Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (2.11)

\begin{equation*}
h_{-r}^{-c}=-{{c}^{-1}}-r{{c}^{r-1}}{\exp(c)}\Gamma \left( -r \right)
\end{equation*} Proof:
To prove (2.11), we need to work out $\underset{n\to \infty } {\mathop{\lim }}\,\sum\limits_{i=0}^{n}{{{w}_{i}}{{\beta }_{i}}}$
\begin{align*}
\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=0}^{n}{{{w}_{i}}{{\beta }_{i}}}&=\left[ \frac{1}{0!}-\frac{\exp \left( -c \right)}{0!} \right]\left[ \frac{0!}{1-r} \right] \\
& +\left[ \frac{1}{0!}-\frac{\exp \left( -c \right)}{0!}-\frac{{{c}^{1}}\exp \left( -c \right)}{1!} \right]\left[ \frac{1!}{\left( 1-r \right)\left( 2-r \right)} \right] \\
& +\left[ \frac{1}{0!}-\frac{\exp \left( -c \right)}{0!}-\frac{{{c}^{1}}\exp \left( -c \right)}{1!}-\frac{{{c}^{2}}\exp \left( -c \right)}{2!} \right]\left[ \frac{2!}{\left( 1-r \right)\left( 2-r \right)\left( 3-r \right)} \right] \\
& +\cdots \\
& =\left[ \frac{1}{0!}-\frac{\exp \left( -c \right)}{0!} \right]\left[ \frac{0!}{1-r}+\frac{1!}{\left( 1-r \right)\left( 2-r \right)}+\frac{2!}{\left( 1-r \right)\left( 2-r \right)\left( 3-r \right)}+\cdots \right] \\
& -\left[ \frac{{{c}^{1}}\exp \left( -c \right)}{1!} \right]\left[ \frac{1!}{\left( 1-r \right)\left( 2-r \right)}+\frac{2!}{\left( 1-r \right)\left( 2-r \right)\left( 3-r \right)}+\cdots \right] \\
& -\left[ \frac{{{c}^{2}}\exp \left( -c \right)}{2!} \right]\left[ \frac{2!}{\left( 1-r \right)\left( 2-r \right)\left( 3-r \right)}+\cdots \right] \\
& -\cdots.
\end{align*}
Here, we know
\begin{align*}
& \frac{0!}{1-r}+\frac{1!}{\left( 1-r \right)\left( 2-r \right)}+\frac{2!}{\left( 1-r \right)\left( 2-r \right)\left( 3-r \right)}+\cdots \\
& =\left( \frac{1}{1-r} \right)+\left( \frac{1}{1-r}-\frac{1}{2-r} \right)+\left( \frac{1}{1-r}-\frac{2}{2-r}+\frac{1}{3-r} \right)+\cdots \\
& =\frac{{{\left( -1 \right)}^{0}}}{1-r}\left[\binom {0} {0}+\binom {1} {0}+\binom {2} {0}+\cdots +\binom {n} {0} \right] \\
& +\frac{{{\left( -1 \right)}^{1}}}{2-r}\left[\binom {1} {1}+\binom {2} {1}+\binom {3} {1}+\cdots +\binom {n} {1} \right] \\
& +\frac{{{\left( -1 \right)}^{2}}}{3-r}\left[\binom {2} {2}+\binom {3} {2}+\binom {4} {2}+\cdots +\binom {n} {2} \right] \\
& +\cdots.
\end{align*}
Therefore,
\begin{align*}
\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=0}^{n}{{{w}_{i}}{{\beta }_{i}}}= \\
\left( \frac{1}{0!}-\frac{\exp \left( -c \right)}{0!} \right)&\left\{ \frac{1}{1-r}\left[\binom {0} {0}+\binom {1} {0}+\binom {2} {0}+\cdots +\binom {n} {0} \right] \right. \\
& -\frac{1}{2-r}\left[ \binom {1} {1}+\binom {2} {1}+\cdots +\binom {n} {1} \right] \\
& +\frac{1}{3-r}\left[ \binom {2} {2}+\cdots +\binom {n} {2} \right] \\
& -\cdots \\
& \left. +\frac{{{\left( -1 \right)}^{n}}}{n+1-r}\left[ \binom {n} {n} \right] \right\} \\
-\left( \frac{{{c}^{1}}\exp \left( -c \right)}{1!} \right)&\left\{ \frac{1}{1-r}\left[ \binom {1} {0}+\binom {2} {0}+\cdots +\binom {n} {0} \right] \right. \\
& -\frac{1}{2-r}\left[ \binom {1} {1}+\binom {2} {1}+\cdots +\binom {n} {1} \right] \\
& +\frac{1}{3-r}\left[ \binom {2} {2}+\cdots +\binom {n} {2} \right] \\
& -\cdots \\
& \left. +\frac{{{\left( -1 \right)}^{n}}}{n+1-r}\left[ \binom {n} {n} \right] \right\} \\
-\left( \frac{{{c}^{2}}\exp \left( -c \right)}{2!} \right)&\left\{ \frac{1}{1-r}\left[ \binom {2} {0}+\cdots +\binom {n} {0} \right] \right. \\
& -\frac{1}{2-r}\left[ \binom {2} {1}+\cdots +\binom {n} {1} \right] \\
& +\frac{1}{3-r}\left[ \binom {2} {2}+\cdots +\binom {n} {2} \right] \\
& -\cdots \\
& \left. +\frac{{{\left( -1 \right)}^{n}}}{n+1-r}\left[ \binom {n} {n} \right] \right\} \\
-\cdots.
