Beta Function of x with y+m+1

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Theorem

Let $\Beta \left({x, y}\right)$ denote the Beta function.

Then:

$\Beta \left({x, y}\right) = \dfrac {\Gamma_m \left({y}\right) m^x} {\Gamma_m \left({x + y}\right)} \Beta \left({x, y + m + 1}\right)$

where $\Gamma_m$ is the partial Gamma function:

$\displaystyle \Gamma_m \left({y}\right) := \frac {m^y m!} {y \left({y + 1}\right) \left({y + 2}\right) \cdots \left({y + m}\right)}$


Proof

\(\displaystyle \Beta \left({x, y}\right)\) \(=\) \(\displaystyle \dfrac {x + y} y \Beta \left({x, y + 1}\right)\) Beta Function of x with y+1 by x+y over y
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\left({x + y}\right)^{\overline {m + 1} } } {y^{\overline {m + 1} } } \Beta \left({x, y + m + 1}\right)\)


Also:

\(\displaystyle \Beta \left({y, m + 1}\right)\) \(=\) \(\displaystyle \Beta \left({y, m}\right) \dfrac m {y + m}\) Beta Function of x with y+1 by x+y over y
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {m!} {\left({y + 1}\right)^{\overline m} } \Beta \left({y, 1}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {m!} {y^{\overline {m + 1} } }\)


and:

\(\displaystyle \Beta \left({x + y, m + 1}\right)\) \(=\) \(\displaystyle \Beta \left({x + y, m}\right) \dfrac m {x + y + m}\) Beta Function of x with y+1 by x+y over y

}}

\(\displaystyle \) \(=\) \(\displaystyle \dfrac {m!} {\left({x + y}\right)^{\overline {m + 1} } }\)


Hence:

\(\displaystyle \Beta \left({x, y}\right)\) \(=\) \(\displaystyle \dfrac {\Beta \left({y, m + 1}\right)} {\Beta \left({x + y, m + 1}\right)} \Beta \left({x, y + m + 1}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\Gamma_m \left({y}\right)} {m^y} \dfrac {m^{x + y} } {\Gamma_m \left({x + y}\right)} \Beta \left({x, y + m + 1}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\Gamma_m \left({y}\right) m^x} {\Gamma_m \left({x + y}\right)} \Beta \left({x, y + m + 1}\right)\)

$\blacksquare$


Sources

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