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Find a general formula

  1. Apr 10, 2008 #1
    Hi,

    I have to find a general formula for the function

    1/(2n)![tex]\int_{-\inf}^{\inf}x^{2n}*e^{-ax^2} [/tex]


    I am a little bit lost in how to proceed - any hints appreciated thanks in advance
     
  2. jcsd
  3. Apr 10, 2008 #2

    tiny-tim

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    Hi greisen! :smile:

    (btw, it's "\infty")

    I assume you've tried integrating by parts, and found that if you start with the x^2n, it just gets worse, and you can't start with the e^{-ax^2}, because it doesn't have a nice anti-derivative?

    Hint: can you split the product in some other way, so that you can get an anti-derivative involving the e^{-ax^2}? :smile:
     
  4. Apr 10, 2008 #3
    Thanks

    yes I have tried integration by parts but that got very messy.

    I don't quite understand how to rewrite the equation involving the anti-derivative?

    Best,
     
  5. Apr 10, 2008 #4
    well, i would say sth like this would put u on the right track

    [tex]\int_{-\infty}^{\infty}x^{2n}*e^{-ax^2} = \int_{-\infty}^{\infty}x^{2n-1}xe^{-ax^2}dx [/tex] then performing integration by parts by letting

    [tex]u=x^{2n-1},v=\int xe^{-ax^2}dx[/tex]

    I think this would work. By performing integ by parts a couple of times, i think you will be able to notice some pattern, then use induction to prove the general case.

    P.S. This is basically what tiny-tim's Hint says...i hope at least...lol...
     
  6. Apr 10, 2008 #5

    tiny-tim

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    problem … solution … it's just a state of mind!

    Yutup, sutupidmath is right, of course. :smile:

    The trick is to recognise the difference between a problem and a solution.

    In this case, the problem is that you look at ∫e^-ax^2
    and think
    "I can't integrate that … if only it had an extra 2ax in front of it!"
    and change that round to
    "I can integrate that if i put an extra 2ax in front of it!"

    Problem … solution … it's just a state of mind! :smile:
     
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