Beta Function of Half with Half

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Theorem

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

where $\Beta$ denotes the Beta function.


Proof 1

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:

\(\ds \map \Beta {\dfrac 1 2, \dfrac 1 2}\) \(=\) \(\ds 2 \int_0^{\pi / 2} \paren {\sin \theta}^{2 \times \frac 1 2 - 1} \paren {\cos \theta}^{2 \times \frac 1 2 - 1} \rd \theta\)
\(\ds \) \(=\) \(\ds 2 \int_0^{\pi / 2} \paren {\sin \theta}^0 \paren {\cos \theta}^0 \rd \theta\)
\(\ds \) \(=\) \(\ds 2 \int_0^{\pi / 2} \rd \theta\)
\(\ds \) \(=\) \(\ds 2 \bigintlimits \theta 0 {\pi / 2}\)
\(\ds \) \(=\) \(\ds 2 \paren {\pi / 2 - 0}\)
\(\ds \) \(=\) \(\ds \pi\)

$\blacksquare$


Proof 2

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:

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


Let $t = \sin^2 \theta$.

Then:

$\rd t = 2 \sin \theta \cos \theta \rd \theta$

and:

\(\ds t\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \sin^2 \theta\) \(=\) \(\ds 0\)
\(\ds t\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \sin^2 \theta\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \theta\) \(=\) \(\ds \pi / 2\)


and so:

\(\ds \map \Beta {\dfrac 1 2, \dfrac 1 2}\) \(=\) \(\ds \int_{\mathop \to 0}^{\mathop \to \pi / 2} \frac {2 \sin \theta \cos \theta \rd \theta} {\sin \theta \sqrt {1 - \sin^2 \theta} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \int_{\mathop \to 0}^{\mathop \to \pi / 2} \frac {2 \sin \theta \cos \theta \rd \theta} {\sin \theta \cos \theta}\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \int_{\mathop \to 0}^{\mathop \to \pi / 2} 2 \rd \theta\)
\(\ds \) \(=\) \(\ds \bigintlimits {2 \theta} 0 {\pi / 2}\)
\(\ds \) \(=\) \(\ds 2 \times \pi / 2 - 0\)
\(\ds \) \(=\) \(\ds \pi\)

$\blacksquare$


Proof 3

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 $\dfrac 1 {\sqrt {1 - u^2} }$
\(\ds \) \(=\) \(\ds 2 \times \pi / 2 - 0\)
\(\ds \) \(=\) \(\ds \pi\)

$\blacksquare$