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