Indefinite Integration of the Gamma Integral
and Related Statistical Applications
1.2 Rsearch Problem
Are the gamma, Gaussian, and exponential integrals truly unsolvable? If we set aside the requirement of the elementary
function, the antiderivative of the three integrals can be expressed as an infinite series. To see why this is so, we first define a
general form of the gamma integral
\begin{equation*}
g\left( s,c,u \right)\equiv \int{{{u}^{s-1}}\exp(cu)du},
\end{equation*}
where $s$ and $c$ can be any real numbers. If we assign particular values to both parameters, the Gaussian and exponential integrals can
be viewed as a special case of the gamma integral. For example, when $s$ is a non-positive integer, it is an exponential integral of
$(1-s)$th order. Specifically, it results in the first-order exponential integral if $s$ is set to $0$
\int{\frac{{\exp(-u)}}{u}du}=g\left( 0,-1,u \right).
\end{equation*}
Moreover, by substitution of variables, we can derive an identity of the Gaussian integral as a gamma integral with $s=1/2$ and $c=-1$
\begin{equation*}
\int{{\exp(-{{u}^{2}})}}du=\frac{1}{2}g\left( \frac{1}{2},-1,u^{2} \right). \tag{1.1}
\end{equation*}
[Proof for (1.1)]
Therefore, all three integrals can be generalized into a gamma integral $g\left( s,c,u \right)$.
Let $c=-1$ and set the lower and upper limits of the integral as $0$ and $\infty $. We will derive the gamma function
$\Gamma \left( s \right)$. Solving a gamma integral is straightforward when we use the Taylor expansion and then integrate the series term by term
\begin{equation*}
g(s,-1,u)=\frac{{{u}^{s}}}{0!\left( s \right)}-\frac{{{u}^{s+1}}}{1!\left( s+1 \right)}+\frac{{{u}^{s+2}}}{2!\left( s+2
\right)}-\frac{{{u}^{s+3}}}{3!\left( s+3 \right)}+\cdots . \tag{1.2}
\end{equation*}
[Proof for (1.2)]
This solution is an infinite series and usually applied to evaluation of the lower incomplete gamma function, which
can be specified as
\begin{equation*}
\gamma \left( s,u \right)=\int_{0}^{u}{{{x}^{s-1}}{\exp(-x)}dx}.
\end{equation*}
We can also express the lower incomplete gamma function as an infinite series of Kummer's confluent hypergeometric function
\begin{equation*}
\gamma \left( s,u \right)=\frac{{{u}^{s}}}{s}M\left( s,s+1,-u \right), \tag{1.3}
\end{equation*}
[Proof for (1.3)]
where the Kummer's confluent hypergeometric function is defined as
\begin{equation*}
M\left( a,b,u \right)=1+\frac{a}{b}u+\frac{a\left( a+1 \right)}{b\left( b+1 \right)}\frac{{{u}^{2}}}{2!}+\frac{a\left( a+1 \right)\left(
a+2 \right)}{b\left( b+1 \right)\left( b+2 \right)}\frac{{{u}^{3}}}{3!}+\cdots.
\end{equation*}
Respecifying $\gamma \left( s,u \right)$ as an indefinite integral with variable $u$, we can regard the infinite series of Kummer's
confluent hypergeometric function as a solution to the indefinite integration of the gamma integral
\begin{equation*}
\int{{{u}^{s-1}}{\exp(-u)}du=}\frac{{{u}^{s}}}{s}M\left( s,s+1,-u \right). \tag{1.4}
\end{equation*}
[Proof for (1.4)]
However, neither (1.2) nor (1.4) is not widely recognized as a general solution for the indefinite integration of the gamma integral.
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