Beta Function of Half with Half/Proof 2
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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:
\(\ds \map \Beta {\dfrac 1 2, \dfrac 1 2}\) | \(=\) | \(\ds \int_{\mathop \to 0}^{\mathop \to 1} t^{\frac 1 2 - 1} \paren {1 - t}^{\frac 1 2 - 1} \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_{\mathop \to 0}^{\mathop \to 1} \dfrac {\rd t} {\sqrt {t \paren {1 - t} } }\) |
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_{\mathop \to 0}^{\mathop \to \pi / 2} 2 \rd \theta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \bigintlimits {2 \theta} 0 {\pi / 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \times \pi / 2 - 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \pi\) |
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 17.7 \ (6)$