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A integral about expotential function

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=36742&stc=1&d=1309100124.jpg

    2. Relevant equations
    Unknown

    3. The attempt at a solution

    attachment.php?attachmentid=36743&stc=1&d=1309100124.jpg

    How to calculate this integral?
    I have tried substitution and by parts. But they fail to get the answer.
     

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  2. jcsd
  3. Jun 26, 2011 #2

    hunt_mat

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    There is a trick for this, square the integral:
    [tex]
    \left(\int_{-\infty}^{\infty}e^{-x^{2}}dx\right)^{2}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^{2}+y^{2})}dxdy
    [/tex]
    Now the idea is to make a change of co-ordinates from cartesian to polar...
     
  4. Jun 26, 2011 #3

    dextercioby

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    Are you taking quantum mechanics before knowing what a gaussian integral is ? Because you shouldn't...
     
  5. Jun 26, 2011 #4
    Oh you are right.
    In fact, I am self-learning quantum mechanics, and I don't know what maths skills I need to have.
     
  6. Jun 27, 2011 #5

    hunt_mat

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    So move on from my explanation:
    [tex]
    \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^{2}+y^{2})}dxdy=\int_{0}^{2\pi}\int_{0}^{\infty}re^{-r^{2}}drd\theta =2\pi\int_{0}^{\infty}re^{-r^{2}}dr
    [/tex]
    I think I will leave the last bit to you as it's standard integration.
     
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