Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (2.9)

$$\begin{align*} h_{-r}^{c}&={{c}^{-1}}{\exp(-c)}\left\{ \left( {\exp(c)}-1 \right)+r\left( \frac{{\exp(c)}-1}{0!} \right)\left( \frac{1}{1-r} \right) \right. \label{eq:q12} \\ \notag & +r\left( \frac{c \cdot {\exp(c)}-{\exp(c)}+1}{1!} \right)\left( \frac{1}{2-r}-\frac{1}{1-r} \right) \\ \notag & \left. +r\left( \frac{{{c}^{2}}\cdot {\exp(c)}-2c \cdot {\exp(c)}+2! \cdot {\exp(c)}-2!}{2!} \right)\left( \frac{1}{3-r}-\frac{2}{2-r}+\frac{1}{1-r} \right)+\cdots \right\}. \end{align*}$$ Proof:
Bringing $k=0$ into (2.8), we know
$$\begin{equation*} h_{-r}^{c}={{c}^{-1}}{\exp(-c)}{{I}^{\left( 0 \right)}}\left( c,0 \right)+{{c}^{-1}}{\exp(-c)}r\sum\limits_{i=0}^{\infty }{\frac{{\left( -1 \right)}^{i}{{I}^{\left( 0 \right)}}\left( c,i \right)}{\prod\limits_{j=0}^{i}{\left( 1+i-r \right)}}}. \end{equation*}$$
Given (2.6), we derive $h_{0}^{c}$ by setting $k=0$ and $r=0$. Therefore,
$$\begin{align*} h_{0}^{c}&=\frac{\exp (-c)}{c}{{I}^{\left( 0 \right)}}\left( c,0 \right) \\ & =\frac{{{(-c)}^{0}}}{1!}+\frac{{{(-c)}^{1}}}{2!}+\frac{{{(-c)}^{2}}}{3!}+\cdots \\ & =\frac{-1}{c}\left[ \frac{{{(-c)}^{1}}}{1!}+\frac{{{(-c)}^{2}}}{2!}+\frac{{{(-c)}^{3}}}{3!}+\cdots \right] \\ & =\frac{-1}{c}\left[ \exp (-c)-1 \right] \\ & =\frac{\exp (-c)}{c}\left[ \exp (c)-1 \right] , \end{align*}$$
and hence,
$$\begin{equation*} {{I}^{\left( 0 \right)}}\left( c,0 \right)=\exp (c)-1. \end{equation*}$$
Using the same method by setting $k=1,2,\cdots$ and $r=0$, we derive
$$\begin{align*} {{I}^{\left( 0 \right)}}\left( c,1 \right) &=c\exp (c)-\exp \left( c \right)+1 \\ {{I}^{\left( 0 \right)}}\left( c,2 \right) &={{c}^{2}}\exp (c)-2c\exp \left( c \right)+2\exp \left( c \right)-2 \\ & \vdots \\ {{I}^{\left( 0 \right)}}\left( c,k \right) &=\left\{ \exp (c)\sum\limits_{i=0}^{k}{\left[ \frac{{{\left( -1 \right)}^{i}}{{c}^{k-i}}k!}{\left( k-i \right)!} \right]} \right\}+{{\left( -1 \right)}^{k+1}}k! . \end{align*}$$
Meanwhile, we know
$$\begin{align*} \frac{1}{0!}\left( \frac{1}{1-r} \right)&=\frac{1}{1-r} \\ \frac{1}{1!}\left( \frac{1}{2-r}-\frac{1}{1-r} \right)&=\frac{-1}{\left( 1-r \right)\left( 2-r \right)} \\ \frac{1}{2!}\left( \frac{1}{3-r}-\frac{2}{2-r}+\frac{1}{1-r} \right)&=\frac{1}{\left( 1-r \right)\left( 2-r \right) \left( 3-r \right)} \\ & \vdots \\ \frac{1}{n!}\left(\sum\limits_{i=0}^{n}{\frac{{{\left(-1\right)}^{n-i}}\binom{n}{i}}{\left( 1+i-r \right)}} \right)&=\frac{{{\left( -1 \right)}^{n}}}{\prod\limits_{i=0}^{n}{\left( 1+i-r \right)}}. \end{align*}$$
Therefore, we can conclude the proof
$$\begin{align*} h_{-r}^{c}&={{c}^{-1}}{\exp(-c)}\left\{ \left( {\exp(c)}-1 \right)+r\left( \frac{{\exp(c)}-1}{0!} \right)\left( \frac{1}{1-r} \right) \right. \\ \notag & +r\left( \frac{c \cdot {\exp(c)}-{\exp(c)}+1}{1!} \right)\left( \frac{1}{2-r}-\frac{1}{1-r} \right) \\ \notag & \left. +r\left( \frac{{{c}^{2}}\cdot {\exp(c)}-2c \cdot {\exp(c)}+2! \cdot {\exp(c)}-2!}{2!} \right)\left( \frac{1}{3-r}-\frac{2}{2-r}+\frac{1}{1-r} \right)+\cdots \right\}. \end{align*}$$
$\square$

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