SF8

$\Gamma(\frac{1}{2}) = \int_0^\infty x^{-\frac{1}{2}}e^{-x}dx\,$

Let $x = z^2, dx=2zdz\,$ , then

$\int_0^\infty z^{-1} e^{-z^2} 2z dz = 2\int_0^\infty e^{-z^2} dz = 2\frac{\sqrt{\pi}}{2} = \sqrt{\pi}\,$