Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (4.4)

$$\begin{equation*} {{E}_{\left( 1 \right)}}\left( x;C,t \right)-{{E}_{\left( 1 \right)}}\left( x;C \right)=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\sum\limits_{s=t+1}^{n}{\left( h_{s}^{x}-h_{s}^{C} \right)}. \end{equation*}$$ Proof:
$$\begin{align*} & {{E}_{\left( 1 \right)}}\left( x;C,t \right)-{{E}_{\left( 1 \right)}}\left( x;C \right) \\ & =\left[ \left( \log \left| -C \right|+\,\sum\limits_{s=0}^{t}{h_{s}^{C}} \right)-\left( \log \left| -x \right|+\,\sum\limits_{s=0}^{t}{h_{s}^{x}} \right) \right] \\ & -\left[ \left( \log \left| -C \right|+\underset{n\to \infty }{\mathop{\lim }}\,\,\sum\limits_{s=0}^{n}{h_{s}^{C}} \right)-\left( \log \left| -x \right|+\underset{n\to \infty }{\mathop{\lim }}\,\,\sum\limits_{s=0}^{n}{h_{s}^{x}} \right) \right] \\ & =\,\sum\limits_{s=0}^{t}{h_{s}^{C}}-\underset{n\to \infty }{\mathop{\lim }}\,\,\sum\limits_{s=0}^{n}{h_{s}^{C}}-\sum\limits_{s=0}^{t}{h_{s}^{x}}+\underset{n\to \infty }{\mathop{\lim }}\,\,\sum\limits_{s=0}^{n}{h_{s}^{x}} \\ & =\underset{n\to \infty }{\mathop{\lim }}\,\,\,\sum\limits_{s=t+1}^{n}{\left( h_{s}^{x}-h_{s}^{C} \right)}. \end{align*}$$
$\square$

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