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*}
$\square$

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