Beta Function expressed using Gamma Functions

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Theorem

Let $\map \Beta {x, y}$ denote the Beta function.

Then:

$\map \Beta {x, y} = \dfrac {\map \Gamma x \, \map \Gamma y} {\map \Gamma {x + y} }$

where $\Gamma$ is the Gamma function:


Proof

From Beta Function of x with y+m+1:

$\map \Beta {x, y} = \dfrac {\map {\Gamma_m} y \, m^x} {\map {\Gamma_m} {x + y} } \map \Beta {x, y + m + 1}$

where $\Gamma_m$ is the partial Gamma function:

$\map {\Gamma_m} y := \dfrac {m^y m!} {y \paren {y + 1} \paren {y + 2} \dotsm \paren {y + m} }$


From Partial Gamma Function expressed as Integral:

\(\ds \map {\Gamma_m} x\) \(=\) \(\ds m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t\)
\(\ds \) \(=\) \(\ds m^x \, \map \Beta {x, m + 1}\)


Thus:

$\ds \lim_{m \mathop \to \infty} m^x \, \map \Beta {x, m + 1} = \map \Gamma x$


As $m^x$ is monotone, it does not matter if $m$ is integer or real.

Thus:

$\ds \lim_{m \mathop \to \infty} \paren {m + y}^x \, \map \Beta {x, m + y + 1} = \map \Gamma x$


Hence:

\(\ds \map \Beta {x, y}\) \(=\) \(\ds \lim_{m \mathop \to \infty} \dfrac {\map {\Gamma_m} y \, m^x} {\map {\Gamma_m} {x + y} } \map \Beta {x, y + m + 1}\)
\(\ds \) \(=\) \(\ds \lim_{m \mathop \to \infty} \dfrac {\map {\Gamma_m} y \, m^x} {\map {\Gamma_m} {x + y} } \frac {\map \Gamma x} {\paren {m + y}^x}\)
\(\ds \) \(=\) \(\ds \lim_{m \mathop \to \infty} \dfrac {\map {\Gamma_m} y \, \map \Gamma x} {\map {\Gamma_m} {x + y} }\) as $\ds \lim_{m \mathop \to \infty} \frac {m^x} {\paren {m + y}^x} = 1$
\(\ds \) \(=\) \(\ds \dfrac {\map \Gamma y \, \map \Gamma x} {\map \Gamma {x + y} }\)

$\blacksquare$


Sources

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