Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (2.4)

$$\begin{equation*} h_{k-r}^{c}={\exp(-c)}\sum\limits_{i=0}^{\infty }{\frac{{{c}^{i}}}{i!\left( k+1+i-r \right)}}. \end{equation*}$$ Proof:
We can construct the following identity by (2.2)
$$\begin{equation*} {{c}^{k+1-r}}{\exp(c)}h_{k-r}^{c}=g\left( k+1-r,1,c \right). \end{equation*}$$
Therefore,
$$\begin{align*} {{c}^{k+1-r}}{\exp(c)}h_{k-r}^{c}&=\frac{{{c}^{k+1-r}}}{0!\left( k+1-r \right)}+\frac{{{c}^{k+2-r}}}{1!\left( k+2-r \right)}+\frac{{{c}^{k+3-r}}}{2!\left( k+3-r \right)}+\cdots \\ h_{k-r}^{c}&={\exp(-c)}\left\{ \frac{{{c}^{0}}}{0!\left( k+1-r \right)}+\frac{{{c}^{1}}}{1!\left( k+2-r \right)}+\frac{{{c}^{2}}}{2!\left( k+3-r \right)}+\cdots \right\}\\ &={\exp(-c)}\sum\limits_{i=0}^{\infty }{\frac{{{c}^{i}}}{i!\left( k+1+i-r \right)}}. \end{align*}$$
$\square$

Welcome all math lovers!