Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (2.6)

\begin{equation*}
h_{k-r}^{c}=\frac{\exp(-c)}{{{c}^{k+1}}}\sum\limits_{i=0}^{\infty }{{{r}^{i}}{{I}^{\left( i \right)}}\left( c,k \right)}
\end{equation*} Proof:
Give (2.5), we can express $h_{k-r}^{c}$ as
\begin{align*}
h_{k-r}^{c}=\exp \left( -c \right)&\left\{ \frac{{{c}^{0}}}{0!}\left[ {{\left( \frac{1}{k+1} \right)}^{1}}{{r}^{0}}+{{\left( \frac{1}{k+1} \right)}^{2}}{{r}^{1}}+{{\left( \frac{1}{k+1} \right)}^{3}}{{r}^{2}}+\cdots \right] \right. \\
& +\frac{{{c}^{1}}}{1!}\left[ {{\left( \frac{1}{k+2} \right)}^{1}}{{r}^{0}}+{{\left( \frac{1}{k+2} \right)}^{2}}{{r}^{1}}+{{\left( \frac{1}{k+2} \right)}^{3}}{{r}^{2}}+\cdots \right] \\
& +\frac{{{c}^{2}}}{2!}\left[ {{\left( \frac{1}{k+3} \right)}^{1}}{{r}^{0}}+{{\left( \frac{1}{k+3} \right)}^{2}}{{r}^{1}}+{{\left( \frac{1}{k+3} \right)}^{3}}{{r}^{2}}+\cdots \right] \\
& \left. +\cdots \right\}.
\end{align*}
Sum all the terms by $r^{i}$, for example $i=0$, and the result is
\begin{align*}
&\exp(-c)\left[ \frac{{{c}^{0}}}{0!}{{\left( \frac{1}{k+1} \right)}^{1}}{{r}^{0}}+\frac{{{c}^{1}}}{1!}{{\left( \frac{1}{k+2} \right)}^{1}}{{r}^{0}}+\frac{{{c}^{2}}}{2!}{{\left( \frac{1}{k+3} \right)}^{1}}{{r}^{0}}+\cdots \right] \\
=&\frac{\exp(-c)}{{{c}^{k+1}}}\left[ \frac{{{c}^{k+1}}}{0!\left( k+1 \right)}+\frac{{{c}^{k+2}}}{1!\left( k+2 \right)}+\frac{{{c}^{k+3}}}{2!\left( k+3 \right)}+\cdots \right] \\
=&\frac{\exp(-c)}{{{c}^{k+1}}}\int{{{c}^{k}}\exp \left( c \right)dc}.
\end{align*}
For $i=1$, we derive
\begin{align*}
& \exp(-c)\left[\frac{{{c}^{0}}}{0!}{{\left( \frac{1}{k+1} \right)}^{2}}r^{1}+\frac{{{c}^{1}}}{1!}{{\left( \frac{1}{k+2} \right)}^{2}}r^{1}+\frac{{{c}^{2}}}{2!}{{\left( \frac{1}{k+3} \right)}^{2}}r^{1}+\cdots \right] \\
=&\frac{\exp(-c)r^{1}}{{{c}^{k+1}}}\left[ \frac{{{c}^{k+1}}}{0!{{\left( k+1 \right)}^{2}}}+\frac{{{c}^{k+2}}}{1!{{\left( k+2 \right)}^{2}}}+\frac{{{c}^{k+3}}}{2!{{\left( k+3 \right)}^{2}}}+\cdots \right] \\
=&\frac{\exp(-c)r^{1}}{{{c}^{k+1}}}\int{\frac{\int{{{c}^{k}}\exp \left( c \right)dc}}{c}}dc.
\end{align*}
Repeating the same operation by changing $i$, we can conclude
\begin{equation*}
h_{k-r}^{c}=\frac{\exp(-c)}{{{c}^{k+1}}}\sum\limits_{i=0}^{\infty }{{{r}^{i}}{{I}^{\left( i \right)}}\left( c,k \right)},
\end{equation*}
where
\begin{equation*}
{{I}^{\left( 0 \right)}}\left( c,k \right)=\int{{{c}^{k}}\exp \left( c \right)dc},
\end{equation*}
and
\begin{equation*}
{{I}^{\left( i \right)}}\left( c,k \right)=\int{\frac{^{(i)}{\iddots}\,\int{\frac{{{I}^{\left( 0 \right)}}\left( c,k \right)}{c}}dc}{c}}dc.
\end{equation*}
$\square$

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