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Expectation value

  1. Sep 23, 2007 #1
    1. The problem statement, all variables and given/known data

    Can somebody help me integrate [tex]\int{x\cdot p(x)}[/tex] where [itex]p(x)[/itex] is the Gaussian distribution (from here http://hyperphysics.phy-astr.gsu.edu/hbase/math/gaufcn.html)

    3. The attempt at a solution

    I can't really get anywhere. It's true that [itex]\int{e^{x^2}}[/itex] has no analytical solution, right?
     
    Last edited: Sep 23, 2007
  2. jcsd
  3. Sep 23, 2007 #2

    HallsofIvy

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    Yes, it is true that you cannot find the indefinite integral
    [tex]\int e^{-x^2}dx[/tex]
    in terms of elementary functions (though you can find the definite integral for some choices of upper and lower bound).

    However, there is a very simple substitution that will give you
    [tex]\int x e^{-x^2}dx[/tex]
     
  4. Sep 23, 2007 #3

    rock.freak667

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    Is [tex]\int e^{-x^2} dx [/tex]

    not

    [tex]\frac{e^{-x^2}}{-2x}[/tex] + K ?
     
  5. Sep 23, 2007 #4
    But I can't use that easy substitution for [tex]\int{x \cdot e^{-(x-x_0)^2} dx[/tex] for some constant [itex]x_0[/itex], can I?
     
    Last edited: Sep 23, 2007
  6. Sep 23, 2007 #5

    rock.freak667

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    No you can't
     
  7. Sep 24, 2007 #6

    HallsofIvy

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    No, it's not. Why in the world would you think it was?
     
  8. Sep 24, 2007 #7

    rock.freak667

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    because [tex]\frac{d}{dx}(\frac{e^{-x^2}}{-2x}) = e^{-x^2}[/tex]
     
  9. Sep 24, 2007 #8

    dextercioby

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    Why would you think that ?
     
  10. Sep 24, 2007 #9
    Oh, don't be so cutting! I can see why the poster would think that and so can you. It's clear that the poster forgot about the derivative of the denominator (in case you hadn't worked that out).
     
  11. Sep 24, 2007 #10

    rock.freak667

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    ah oh my...stupid me...sorry...my bad
     
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