Reciprocal of Beta Function

Asked Aug 23 '20 at 00:08

Active Aug 23 '20 at 02:27

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We can write, using the definition of Beta function,

$\Gamma(x) \Gamma(y) = \operatorname{B}(x, y) \Gamma(x+y) $

So I was wondering if it is possible to write it the other way, $\Gamma(x+y) =\operatorname{A}(x, y) \Gamma(x) \Gamma(y)$

where I call $\operatorname{A}(x, y)$ as the reciprocal to beta function $\operatorname{B}(x, y)$. My question is what will be the expression for $\operatorname{A}(x,y)$?

edited Aug 23 '20 at 02:27

asked Aug 23 '20 at 00:08

Epsilon

are you looking for an expression in terms of inverse Gamma function? – jimjim Aug 23 '20 at 00:29
If you write$\Gamma(x+y) =\operatorname{A}(x, y) \Gamma(x) \Gamma(y)$, it seems that you look for the reciprocal and not the inverse. – Claude Leibovici Aug 23 '20 at 02:18
Yes, it is reciprocal. Like beta function B(x, y) which has several integral representation. Is there any such representation for A(x, y)? – Epsilon Aug 23 '20 at 02:25
An example is here. – metamorphy Aug 23 '20 at 02:50

Reciprocal of Beta Function

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