Beta Function is Defined for Positive Reals
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Theorem
Let $x, y \in \R$ be real numbers.
Let $\map \Beta {x, y}$ be the Beta function:
- $\ds \map \Beta {x, y} := \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \paren {1 - t}^{y - 1} \rd t$
Then $\map \Beta {x, y}$ exists provided that $x, y > 0$.
Proof
Consider the following inequalities, valid for $0 < t < 1$:
\(\ds t^{x - 1} \paren {1 - t}^{y - 1}\) | \(<\) | \(\ds t^{x - 1}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds t^{x - 1} \paren {1 - t}^{y - 1}\) | \(<\) | \(\ds \paren {1 - t}^{y - 1}\) |
Then:
\(\ds \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \rd t\) | \(=\) | \(\ds \intlimits {\frac {t^x} x} {\mathop \to 0} {\mathop \to 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x - 0\) |
and similarly:
\(\ds \int_{\mathop \to 0}^{\mathop \to 1} \paren {1 - t}^{y - 1} \rd t\) | \(=\) | \(\ds -\intlimits {\frac {\paren {1 - t}^y} y} {\mathop \to 0} {\mathop \to 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds y - 0\) |
The result follows from the Comparison Test for Improper Integral.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 17.7 \ (4)$