Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (4.3)

The upper limit of the error for using $C$ to evaluated ${{E}_{(n)}}$ is
\begin{equation*}
\int_{C}^{\infty }{{{t}^{-1}}{\exp(-t)}dt}.
\end{equation*}Proof:
\begin{align*}
{{E}_{\left( 1 \right)}}\left( x \right)&=\int_{x}^{\infty }{{{t}^{-1}}\exp \left( -t \right)dt} \\
& =\int_{x}^{C}{{{t}^{-1}}\exp \left( -t \right)dt}+\int_{C}^{\infty }{{{t}^{-1}}\exp \left( -t \right)dt}
\end{align*}

Given the fact that $C\gg 1$ and ${{E}_{(1)}}$ has the heavest tail for all ${{E}_{(n)}}$, the error will be $\int_{C}^{\infty }{{{t}^{-1}}\exp \left( -t \right)dt}$ if we use $C$ in replace of $\infty $.

$\square$

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