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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