Gamma of one half

Remarks

By definition $\Gamma(\tfrac12) = \int_0^\infty t^{-1/2} e^{-t}\, dt$. The substitution $t = x^2$, $dt = 2x\,dx$, turns this into $2\int_0^\infty e^{-x^2}\,dx$, and the Gaussian integral gives $2 \cdot \tfrac{\sqrt{\pi}}{2} = \sqrt{\pi}$. Another proof uses the reflection formula at $z = \tfrac12$: $\Gamma(\tfrac12)^2 = \pi/\sin(\tfrac{\pi}{2}) = \pi$, and $\Gamma(\tfrac12) > 0$. With the functional equation it gives all half-integer values, for example $\Gamma(\tfrac32) = \tfrac{\sqrt\pi}{2}$ and $\Gamma(-\tfrac12) = -2\sqrt{\pi}$. In Mathlib the statement is Complex.Gamma_one_half_eq.

Going deeper

For a student

The value is the gamma-function form of the Gaussian integral $\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}$. It is the reason $\pi$ appears in the volume of balls in odd dimension and in the normalisation of the normal and chi-squared distributions.