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\begin{document}
\title{The gamma-integral formula}
\author{John Kerl}
\maketitle

This is an often-recurring formula which I thought I'd finally jot down for
handy reference.  It's a special case of the gamma function which comes up in
probability and statistics, as well as introductory calculus.

The claim is that
$$
	\int_{0}^\infty x^n e^{-x}\,dx = n!.
$$

\begin{proof}
First take $n=0$.  Then
$$
	\int_{0}^\infty e^{-x}\,dx = \left[-e^{-x}\right]_0^\infty = 1.
$$

Now define
$$
	I_n = \int_{0}^\infty x^n e^{-x}\,dx.
$$
For $n \ge 1$, integrate by parts.  Put
\begin{align*}
	u  &=  x^n       &   dv &=  e^{-x}\,dx \\
	du &= nx^{n-1}   &    v &= -e^{-x}.
\end{align*}
Then
\begin{align*}
	I_n &= \int_{0}^\infty u\,dv \\
	&= \left[uv\right]_0^\infty - \int_{0}^\infty v\,du \\
	&= \left[-x^ne^{-x}\right]_0^\infty + n \int_{0}^\infty x^{n-1} e^{-x}\,dx\\
	&= 0 + n I_{n-1}.
\end{align*}
This recurrence relation defines $n!$ so we have
$$
	\int_{0}^\infty x^n e^{-x}\,dx = n!.
$$
\end{proof}

The \textbf{gamma function} is defined to be the value of this integral, even
when $n$ is not necessarily an integer:
$$
	\Gamma(z) := \int_0^\infty x^{z-1} e^{-x}\,dx.
$$
(This blows up when $z$ is zero or a negative integer.) So, we have
$$
	\Gamma(n) = (n-1)!.
$$

\end{document}
