Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (3.10)

$$\begin{equation*} \int{h_{s}^{c}dc=-\sum\limits_{i=0}^{\infty }{h_{s+i}^{c}}}. \end{equation*}$$ Proof:
Given the first-order partial derivative formula of $h_{s}^{c}$ with respect to $c$, we know
$$\begin{equation*} \int{h_{s}^{c}dc}=\int{h_{s+1}^{c}dc-}h_{s}^{c}. \end{equation*}$$
We can apply this formula infinite times and sum all the terms in the left-hand and right-hand sides. All the integral terms will cancel out, except $\int{h_{s}^{c}dc}$ and $\int{h_{s+n+1}^{c}dc}$.
$$\begin{align*} \int{h_{s}^{c}dc}=&\int{h_{s+1}^{c}dc-}h_{s}^{c} \\ \int{h_{s+1}^{c}dc}=&\int{h_{s+2}^{c}dc-}h_{s+1}^{c} \\ \int{h_{s+2}^{c}dc}=&\int{h_{s+3}^{c}dc-}h_{s+2}^{c} \\ &\vdots \\ \int{h_{s+n}^{c}dc}=&\int{h_{s+n+1}^{c}dc-}h_{s+n}^{c}. \end{align*}$$
, where $n\to \infty$.

Since $\underset{n\to \infty}{\mathop{\lim }}\,\int{h_{s+n+1}^{c}dc}\to 0$,

$$\begin{equation*} \int{h_{s}^{c}dc=-\sum\limits_{i=0}^{\infty }{h_{s+i}^{c}}}. \end{equation*}$$
$\square$

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