area under Gaussian curveAreaUnderGaussianCurve2005-05-17 14:23:552008-08-15 17:18:33Theoremimproper integraldouble integralpolar coordinates% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
\usepackage{graphicx}
% for neatly defining theorems and propositions
\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here
\newcommand{\sijoitus}[2]%
{\operatornamewithlimits{\Big/}_{\!\!\!#1}^{\,#2}}
\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem}\begin{thmplain}
The area between the curve\,\, $y = e^{-x^2}$\, and the $x$-axis equals $\sqrt{\pi}$,\, i.e.
$$\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}.$$
\end{thmplain}
\begin{figure}[!htb]
\begin{center}
\includegraphics{gaussian-curve.eps}
\end{center}
\end{figure}
{\em Proof.}\, The square of the area is
\begin{align*}
\bigg(\int_{-\infty}^\infty e^{-x^2}\,dx\bigg)^2
& = \lim_{a\to\infty}\bigg(\int_{-a}^a e^{-x^2}\,dx\bigg)^2\\
& = \lim_{a\to\infty}\int_{-a}^a e^{-x^2}\,dx\,\cdot\int_{-a}^a e^{-y^2}\,dy\\
& = \lim_{a\to\infty}\int_{-a}^a \int_{-a}^a e^{-(x^2+y^2)}\,dx\,dy\\
& = \lim_{R\to\infty}\int_0^R\!\int_0^{2\pi}e^{-r^2}r\,dr\,d\varphi\\
& = \lim_{R\to\infty}2\pi\!\int_0^R e^{-r^2}r\,dr\\
& = -\pi\!\lim_{R\to\infty}\!\sijoitus{0}{\quad\,\,R}e^{-r^2}\\
& = \pi\!\lim_{R\to\infty}(1-e^{-R^2}) \;=\; \pi.
\end{align*}
Here, the limit of the double integral over a square has been replaced by the limit of the double integral over a disc, because both limits are equal.\, That both limits are equal can be demonstrated by the elementary \PMlinkescapetext{estimate}
$$0 \leq \int_{-a}^a \int_{-a}^a\!e^{-(x^2 + y^2)}\,dx\,dy
- \int_0^a\!\int_0^{2\pi}\!e^{-r^2} r\,dr\,d\varphi
\leq \underbrace{e^{-a^2}}_{greatest\,value}\!\cdot\,\underbrace{(4a^2\!-\!\pi a^2)}_{area}
= (4\!-\!\pi)\!\cdot\!\frac{a^2}{e^{a^2}},$$
and\, $\frac{a^2}{e^{a^2}} \to 0$\, when\, $a \to \infty$\, (see growth of exponential function).\\
\textbf{Remark.}\, Since $e^{-x^2}$ is an even function,
\begin{align*}
\int_0^\infty e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}\:\cdot
\end{align*}