Indefinite Integration of the Gamma Integral
and Related Statistical Applications

2.2 Connecting "$h$" to the Gamma Function

When the integration interval is set from $0$ to $\infty$, the gamma integral $g\left( s+1, -1,u \right)$ becomes $\Gamma \left( s+1 \right)$. If $s$ is a nonnegative integer, it defines the factorial function $s!$. If $s$ is not an integer, due to its recursive regularity, all gamma functions can be calculated as a function of $\Gamma \left( \bmod (s,1) \right)$, where $0<\bmod \left( s,1 \right)<1$.

$$\begin{equation*} \Gamma \left( s+1 \right)= \begin{cases} \Gamma \left( \bmod \left( s,1 \right) \right)\prod\limits_{i=0}^{\left\lfloor s \right\rfloor }{\left( s-i \right)}, &\text{if $s>-1$ & $s \notin {{\mathbb{Z}}^{+}_{0}}$;} \\ \Gamma \left( \bmod \left( s,1 \right) \right)\prod\limits_{i=0}^{\left| \left\lceil s \right\rceil \right|-1}{\left( s+1+i \right)^{-1}}, &\text{if $s<-1$ & $s \notin {{\mathbb{Z}}^{-}}$;} \\ s!, &\text{if $s\in {{\mathbb{Z}}^{+}_{0}}$.} \end{cases} \tag{2.3} \end{equation*}$$
[Examples of (2.3)]

For any non-integer $s$, the remainder for one, $\bmod (s,1)$, is bounded within the interval $\left( 0,1 \right)$. Thus,all gamma functions can be calculated if $\Gamma \left( \bmod (s,1) \right)$ is known. Given that the "$h$" function is derived from factorizing the gamma integral, the investigation of "$h$" will focus on the same integral limit as $\Gamma \left( \bmod (s,1) \right)$ .

An identity of the "$h$" function can be respecified from (2.2)

$$\begin{equation*} h_{k-r}^{c}={\exp(-c)}\sum\limits_{i=0}^{\infty }{\frac{{{c}^{i}}}{i!\left( k+1+i-r \right)}}. \tag{2.4} \end{equation*}$$
[Proof for (2.4)]

Here, $k$ is a non-negative integer and $0<r<1$. Performing long division for ${1}/{\left[ \left( k+1+i \right)-r \right]}$, we can derive

$$\begin{equation*} h_{k-r}^{c}={\exp(-c)}\sum\limits_{i=0}^{\infty }{\left\{ \frac{{{c}^{i}}}{i!}\left[ \sum\limits_{j=0}^{\infty } {\frac{{{r}^{j}}}{{{\left( k+i+1 \right)}^{j+1}}}} \right] \right\}}. \tag{2.5} \end{equation*}$$
[Proof for (2.5)]

If we rearrange (2.5) and sum all the terms by $i$ given $j$, we derive

$$\begin{equation*} h_{k-r}^{c}=\frac{\exp(-c)}{{{c}^{k+1}}}\sum\limits_{i=0}^{\infty }{{{r}^{i}}{{I}^{\left( i \right)}}\left( c,k \right)}, \tag{2.6} \end{equation*}$$

where

$$\begin{equation*} {{I}^{\left( 0 \right)}}\left( c,k \right)=\int{{{c}^{k}}\exp \left( c \right) dc}, \end{equation*}$$
and
$$\begin{equation*} {{I}^{\left( i \right)}}\left( c,k \right)=\int{\frac{^{(i)}{\iddots}\,\int{\frac{{{I}^{\left( 0 \right)}} \left( c,k \right)}{c}}dc}{c}}dc. \end{equation*}$$ [Proof for (2.6)]

We can further simplify ${{I}^{\left( n \right)}}\left( c,k \right)$ as

$$\begin{equation*} {{I}^{(n)}}\left( c,k \right)=\sum\limits_{i=0}^{\infty }{\left\{ \frac{{{\left( -1 \right)}^{n-1+i}}}{\left( n-1 \right)!}\frac{{{d}^{(n-1)}}}{dk}\left[ \frac{1}{\prod\limits_{j=0}^{i}{\left( k+1+j \right)}} \right]{{I}^{(0)}}\left( c,k+i \right) \right\}}.\tag{2.7} \end{equation*}$$
[Proof for (2.7)]

