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Density function for a normal distribution

  1. Mar 9, 2015 #1
    1. The problem statement, all variables and given/known data
    I have to prove that ## \int e^{\frac{x^2}{-2}}dx ## from +∞ to -∞ = ##\sqrt{2\pi} ##

    2. Relevant equations
    N/A

    3. The attempt at a solution
    My GSI went from
    1) ## \int e^{\frac{x^2}{-2}}dx ## from +∞ to -∞ = ##\sqrt{2\pi} ##
    to
    2) ## (\int e^{\frac{x^2}{-2}}dx)(\int e^{\frac{y^2}{-2}}dy) ## is equal to ## 2\pi ##
    Where did the red part of the function come from?
    This leads to another question, how do we convert a single integral to multiple integrals? Thank you.
     
  2. jcsd
  3. Mar 9, 2015 #2

    statdad

    User Avatar
    Homework Helper

    If you start with this:
    [tex]
    I = \int_{-\infty}^\infty e^{-\frac{x^2} 2} \, dx
    [/tex]
    you can also write it as
    [tex]
    I = \int_{-\infty}^\infty e^{-\frac{y^2} 2} \, dy
    [/tex]
    since the variable of integration is immaterial. Multiplying I with itself gives
    [tex]
    I^2 = \left(\int_{-\infty}^\infty e^{-\frac{x^2} 2} \, dx\right) \left(\int_{-\infty}^\infty e^{-\frac{y^2} 2} \, dy\right)
    [/tex]

    This product of two integrals can be written as a double integral. If you can do that, and then show that [itex] I^2 = 2 \pi [/itex], you will
    essentially be done, no?
     
  4. Mar 9, 2015 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Google is your friend. See, eg.,
    http://mathworld.wolfram.com/GaussianIntegral.html or
    http://en.wikipedia.org/wiki/Gaussian_integral
     
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