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
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)]
\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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