Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (4.5)

$$\begin{equation*} {{E}_{\left(1\right)}}\left(x;C,t\right)=\log\left|\frac{C}{x} \right|+\sum\limits_{s=0}^{t}{\left( h_{s}^{C}-h_{s}^{x} \right)}. \end{equation*}$$ Proof:

Given we know the fact

$$\begin{align*} {{E}_{\left( 1 \right)}}\left( x;C \right)&=\int_{x}^{C}{{{t}^{-1}}\exp \left( -t \right)}dt \\ & =\exp \left( -t \right)h_{-1}^{-t}|_{x}^{C} \\ & =\exp \left( -C \right)h_{-1}^{-C}-\exp \left( -x \right)h_{-1}^{-x} \\ & =\left( \log \left| -C \right|+\sum\limits_{s=0}^{\infty }{h_{s}^{C}} \right)-\left( \log \left| -x \right|+\sum\limits_{s=0}^{\infty }{h_{s}^{x}} \right), \\ \end{align*}$$
if we use $t$ negative subtracting terms to evaluate ${{E}_{\left( 1 \right)}}\left( x;C \right)$,
$$\begin{align*} {{E}_{\left( 1 \right)}}\left( x;C,t \right)&=\left( \log \left| -C \right|+\sum\limits_{s=0}^{t}{h_{s}^{C}} \right)-\left( \log \left| -x \right|+\sum\limits_{s=0}^{t}{h_{s}^{x}} \right) \\ & =\log \left| \frac{C}{x} \right|+\sum\limits_{s=0}^{t}{\left( h_{s}^{C}-h_{s}^{x} \right)}. \end{align*}$$
$\square$

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