Also, we can reduce $h_{k-r}^{c}$ to

$$\begin{equation} h_{k-r}^{c}={{c}^{-k-1}}{\exp(-c)}{{I}^{\left( 0 \right)}}\left( c,k \right)+{{c}^{-k-1}}{\exp(-c)}r\sum\limits_{i=0}^{\infty }{\frac{{\left( -1 \right)}^{i}{{I}^{\left( 0 \right)}}\left( c,k+i \right)}{\prod\limits_{j=0}^{i}{\left( k+1+j-r \right)}}}. \tag{2.8} \end{equation}$$
[Proof for (2.8)]

As mentioned earlier, we are interested in the gamma integral where the base parameter $s$ is bounded within the interval $\left( 0,1 \right)$. Applying the "$h$" factorization as shown in (2.2), we derive $-1<k-r< 0$ and $k=0$, given $0< r<1$. Thus,

$$\begin{align} h_{-r}^{c}&={{c}^{-1}}{\exp(-c)}\left\{ \left( {\exp(c)}-1 \right)+r\left( \frac{{\exp(c)}-1}{0!} \right)\left( \frac{1}{1-r} \right) \right. \tag{2.9} \\ \notag & +r\left( \frac{c \cdot {\exp(c)}-{\exp(c)}+1}{1!} \right)\left( \frac{1}{2-r}-\frac{1}{1-r} \right) \\ \notag & \left. +r\left( \frac{{{c}^{2}}\cdot {\exp(c)}-2c \cdot {\exp(c)}+2! \cdot {\exp(c)}-2!}{2!} \right)\left( \frac{1}{3-r}-\frac{2}{2-r}+\frac{1}{1-r} \right)+\cdots \right\}. \end{align}$$
[Proof for (2.9)]

Since $\Gamma \left( -r+1 \right)={{c}^{-r+1}}{\exp(-c)}h_{-r}^{-c}{{\Bigr|}_{c\to \infty }}$, we can respecify (2.9) and replace the power-term parameter $c$ with $-c$

$$\begin{equation} h_{-r}^{-c}={{c}^{-1}}{\exp(c)}\left[1-{\exp(-c)} \right]+{{c}^{-1}}{\exp(c)}r\sum\limits_{i=0}^{n }{{{w}_{i}}{{\beta }_{i}}}, \tag{2.10} \end{equation}$$

where ${{w}_{i}}=1-\sum\limits_{j=0}^{i}{\frac{{{c}^{j}}{\exp(-c)}}{j!}}$, ${{\beta }_{i}}=\frac{i!}{\prod\limits_{j=0}^{i}{\left( j+1-r \right)}}$, and $n\to \infty$.
[Proof for (2.10)]

As demonstrated below, we can reduce $h_{-r}^{-c}$ to its simplest form, which directly links to the gamma function of $\Gamma \left( -r \right)$

$$\begin{equation} h_{-r}^{-c}=-{{c}^{-1}}-r{{c}^{r-1}}{\exp(c)}\Gamma \left( -r \right). \tag{2.11} \end{equation}$$
[Proof for (2.11)]

Therefore, "$h$" explains the recursive rule of the gamma function

$$\begin{align*} \Gamma \left( -r+1 \right)&={{c}^{-r+1}}{\exp(-c)}h_{-r}^{-c}{{|}_{c\to \infty }} \tag{2.12} \\ \notag & =-r\Gamma \left( -r \right) \notag. \end{align*}$$
[Proof for (2.12)]

This finding not only defines the algebraic meaning of the "$h$" function by connecting to the gamma function, but also generalizes the factorial function to non-integer cases. As is evident in (2.12), we can explicitly define $\Gamma \left( s \right)$ as a function of $h_{s-1}^{-c}{{|}_{c\to \infty }}$, where $0<s<1$. Hence, all gamma functions defined in (2.3) can be expressed as an "$h$" function for all real arguments, except for the case of non-positive integers.

 

 

 

 

 

 

 

 

 

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