Beta Function of Half with Half/Proof 3
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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:
\(\ds \map \Beta {\dfrac 1 2, \dfrac 1 2}\) | \(=\) | \(\ds \int_{\mathop \to 0}^{\mathop \to 1} t^{\frac 1 2 - 1} \paren {1 - t}^{\frac 1 2 - 1} \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_{\mathop \to 0}^{\mathop \to 1} \dfrac {\rd t} {t^{1/2} \paren {1 - t}^{1/2} }\) |
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 $\dfrac 1 {\sqrt {1 - u^2} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \times \pi / 2 - 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \pi\) |
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $43$