Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (4.10)

$$\begin{equation*} \begin{split} {{B}_{x}}\left( \alpha ,\beta \right)={{x}^{\alpha }}{\exp(-x)}h_{\alpha -1}^{-x}+\sum\limits_{i=1}^{\infty }&\Bigg\{ \left[ \prod\limits_{j=1,j\ne i}^{i+1}{\left( \beta -j \right)} \right] \\& \left[ {{x}^{\alpha }}{\exp\left(-x\right)}h_{\alpha -1}^{-x}-\sum\limits_{k=0}^{i-1}{{{\left( -1 \right)}^{k}}\frac{{{x}^ {\alpha +k}}}{k!\left( \alpha +k \right)}} \right] \Bigg\}. \end{split} \end{equation*}$$ Proof:
According to Dutka (1981, p.17), Gauss derived an idenity of the incomplete beta function as
$$\begin{equation*} {{B}_{x}}\left( \alpha ,\beta \right)=\frac{{{x}^{\alpha }}}{0!\alpha }-\frac{\left( \beta -1 \right){{x}^{\alpha +1}}} {1!\left( \alpha +1 \right)}+\frac{\left( \beta -1 \right)\left( \beta -2 \right){{x}^{\alpha +2}}}{2!\left( \alpha +2 \right)}-\cdots . \end{equation*}$$
We can work with this form by rearranging it as an infinite series of the ``$h$" function.
$$\begin{align*} {{B}_{x}}\left( \alpha ,\beta \right) &=\frac{{{x}^{\alpha }}}{0!\alpha }-\frac{\left( \beta -1 \right){{x}^{\alpha +1}}}{1!\left( \alpha +1 \right)}+\frac{\left( \beta -1 \right)\left( \beta -2 \right){{x}^{\alpha +2}}}{2!\left( \alpha +2 \right)}-\cdots \\ & =\left[ \frac{{{x}^{\alpha }}}{0!\alpha }-\frac{{{x}^{\alpha +1}}}{1!\left( \alpha +1 \right)}+\frac{{{x}^{\alpha +2}}}{2!\left( \alpha +2 \right)}-\cdots \right] \\ & +\left\{ -\frac{{{x}^{\alpha +1}}}{1!\left( \alpha +1 \right)}\left( \beta -2 \right)+\frac{{{x}^{\alpha +2}}}{2!\left( \alpha +2 \right)}\left[ \left( \beta -1 \right)\left( \beta -2 \right)-1 \right]-\cdots \right\} \\ & =\int{{{x}^{\alpha -1}}\exp \left( -x \right)dx}+\left( \beta -2 \right)\left[ -\frac{{{x}^{\alpha +1}}}{1!\left( \alpha +1 \right)}+\frac{{{x}^{\alpha +2}}}{2!\left( \alpha +2 \right)}-\cdots \right] \\ & +\left\{ \frac{{{x}^{\alpha +2}}}{2!\left( \alpha +2 \right)}\left( \beta -1 \right)\left( \beta -3 \right)-\frac{{{x}^ {\alpha +3}}}{3!\left( \alpha +3 \right)}\left[ \left( \beta -1 \right)\left( \beta -2 \right)\left( \beta -3 \right)-\beta +1 \right]+\cdots \right\} \\ & ={{x}^{\alpha }}{\exp(-x)}h_{\alpha -1}^{-x}+\left( \beta -2 \right)\left( {{x}^{\alpha }}{\exp(-x)} h_{\alpha -1}^{-x}-\frac{{{x}^{\alpha }}}{0!\alpha } \right) \\ & +\left( \beta -1 \right)\left( \beta -3 \right)\left( {{x}^{\alpha }}{\exp(-x)}h_{\alpha -1}^{-x}-\frac{{{x}^{\alpha }}} {0!\alpha }+\frac{{{x}^{\alpha +1}}}{1!\left( \alpha +1 \right)} \right) \\ & +\left( \beta -1 \right)\left( \beta -2 \right)\left( \beta -4 \right)\left( {{x}^{\alpha }}{\exp(-x)} h_{\alpha -1}^{-x}-\frac{{{x}^{\alpha }}}{0!\alpha }+\frac{{{x}^{\alpha +1}}}{1!\left( \alpha +1 \right)} -\frac{{{x}^{\alpha +2}}}{2!\left( \alpha +2 \right)} \right) \\ & +\cdots. \end{align*}$$
The result concludes the proof.
$\square$

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