Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (3.2)

$$\begin{align*} {}^{-1}{{h}_{s}}\left( \delta \right) =&-\left[ \left( s+1 \right)\left( s+2 \right)\delta-\left( s+2 \right) \right] \label{eq:q19} \\ \notag &+\frac{1}{\left( s+3 \right)}{{\left[ \left( s+1 \right)\left( s+2 \right)\delta-\left( s+2 \right) \right]}^{2}}\\ \notag &-\frac{s+5}{{{\left( s+3 \right)}^{2}}\left( s+4 \right)}{{\left[ \left( s+1 \right)\left( s+2 \right)\delta-\left( s+2 \right) \right]}^{3}} \\ \notag &+\frac{{{s}^{2}}+11s+34}{{{\left( s+3 \right)}^{3}}\left( s+4 \right)\left( s+5 \right)}{{\left[ \left( s+1 \right)\left( s+2 \right)\delta-\left( s+2 \right) \right]}^{4}}+\cdots. \notag \end{align*}$$ Proof:
Given (2.1), we know
$$\begin{equation*} \delta =h_{s}^{c}=\frac{1}{\left( s+1 \right)}+\frac{{{\left( -c \right)}^{1}}}{\left( s+1 \right)\left( s+2 \right)}+\frac{{{\left( -c \right)}^{2}}}{\left( s+1 \right)\left( s+2 \right)\left( s+3 \right)}+\cdots. \end{equation*}$$
Therefore,
$$\begin{equation*} \left( s+1 \right)\left( s+2 \right)\delta -\left( s+2 \right)={{\left( -c \right)}^{1}}+\frac{{{\left( -c \right)}^{2}}}{\left( s+3 \right)}+\frac{{{\left( -c \right)}^{3}}}{\left( s+3 \right)\left( s+4 \right)}\cdots. \end{equation*}$$
To find the inverse function of "$h$", ${}^{-1}{{h}_{s}}\left( \delta \right)$, we need to use $\left[\left( s+1 \right)\left( s+2 \right)\delta -\left( s+2 \right)\right]$ as the core function to construct a power function, with which the final results is $c$.
For instance, the coefficient of the first term of the power function is $-1$, since the result is $c$ plus the remaining terms
$$\begin{equation*} -\left[ \left( s+1 \right)\left( s+2 \right)\delta -\left( s+2 \right) \right]={c}-\frac{{{c}^{2}}}{\left( s+3 \right)}+\frac{{{c}^{3}}}{\left( s+3 \right)\left( s+4 \right)}\cdots. \end{equation*}$$
Next, we will construct a second-order power term with the core function to eliminate the remaining term which has the same order. The result is
$$\begin{equation*} \frac{1}{\left( s+3 \right)}{{\left[ \left( s+1 \right)\left( s+2 \right)\delta -\left( s+2 \right) \right]}^{2}}=\frac{{{c}^{2}}}{\left( s+3 \right)}-\frac{2{{c}^{3}}}{{{\left( s+3 \right)}^{2}}}+\left[ \frac{2{{c}^{4}}}{{{\left( s+3 \right)}^{2}}\left( s+4 \right)}+\frac{{{c}^{4}}}{{{\left( s+3 \right)}^{3}}} \right]+\cdots \end{equation*}$$
The leading term will cancel out the first remaining term and generates new remaining terms
$$\begin{align*} & -\left[ \left( s+1 \right)\left( s+2 \right)\delta -\left( s+2 \right) \right]+\frac{1}{\left( s+3 \right)}{{\left[ \left( s+1 \right)\left( s+2 \right)\delta -\left( s+2 \right) \right]}^{2}} \\ & ={{c}^{1}}+\left[ -\frac{2{{c}^{3}}}{{{\left( s+3 \right)}^{2}}}+\frac{{{c}^{3}}}{\left( s+3 \right)\left( s+4 \right)} \right]\\ &+\left[ \frac{2{{c}^{4}}}{{{\left( s+3 \right)}^{2}}\left( s+4 \right)}+\frac{{{c}^{4}}}{{{\left( s+3 \right)}^{3}}}-\frac{{{c}^{4}}}{\left( s+3 \right)\left( s+4 \right)\left( s+5 \right)} \right]+\cdots. \end{align*}$$
Repeating the same process, we can derive the inverse function of "$h$"
$$\begin{align*} {}^{-1}{{h}_{s}}\left( \delta \right) =&-\left[ \left( s+1 \right)\left( s+2 \right)\delta-\left( s+2 \right) \right] \\ \notag &+\frac{1}{\left( s+3 \right)}{{\left[ \left( s+1 \right)\left( s+2 \right)\delta-\left( s+2 \right) \right]}^{2}}\\ \notag &-\frac{s+5}{{{\left( s+3 \right)}^{2}}\left( s+4 \right)}{{\left[ \left( s+1 \right)\left( s+2 \right)\delta-\left( s+2 \right) \right]}^{3}} \\ \notag &+\frac{{{s}^{2}}+11s+34}{{{\left( s+3 \right)}^{3}}\left( s+4 \right)\left( s+5 \right)}{{\left[ \left( s+1 \right)\left( s+2 \right)\delta-\left( s+2 \right) \right]}^{4}}+\cdots. \notag \end{align*}$$
$\square$

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