I am reading this, and don't understand how this is reached $$\frac{\Gamma(\alpha)}{\Gamma(\alpha+n)}=\frac{(\alpha+n)\beta(\alpha+1,n)}{\alpha\Gamma(n)}$$

The following relation mentioned on the Wikipedia,$$\beta(\alpha,n)=\frac{\Gamma(\alpha)\Gamma(n)}{\Gamma(\alpha+n)} \Rightarrow \frac{\Gamma(\alpha)}{\Gamma(\alpha+n)}=\frac{\beta(\alpha,n)}{\Gamma(n)}$$

so given the above two, this must be true, $$\frac{(\alpha+n)\beta(\alpha+1,n)}{\alpha\Gamma(n)}=\frac{\beta(\alpha,n)}{\Gamma(n)}$$ which means this also must be true, $$\frac{(\alpha+n)\beta(\alpha+1,n)}{\alpha}=\beta(\alpha,n)$$. I can't see why this is true, given the definition of the $\beta$ function.

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    $\begingroup$ Use the fact that $\Gamma(n)=(n-1)\Gamma(n-1)$ on the expanded $\beta(\alpha+1,n)$ $\endgroup$ Jul 26, 2016 at 15:26
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    $\begingroup$ This could be on topic on the Mathematics SE site, but is also on topic here, IMO. We can defer to the OP's preferences. $\endgroup$ Jul 26, 2016 at 16:13

1 Answer 1


The last statement you have \begin{equation} \frac{(\alpha+n)\beta(\alpha+1,n)}{\alpha} = \beta(\alpha,n) \end{equation} is true if you expand $\beta(\alpha+1,n)$ as $\frac{\Gamma(\alpha+1)\Gamma(n)}{\Gamma(\alpha+1+n)}$ then follow @ArtificialBreeze's suggestion. Here it is

\begin{split} \frac{(\alpha+n)\beta(\alpha+1,n)}{\alpha} & = \frac{(\alpha+n)\color{red}{\Gamma(\alpha+1)}\Gamma(n)}{\alpha\color{blue}{\Gamma(\alpha+1+n)}} \\ &= \frac{(\alpha+n)\color{red}{[\alpha\times\Gamma(\alpha)]}\Gamma(n)}{\alpha\color{blue}{[(\alpha+n)\Gamma(\alpha+n)]}} \\ \end{split}

Cancelling out terms will result in $\frac{\Gamma(\alpha)\Gamma(n)}{\Gamma(\alpha+n)}$, which is what you have above.


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