Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (3.11)

$$\begin{equation*} \int_{{}}^{(n)}h_{s}^{c}dc={{\left( -1 \right)}^{n}}\sum\limits_{i=0}^{\infty }{{i+n-1 \choose i} h_{s+i}^{c}}. \end{equation*}$$ Proof:
Given the first-order antiderivative formula of $h_{s}^{c}$ with respect to $c$, we can derive the second-order furmula as
$$\begin{align*} \int_{{}}^{\left( 2 \right)}{h_{s}^{c}dc}&=-\int{h_{s}^{c}dc}-\int{h_{s+1}^{c}dc}-\int{h_{s+2}^{c}dc}-\cdots \\ & =\left( h_{s}^{c}+h_{s+1}^{c}+h_{s+2}^{c}+h_{s+3}^{c}+\cdots \right) \\ & +\left( h_{s+1}^{c}+h_{s+2}^{c}+h_{s+3}^{c}+h_{s+4}^{c}+\cdots \right) \\ & +\left( h_{s+2}^{c}+h_{s+3}^{c}+h_{s+4}^{c}+h_{s+5}^{c}+\cdots \right) \\ & +\cdots \\ & ={{\left( -1 \right)}^{2}}\sum\limits_{i=0}^{\infty }\binom {i+1}{i}h_{s+i}^{c}. \end{align*}$$
Repeating the first-order antiderivative formula $n$ times, we derive
$$\begin{equation*} \int_{{}}^{(n)}h_{s}^{c}dc={{\left( -1 \right)}^{n}}\sum\limits_{i=0}^{\infty }{{i+n-1 \choose i} h_{s+i}^{c}}. \end{equation*}$$
$\square$

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