Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (2.7)

\begin{equation*}
{{I}^{(n)}}\left( c,k \right)=\sum\limits_{i=0}^{\infty }{\left\{ \frac{{{\left( -1 \right)}^{n-1+i}}}{\left( n-1 \right)!}\frac{{{d}^{(n-1)}}}{dk}\left[ \frac{1}{\prod\limits_{j=0}^{i}{\left( k+1+j \right)}} \right]{{I}^{(0)}}\left( c,k+i \right) \right\}}
\end{equation*} Proof:
By definition, ${{I}^{\left( 0 \right)}}\left( c,k \right)=\int{{{c}^{k}}\exp \left( c \right)dc}$. Therefore,
\begin{equation*}
{{I}^{\left( 0 \right)}}\left( c,k \right)={{c}^{k+1}}\exp \left( c \right)h_{k}^{c}.
\end{equation*}
Using integration by parts, we can solve ${{I}^{\left( 1 \right)}}\left( c,k \right)$
\begin{align*}
& {{I}^{\left( 1 \right)}}\left( c,k \right)\\
&=\int{\frac{{{I}^{\left( 0 \right)}}\left( c,k \right)}{c}dc} \\
& =\int{{{c}^{k}}\exp \left( c \right)h_{k}^{c}dc} \\
& =\frac{{{c}^{k+1}}}{k+1}\exp \left( c \right)h_{k}^{c}-\int{\frac{{{c}^{k+1}}}{k+1}\exp \left( c \right)h_{k+1}^{c}dc} \\
& =\frac{{{c}^{k+1}}}{k+1}\exp \left( c \right)h_{k}^{c}-\frac{{{c}^{k+2}}}{\left( k+1 \right)\left( k+2 \right)}\exp \left( c \right)h_{k+1}^{c}+\int{\frac{{{c}^{k+2}}}{\left( k+1 \right)\left( k+2 \right)}\exp \left( c \right)h_{k+2}^{c}dc} \\
& =\frac{{{c}^{k+1}}}{k+1}\exp \left( c \right)h_{k}^{c}-\frac{{{c}^{k+2}}}{\left( k+1 \right)\left( k+2 \right)}\exp \left( c \right)h_{k+1}^{c}+\frac{{{c}^{k+3}}}{\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)}\exp \left( c \right)h_{k+2}^{c}-\cdots \\
& =\frac{{{I}^{\left( 0 \right)}}\left( c,k \right)}{0!(k+1)}-\frac{{{I}^{\left( 0 \right)}}\left( c,k+1 \right)}{0!\left( k+1 \right)\left( k+2 \right)}+\frac{{{I}^{\left( 0 \right)}}\left( c,k+2 \right)}{0!\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)}-\cdots.
\end{align*}
By the same way, we derive
\begin{align*}
{{I}^{\left( 2 \right)}}\left( c,k \right)=\frac{1}{1!{{\left( k+1 \right)}^{2}}}{{I}^{\left( 0 \right)}}\left( c,k \right)&-\Bigg[ \frac{1}{1!{{\left( k+1 \right)}^{2}}\left( k+2 \right)}\\&+\frac{1}{1!\left( k+1 \right){{\left( k+2 \right)}^{2}}} \Bigg]{{I}^{\left( 0 \right)}}\left( c,k+1 \right)+\cdots.
\end{align*}
Repeating the same operations for $n$ times, we derive
\begin{align*}
{{I}^{(n)}}\left( c,k \right)&=\frac{{{\left( -1 \right)}^{n-1}}}{\left( n-1 \right)!}\frac{{{d}^{(n-1)}}}{dk}\left[ \frac{1}{\left( k+1 \right)} \right]{{I}^{(0)}}\left( c,k \right) \\&+\frac{{{\left( -1 \right)}^{n}}}{\left( n-1 \right)!}\frac{{{d}^{(n-1)}}}{dk}\left[ \frac{1}{\left( k+1 \right)\left( k+2 \right)} \right]{{I}^{(0)}}\left( c,k+1 \right) \\ & +\frac{{{\left( -1 \right)}^{n+1}}}{\left( n-1 \right)!}\frac{{{d}^{(n-1)}}}{dk}\left[ \frac{1}{\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)} \right]{{I}^{(0)}}\left( c,k+2 \right) \\
& + \cdots \\
&=\sum\limits_{i=0}^{\infty }{\left\{ \frac{{{\left( -1 \right)}^{n-1+i}}}{\left( n-1 \right)!}\frac{{{d}^{(n-1)}}}{dk}\left[ \frac{1}{\prod\limits_{j=0}^{i}{\left( k+1+j \right)}} \right]{{I}^{(0)}}\left( c,k+i \right) \right\}}.
\end{align*}
$\square$

Welcome all math lovers!