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*}
$\square$

Welcome all math lovers!