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.

$\blacksquare$


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