Beta Function of Half with Half/Proof 3

Theorem

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

where $\Beta$ denotes the Beta function.

Proof

By definition of the Beta function:

$\ds \map \Beta {x, y} := \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \paren {1 - t}^{y - 1} \rd t$

Thus:

 $\ds \map \Beta {\dfrac 1 2, \dfrac 1 2}$ $=$ $\ds \int_{\mathop \to 0}^{\mathop \to 1} t^{\frac 1 2 - 1} \paren {1 - t}^{\frac 1 2 - 1} \rd t$ $\ds$ $=$ $\ds \int_{\mathop \to 0}^{\mathop \to 1} \dfrac {\rd t} {t^{1/2} \paren {1 - t}^{1/2} }$

Let $t = u^2$.

Then:

$\rd t = 2 u \rd u$

and:

 $\ds t$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds u^2$ $=$ $\ds 0$ $\ds t$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds u^2$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds u$ $=$ $\ds 1$

and so:

 $\ds \map \Beta {\dfrac 1 2, \dfrac 1 2}$ $=$ $\ds \int_{\mathop \to 0}^{\mathop \to 1} \frac {2 u \rd u} {u \paren {1 - u^2}^{1/2} }$ Integration by Substitution $\ds$ $=$ $\ds 2 \int_{\mathop \to 0}^{\mathop \to 1} \frac {\rd u} {\sqrt {\paren {1 - u^2} } }$ $\ds$ $=$ $\ds 2 \bigintlimits {\arcsin u} 0 1$ Primitive of $\dfrac 1 {\sqrt {1 - u^2} }$ $\ds$ $=$ $\ds 2 \times \pi / 2 - 0$ $\ds$ $=$ $\ds \pi$

$\blacksquare$