Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (2.8)

\begin{equation*}
h_{k-r}^{c}={{c}^{-k-1}}{\exp(-c)}{{I}^{\left( 0 \right)}}\left( c,k \right)+{{c}^{-k-1}}{\exp(-c)}r\sum\limits_{i=0}^{\infty }{\frac{{\left( -1 \right)}^{i}{{I}^{\left( 0 \right)}}\left( c,k+i \right)}{\prod\limits_{j=0}^{i}{\left( k+1+j-r \right)}}}
\end{equation*} Proof:
Given (2.6) and (2.7) , we know
\begin{align*}
h_{k-r}^{c}={{c}^{-k-1}}{\exp(-c)}&\left\{ {{r}^{0}}{{I}^{(0)}}\left( c,k \right)+{{r}^{1}}\sum\limits_{i=0}^{\infty }{\left[ \frac{{{\left( -1 \right)}^{i}}}{0!}\frac{{{d}^{(0)}}}{dk}\left( \frac{1}{\prod\limits_{j=0}^{i}{\left( k+1+j \right)}} \right){{I}^{(0)}}\left( c,k+i \right) \right]} \right. \\
& \left. +{{r}^{2}}\sum\limits_{i=0}^{\infty }{\left[ \frac{{{\left( -1 \right)}^{i+1}}}{\left( 1 \right)!}\frac{{{d}^{(1)}}}{dk}\left( \frac{1}{\prod\limits_{j=0}^{i}{\left( k+1+j \right)}} \right){{I}^{(0)}}\left( c,k+i \right) \right]}+\cdots \right\}.
\end{align*}
Summing the series by $i$, for example $i=0$, we derive
\begin{align*}
& {{c}^{-k-1}}\exp \left( -c \right)r{{I}^{\left( 0 \right)}}\left( c,k \right)\left[ \frac{{{\left( -r \right)}^{0}}}{0!}\frac{{{d}^{\left( 0 \right)}}}{dk}\left( \frac{1}{k+1} \right)+\frac{{{\left( -r \right)}^{1}}}{1!}\frac{{{d}^{\left( 1 \right)}}}{dk}\left( \frac{1}{k+1} \right)+\cdots \right] \\
=&{{c}^{-k-1}}\exp \left( -c \right)r{{I}^{\left( 0 \right)}}\left( c,k \right)\left[ \frac{{{\left( -r \right)}^{0}}}{0!}\frac{{{\left( -1 \right)}^{0}}0!}{k+1}+\frac{{{\left( -r \right)}^{1}}}{1!}\frac{{{\left( -1 \right)}^{1}}1!}{{{\left( k+1 \right)}^{2}}}+\frac{{{\left( -r \right)}^{2}}}{2!}\frac{{{\left( -1 \right)}^{2}}2!}{{{\left( k+1 \right)}^{2}}}+\cdots \right]\\
=&\frac{{{c}^{-k-1}}\exp \left( -c \right)r{{I}^{\left( 0 \right)}}\left( c,k \right)}{k+1}\left[ {{\left( \frac{r}{k+1} \right)}^{0}}+{{\left( \frac{r}{k+1} \right)}^{1}}+{{\left( \frac{r}{k+1} \right)}^{2}}+\cdots \right] \\
=&\frac{{{c}^{-k-1}}\exp \left( -c \right)r{{I}^{\left( 0 \right)}}\left( c,k \right)}{k+1}\frac{k+1}{k+1-r} \\
=&\frac{{{c}^{-k-1}}\exp \left( -c \right)r{{I}^{\left( 0 \right)}}\left( c,k \right)}{k+1-r}.
\end{align*}
For $i=1$ and $2$, we derive
\begin{align*}
&{{\left( -1 \right)}^{1}}{{c}^{-k-1}}\exp \left( -c \right)r{{I}^{\left( 0 \right)}}\left( c,k+1 \right)\left[ \frac{1}{1!\left( k+1-r \right)}-\frac{1}{1!\left( k+2-r \right)} \right], \\
&{{\left( -1 \right)}^{2}}{{c}^{-k-1}}\exp \left( -c \right)r{{I}^{\left( 0 \right)}}\left( c,k+2 \right)\left[ \frac{1}{2!\left( k+1-r \right)}-\frac{2}{2!\left( k+2-r \right)}+\frac{1}{2!\left( k+3-r \right)} \right],
\end{align*}
respectively.

Bringing the above results back to $h_{k-r}^{c}$,
\begin{align*}
h_{k-r}^{c}=&{{c}^{-k-1}}{\exp(-c)}{{I}^{\left( 0 \right)}}\left( c,k \right)+{{c}^{-k-1}}{\exp(-c)}r\left\{ {{I}^{\left( 0 \right)}}\left( c,k \right)\left( \frac{1}{0!\left( k+1-r \right)} \right)\right. \label{eq:aq1}\\
-&{{I}^{\left( 0 \right)}}\left( c,k+1 \right)\left( \frac{1}{1!\left( k+1-r \right)}-\frac{1}{1!\left( k+2-r \right)} \right) \notag \\
+& \left.{{I}^{\left( 0 \right)}}\left( c,k+2 \right)\left( \frac{1}{2!\left( k+1-r \right)}-\frac{2}{2!\left( k+2-r \right)}+\frac{1}{2!\left( k+3-r \right)} \right)+\cdots \right\}, \notag
\end{align*}
where
\begin{align*}
& \frac{1}{1!\left( k+1-r \right)}-\frac{1}{1!\left( k+2-r \right)}=\frac{1}{\left( k+1-r \right)\left( k+2-r \right)}, \\
& \frac{1}{2!\left( k+1-r \right)}-\frac{2}{2!\left( k+2-r \right)}+\frac{1}{2!\left( k+3-r \right)}=\frac{1}{\left( k+1-r \right)\left( k+2-r \right)\left( k+3-r \right)}.
\end{align*}
We can simplify the expression
\begin{equation*}
h_{k-r}^{c}={{c}^{-k-1}}{\exp(-c)}{{I}^{\left( 0 \right)}}\left( c,k \right)+{{c}^{-k-1}}{\exp(-c)}r\sum\limits_{i=0}^{\infty }{\frac{{\left( -1 \right)}^{i}{{I}^{\left( 0 \right)}}\left( c,k+i \right)}{\prod\limits_{j=0}^{i}{\left( k+1+j-r \right)}}},
\end{equation*}
and conclude the proof.
$\square$

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