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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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