Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (2.16)

$$\begin{equation*} {{\beta }_{0}}>{{\beta }_{1}}>\cdots >{{\beta }_{n}}\to \frac{\Gamma \left( 1-r \right)}{{{n}^{1-r}}} \end{equation*}$$ Proof:
$$\begin{align*} {{\beta }_{n}}&=\frac{n!}{\left( 1-r \right)\left( 2-r \right)\cdots \left( n+1-r \right)} \\ & >\frac{n!}{\left( 1-r \right)\left( 2-r \right)\cdots \left( n+1-r \right)}\left( \frac{1}{1+\tfrac{1-r}{n+1}} \right) \\ & =\frac{n!}{\left( 1-r \right)\left( 2-r \right)\cdots \left( n+1-r \right)}\left[ \frac{n+1}{\left( n+1 \right)+\left( 1-r \right)} \right] \\ & =\frac{\left( n+1 \right)!}{\left( 1-r \right)\left( 2-r \right)\cdots \left( n+2-r \right)} \\ & ={{\beta }_{n+1}} \end{align*}$$
Thus,
$$\begin{equation*} {{\beta }_{0}}>{{\beta }_{1}}>\cdots >{{\beta }_{n}}\to \frac{\Gamma \left( 1-r \right)}{{{n}^{1-r}}}. \end{equation*}$$
$\square$

Welcome all math lovers!