Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (1.4)

\begin{equation*}
\int{{{u}^{s-1}}{\exp(-u)}du=}\frac{{{u}^{s}}}{s}M\left( s,s+1,-u \right).
\end{equation*} Proof:

\begin{align*}
& \int{{{u}^{s-1}}\exp (-u)du} \\
=&\int{\frac{{{u}^{s-1}}}{0!}du}-\int{\frac{{{u}^{s}}}{1!}du}+\int{\frac{{{u}^{s+1}}}{2!}du}-\int{\frac{{{u}^{s+2}}}{3!}du}+\cdots\\
=&\frac{{{u}^{s}}}{0!s}-\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  \\
=&\frac{{{u}^{s}}}{s}\left\{ 1-\frac{s}{\left( s+1 \right)}\cdot u+\frac{s\left( s+1 \right)}{\left( s+1 \right)\left( s+2 \right)}\cdot \frac{{{u}^{2}}}{2!}-\frac{s\left( s+1 \right)\left( s+2 \right)}{\left( s+1 \right)\left( s+2 \right)\left( s+3 \right)}\cdot \frac{{{u}^{3}}}{3!}+\cdots  \right\} \\
=&\frac{{{u}^{s}}}{s}M\left( s,s+1,-u \right) \\
\end{align*} $\square$

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