Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (2.13)

$$\begin{equation*} \Gamma \left( -r+1 \right)=\frac{p}{p-q}\int_{0}^{\infty }{\exp (-{{x}^{\frac{p}{p-q}}})dx} \end{equation*}$$ Proof:
Given $r \in \left( 0,1 \right)$, $r={q}/{p}$, and $0<q<p$, we know
$$\begin{align*} \Gamma \left( -r+1 \right) &=\Gamma \left( \frac{p-q}{p} \right) \\ & =\int_{0}^{\infty }{{{t}^{\tfrac{-q}{p}}}\exp \left( -t \right)}dt. \end{align*}$$
Let $t={{x}^{\tfrac{p}{p-q}}}$, and we derive
$$\begin{equation*} dt=\frac{p}{p-q}{{x}^{\tfrac{q}{p-q}}}dx. \end{equation*}$$
By the transformation of variable, we can derive
$$\begin{equation*} \Gamma \left( -r+1 \right)=\frac{p}{p-q}\int_{0}^{\infty }{\exp \left( -{{x}^{\tfrac{p}{p-q}}} \right)}dx. \end{equation*}$$
$\square$

Welcome all math lovers!