^2+v^2)}du\,dv\\
&=4\int_0^{\pi/2}\int_0^\infty e^{-r^2}r\,dr\,d\theta=\pi.
\end{aligned}
$$
The defining <integral> is positive, fixing the sign of the square root. Therefore
$$
\boxed{\Gamma(1/2)^2=\pi,\qquad\Gamma(1/2)=\sqrt\pi.}
$$
This is the positive-quadrant proof of the <Gaussian integral>.
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