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Proof of Equation (2.14)

\begin{equation*}
{{w}_{i}}=\frac{\int_{0}^{c}{{{x}^{i}}\exp (-x)dx}}{\int_{0}^{\infty }{{{x}^{i}}\exp (-x)dx}}
\end{equation*} Proof:
From (2.10), we know ${{w}_{i}}=1-\sum\limits_{j=0}^{i}{\frac{{{c}^{j}}\exp (-c)}{j!}}$. We can respecify ${{w}_{i}}$ as
\begin{equation*}
{{w}_{i}}=\frac{i!-\sum\limits_{j=0}^{i}{\left( \frac{i!{{c}^{j}}\exp (-c)}{j!} \right)}}{i!}.
\end{equation*}
The numerator is in fact the solution of the definite gamma integral with the lower and uppoer limits $\left[ 0,c \right]$
\begin{equation*}
\int_{0}^{c}{{{x}^{i}}\exp (-x)dx}=i!-\sum\limits_{j=0}^{i}{\left( \frac{i!{{c}^{j}}\exp (-c)}{j!} \right)}.
\end{equation*}
The factorial denominator is also the solution of the same integral, but with an infinite upper limit rather than $c$
\begin{equation*}
\int_{0}^{\infty }{{{x}^{i}}\exp (-x)dx}=i!.
\end{equation*}
Thus,
\begin{equation*}
{{w}_{i}}=\frac{\int_{0}^{c}{{{x}^{i}}\exp (-x)dx}}{\int_{0}^{\infty }{{{x}^{i}}\exp (-x)dx}}.
\end{equation*}
$\square$

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