Indefinite Integration of the Gamma Integral
and Related Statistical Applications

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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