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