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Integration Trickiness

  1. May 6, 2007 #1
    Hi,

    This integral seems to be coming up a fair bit in questions involving normalisation of wave functions

    [tex]\int x^2\exp{(-ax^2)}dx[/tex]

    and my tutor and lecturer both say to just use the fact that

    [tex]x^2\exp{(-ax^2)}=-\frac{d}{da}\exp{(-ax^2)}[/tex]

    my question simply is how do we use this. it may be obvious can you say

    [tex]\int x^2\exp{(-ax^2)}dx=\int-\frac{d}{da}\exp{(-ax^2)}dx=\int-\frac{dx}{da}\exp{(-ax^2)}d=-\frac{dx}{da}\exp{(-ax^2)}[/tex]

    ...?

    cheers for any help,

    Bart
     
    Last edited: May 6, 2007
  2. jcsd
  3. May 6, 2007 #2

    quasar987

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    It's like this:

    [tex]\int_{-\infty}^{+\infty}x^2\exp{(-ax^2)}dx=\int_{-\infty}^{+\infty}-\frac{d}{da}\exp{(-ax^2)}dx=-\frac{d}{da}\int_{-\infty}^{+\infty}\exp{(-ax^2)}dx[/tex]

    And he assumes that you know what [itex]\int_{-\infty}^{+\infty}\exp{(-ax^2)}dx[/itex] equals to.
     
  4. May 6, 2007 #3

    well i do know that

    [tex]\int_{-\infty}^{+\infty}\exp{(-ax^2)}dx=\sqrt{\frac{\pi}{a}}[/tex]

    so then my next question is how do we work that out (rather than me just knowing it). i started off with integration by parts and then kept on going in circles gettin back to an integral similar to

    [tex]\int_{-\infty}^{+\infty}x^2\exp{(-ax^2)}dx[/tex]

    this isnt an essential question now to what im doing, just out of interest really....


    cheers

    bart
     
  5. May 6, 2007 #4

    quasar987

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  6. May 6, 2007 #5

    quasar987

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    Talking about the identity [itex]\int_{-\infty}^{+\infty}\exp{(-ax^2)}dx=\sqrt{\frac{\pi}{a}}[/itex], Lord Kelvin once said "A mathematician is one to whom that is as obvious as two twice two makes four to you. Liouville was a mathematician." (quote taken from Spivak's calc. on manifolds)

    Welcome to the club of the non-mathematicians.
     
  7. May 6, 2007 #6
    haha....

    looks like i was maybe taking a little too simple approach trying to use integratoin by parts.

    cheers for the help

    bart
     
  8. May 7, 2007 #7

    HallsofIvy

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    Lord Kelvin may have said "twice two is four" but he surely didn't say "two twice two"!
     
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