Beta Function of Half with Half/Proof 1

Theorem

$\map \Beta {\dfrac 1 2, \dfrac 1 2} = \pi$

where $\Beta$ denotes the Beta function.

By definition of the Beta function:

$\ds \map \Beta {x, y} := 2 \int_0^{\pi / 2} \paren {\sin \theta}^{2 x - 1} \paren {\cos \theta}^{2 y - 1} \rd \theta$

Thus:

$\ds \map \Beta {\dfrac 1 2, \dfrac 1 2}$

$=$

$\ds 2 \int_0^{\pi / 2} \paren {\sin \theta}^{2 \times \frac 1 2 - 1} \paren {\cos \theta}^{2 \times \frac 1 2 - 1} \rd \theta$

$\ds $

$=$

$\ds 2 \int_0^{\pi / 2} \paren {\sin \theta}^0 \paren {\cos \theta}^0 \rd \theta$

$\ds $

$=$

$\ds 2 \int_0^{\pi / 2} \rd \theta$

$\ds $

$=$

$\ds 2 \bigintlimits \theta 0 {\pi / 2}$

$\ds $

$=$

$\ds 2 \paren {\pi / 2 - 0}$

$\ds $

$=$

$\ds \pi$

$\blacksquare$