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Using only the definition

$$B(x, y) = \int_0^1 t^{x-1}(1-t)^{y-1}dt$$

for the Beta function $B(x, y)$, it's symmetry $B(x,y) = B(y,x)$ aswell as the fact that $(x + y)B(x + 1, y) = xB(x, y) \space\space \forall x, y > 0$ , is there a way to show that:

$$ B(m, n) = \frac{(n-1)!(m-1)!}{(n+m-1)!} \space\space\space \forall m, n \in \mathbb{N}$$

When we already presuppose other formulas and relations between the Beta function and the Gamma function which interpolates $n!$, it's probably easy (or easier at least) to argue that this is the case. But how can this be shown when only using said presumptions? Thanks in advance!

Américo Tavares
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moran
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1 Answers1

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Yes, you can do it directly using integration by parts to get to $(x+y)B(x+1,y)=x B(x,y)$.

With this, you then actually want to go the other way: write the formula as $$ B(x,y) = \frac{x-1}{x+y-1} B(x-1,y). $$ Now you iterate/use induction to get to the formula $$ B(m,n) = \frac{(m-1)(m-2) \dotsm (m-(m-1))}{(m+n-1)(m+n-2)\dotsm (m+n-(m-1))} B(0,n) = \frac{(m-1)!}{(m+n-1)!/(n-2)!} B(0,n), $$ and you can do $B(0,n)$ directly to get the final $(n-1)$.

Chappers
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