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[parent] area under Gaussian curve (Theorem)
Theorem 1   The area between the curve $ y = e^{-x^2}$ and the $ x$-axis equals $ \sqrt{\pi}$, i.e.
$\displaystyle \int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}.$
\includegraphics{gaussian-curve.eps}

Proof. The square of the area is

$\displaystyle \bigg(\int_{-\infty}^\infty e^{-x^2} dx\bigg)^2$ $\displaystyle = \lim_{a\to\infty}\bigg(\int_{-a}^a e^{-x^2} dx\bigg)^2$    
  $\displaystyle = \lim_{a\to\infty}\int_{-a}^a e^{-x^2} dx \cdot\int_{-a}^a e^{-y^2} dy$    
  $\displaystyle = \lim_{a\to\infty}\int_{-a}^a \int_{-a}^a e^{-(x^2+y^2)} dx dy$    
  $\displaystyle = \lim_{R\to\infty}\int_0^R\!\int_0^{2\pi}e^{-r^2}r dr d\varphi$    
  $\displaystyle = \lim_{R\to\infty}2\pi\!\int_0^R e^{-r^2}r dr$    
  $\displaystyle = -\pi\!\lim_{R\to\infty}\!\operatornamewithlimits{\Big/}_{\!\!\!0}^{ \quad R}e^{-r^2}$    
  $\displaystyle = \pi\!\lim_{R\to\infty}(1-e^{-R^2}) \;=\; \pi.$    

Here, the limit of the double integral over a square has been replaced by the limit of the double integral over a disc, because both limits are equal.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 estimate

$\displaystyle 0 \leq \int_{-a}^a \int_{-a}^a\!e^{-(x^2 + y^2)} dx dy - \int_0... ...nderbrace{(4a^2\!-\!\pi a^2)}_{area} = (4\!-\!\pi)\!\cdot\!\frac{a^2}{e^{a^2}},$
and $ \frac{a^2}{e^{a^2}} \to 0$ when $ a \to \infty$ (see growth of exponential function).

Remark. Since $ e^{-x^2}$ is an even function,

$\displaystyle \int_0^\infty e^{-x^2} dx=\frac{\sqrt{\pi}}{2}\:\cdot$    



"area under Gaussian curve" is owned by pahio. [ full author list (4) ]
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See Also: substitution notation, proof that normal distribution is a distribution, probability distribution function, error function, evaluating the gamma function at 1/2, normal random variable, table of probabilities of standard normal distribution

Other names:  area under the bell curve
Keywords:  double integral, polar coordinates

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Cross-references: even function, growth of exponential function, disc, integral, limit, square, curve, area
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This is version 18 of area under Gaussian curve, born on 2005-05-17, modified 2008-04-19.
Object id is 7065, canonical name is AreaUnderGaussianCurve.
Accessed 11523 times total.

Classification:
AMS MSC26A36 (Real functions :: Functions of one variable :: Antidifferentiation)
 26B15 (Real functions :: Functions of several variables :: Integration: length, area, volume)

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request "integration" 2008-3-5 by pahio on 2008-03-05 12:17:38
itsiffi,
For convergence of the integral the coefficient (ia) of y^2 must be negative. It were clearer that instead of "ia" you write e.g. -a.
See the entry
http://planetmath.org/encyclopedia/AreaUnderGaussianCurve.html
Perform the substitution
y = t/\sqrt{a}, dy = dt/\sqrt{a}.
You will easily obtain the result \sqrt{\pi/a}.
Jussi
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