Beta Function of x with y+m+1

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Theorem

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

Then:

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

where $\Gamma_m$ is the partial Gamma function:

\(\ds \map {\Gamma_m} y\) \(=\) \(\ds \dfrac {m^y m!} {y \paren {y + 1} \paren {y + 2} \cdots \paren {y + m} }\)
\(\ds \) \(=\) \(\ds \dfrac {m^y m!} { y^{\overline {m + 1} } }\)

Proof

\(\ds \map \Beta {x, y}\) \(=\) \(\ds \dfrac {x + y} y \map \Beta {x, y + 1}\) Beta Function of x with y+1 by x+y over y
\(\ds \) \(=\) \(\ds \paren {\dfrac {x + y} y } \paren {\dfrac {x + y + 1} {y + 1} } \map \Beta {x, y + 1 + 1}\) applying Beta Function of x with y+1 by x+y over y a 2nd time
\(\ds \) \(=\) \(\ds \dfrac {\paren {x + y}^{\overline {m + 1} } } {y^{\overline {m + 1} } } \map \Beta {x, y + m + 1}\) after $m + 1$ times: rising factorial

Hence:

\(\ds \map \Beta {x, y}\) \(=\) \(\ds \dfrac {\paren {x + y}^{\overline {m + 1} } } {y^{\overline {m + 1} } } \map \Beta {x, y + m + 1}\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {\dfrac {m! m^x m^y} {y^{\overline {m + 1} } } } } {\paren {\dfrac {m! m^x m^y} {\paren {x + y}^{\overline {m + 1} } } } }\) multiplying by $1$ and rearranging
\(\ds \) \(=\) \(\ds \dfrac {\paren {\dfrac {m! m^y} {y^{\overline {m + 1} } } } m^x } {\paren {\dfrac {m! m^x m^y} {\paren {x + y}^{\overline {m + 1} } } } }\) isolating $m^x$
\(\ds \) \(=\) \(\ds \dfrac {\map {\Gamma_m} y m^x} {\map {\Gamma_m} {x + y} } \map \Beta {x, y + m + 1}\)

$\blacksquare$

Sources