\end{align*}
Summing up all the items by $1/(i+1-r)$, where $i=0,1,2,\cdots,n$. For $i=0$ and $1/(1-r)$, we derive
\begin{align*}
& \frac{1}{1-r}\left[ \frac{\binom {0}{0}}{0!}+\frac{\binom {1}{0}}{0!}+\cdots +\frac{\binom {n}{0}}{0!} \right] \\
&-\frac{\exp \left( -c \right)}{1-r}\left[ \frac{\binom {0}{0}}{0!}+\frac{\binom {1}{0}}{0!}+\cdots +\frac{\binom {n}{0}}{0!} \right] \\
&-\frac{{{c}^{1}}\exp \left( -c \right)}{1-r}\left[ \frac{\binom {1}{0}}{1!}+\cdots +\frac{\binom {n}{0}}{1!} \right] \\
&-\cdots \\
=&\left[ \frac{1}{1-r}\right]\left[\frac{\left( n+1 \right)-0}{0!} \right]-\left[ \frac{\exp \left( -c \right)}{1-r}\right]\left[\frac{\left( n+1 \right)-0}{0!} \right]-\left[ \frac{{{c}^{1}}\exp \left( -c \right)}{1-r}\right]\left[\frac{\left( n+1 \right)-1}{1!} \right] \\
&- \cdots -\left[ \frac{{{c}^{n}}\exp \left( -c \right)}{1-r}\right]\left[\frac{\left( n+1 \right)-n}{n!} \right] \\
=& -\frac{\exp \left( -c \right)}{1-r}\left\{ \frac{\left( n+1 \right)}{0!}+\frac{\left( n+1 \right){{c}^{1}}}{1!}+\cdots \frac{\left( n+1 \right){{c}^{n}}}{n!} \right\}+\frac{1}{1-r}\frac{\left( n+1 \right)}{0!} \\
&+\frac{\exp \left( -c \right)}{1-r}\left\{ \frac{{{c}^{1}}}{0!}+\frac{{{c}^{2}}}{1!}+\frac{{{c}^{n}}}{\left( n-1 \right)!} \right\} \\
=&-\frac{\left( n+1 \right)}{1-r}+\frac{\left( n+1 \right)}{1-r}+\frac{c\exp \left( -c \right)}{1-r}\left( \frac{{{c}^{0}}}{0!}+\frac{{{c}^{1}}}{1!}+\cdots \frac{{{c}^{n-1}}}{\left( n-1 \right)!} \right) \\
=&\frac{c}{1-r},
\end{align*}
where $n\to \infty $.
As a result, we can conclude a general result: for $1/(i+1-r)$, the sum of all the relevant items is
\begin{equation*}
\frac{{{\left( -1 \right)}^{i}}{{c}^{i+1}}}{\left( i+1 \right)!\left( i+1-r \right)}.
\end{equation*}
Therefore,
\begin{align*}
& \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=0}^{n}{{{w}_{i}}{{\beta }_{i}}}=\frac{{{c}^{1}}}{\left( 1-r \right)1!}-\frac{{{c}^{2}}}{\left( 2-r \right)2!}+\frac{{{c}^{3}}}{\left( 3-r \right)3!}-\cdots \\
=&{{c}^{r}}\left( \frac{{{c}^{1-r}}}{\left( 1-r \right)1!}-\frac{{{c}^{2-r}}}{\left( 2-r \right)2!}+\frac{{{c}^{3-r}}}{\left( 3-r \right)3!}-\cdots \right) \\
=&{{c}^{r}}\left( \frac{{{c}^{-r}}}{\left( -r \right)0!}-\Gamma \left( -r \right) \right) \\
=&-{{c}^{r}}\Gamma \left( -r \right)-\frac{1}{r}.
\end{align*}
Bringing the above result back to (2.10), we can conlude the proof
\begin{equation*}
h_{-r}^{-c}=-{{c}^{-1}}-r{{c}^{r-1}}{\exp(c)}\Gamma \left( -r \right).
\end{equation*}
$\square$

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