Gamma Function of One Half/Proof 2

Theorem

$\map \Gamma {\dfrac 1 2} = \sqrt \pi$

Proof

From Euler's Reflection Formula:

$\forall z \notin \Z: \map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$

Setting $z = \dfrac 1 2$:

\(\ds \map \Gamma {\frac 1 2} \map \Gamma {\frac 1 2}\)

\(=\)

\(\ds \frac \pi {\map \sin {\frac \pi 2} }\)

\(\ds \)

\(=\)

\(\ds \frac \pi 1\)

Sine of Right Angle

\(\ds \leadsto \ \ \)

\(\ds \map \Gamma {\frac 1 2}\)

\(=\)

\(\ds \pm \sqrt \pi\)

By definition of the gamma function:

$\forall z \in \R_{\ge 0}: \map \Gamma z > 0$

and so the negative square root can be discarded.

Hence:

$\map \Gamma {\dfrac 1 2} = \sqrt \pi$

as required.

$\blacksquare$