Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (3.8)

\begin{equation*}
\frac{{{\partial }^{n}}\left( h_{s}^{c} \right)}{\partial {{c}^{n}}}=\Delta _{1}^{n}\left[ {{h}^{c}} \right]\left( s \right).
\end{equation*}Proof:
Applying the first partial derivative formula, we can derive the second-order partial derivative of $h_{s}^{c}$ as the second-order forward difference on $s$
\begin{align*}
\frac{{{\partial }^{2}}\left( h_{s}^{c} \right)}{\partial {{c}^{2}}}&=\frac{\partial }{\partial c}\left( h_{s+1}^{c}-h_{s}^{c} \right) \\
& =h_{s+2}^{c}-2h_{s+1}^{c}+h_{s}^{c} \\
& =\Delta _{1}^{2}\left[ {{h}^{c}} \right]\left( s \right).
\end{align*}
For the $n$th-order partial derivative, repeating the first-order forluma $n$ times will result in the $n$th-order forward difference on $s$
\begin{equation*}
\frac{{{\partial }^{n}}\left( h_{s}^{c} \right)}{\partial {{c}^{n}}}=\Delta _{1}^{n}\left[ {{h}^{c}} \right]\left( s \right).
\end{equation*}
$\square$

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