Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (2.10)

\begin{equation*}
h_{-r}^{-c}={{c}^{-1}}{\exp(c)}\left[1-{\exp(-c)} \right]+{{c}^{-1}}{\exp(c)}r\sum\limits_{i=0}^{n }{{{w}_{i}}{{\beta }_{i}}},
\end{equation*}
where ${{w}_{i}}=1-\sum\limits_{j=0}^{i}{\frac{{{c}^{j}}{\exp(-c)}}{j!}}$, ${{\beta }_{i}}=\frac{i!}{\prod\limits_{j=0}^{i}{\left( j+1-r \right)}}$, and $n\to \infty $.
Proof:
Replacing $c$ with $-c$ into (2.9), we derive
\begin{align*}
h_{-r}^{-c}&={{c}^{-1}}{\exp(c)}\left\{ \left( {-\exp(-c)+1} \right)+r\left( \frac{{1-\exp(-c)}}{0!} \right)\left( \frac{1}{1-r} \right) \right. \label{eq:q12} \\ \notag
& \left. +r\left( \frac{-c \cdot {\exp(-c)}-{\exp(-c)}+1}{1!} \right)\left( \frac{1}{1-r}-\frac{1}{2-r} \right) +\cdots \right\}.
\end{align*}
We can further specify
\begin{align*}
{{w}_{0}}&=\frac{-\exp (-c)+1}{0!}=1-\sum\limits_{j=0}^{0}{\frac{{{c}^{j}}\exp (-c)}{j!}} \\
{{w}_{1}}&=\frac{-c\exp (-c)-\exp (-c)+1}{1!}=1-\sum\limits_{j=0}^{1}{\frac{{{c}^{j}}\exp (-c)}{j!}} \\
& \vdots \\
{{w}_{i}}&=1-\sum\limits_{j=0}^{i}{\frac{{{c}^{j}}\exp (-c)}{j!}},
\end{align*}
and
\begin{align*}
{{\beta }_{0}}&=\frac{1}{1-r}=\frac{0!}{\prod\limits_{j=0}^{0}{\left( j+1-r \right)}} \\
{{\beta }_{1}}&=\frac{1}{1-r}-\frac{1}{2-r}=\frac{1!}{\prod\limits_{j=0}^{1}{\left( j+1-r \right)}} \\
& \vdots \\
{{\beta }_{i}}&=\frac{i!}{\prod\limits_{j=0}^{i}{\left( j+1-r \right)}}.
\end{align*}
Therefore, we conclude the proof.

$\square$

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