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*}
$\square$

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