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Expectation value

  • Thread starter cscott
  • Start date
  • #1
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Homework Statement



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)

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:

Answers and Replies

  • #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]
 
  • #3
rock.freak667
Homework Helper
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Is [tex]\int e^{-x^2} dx [/tex]

not

[tex]\frac{e^{-x^2}}{-2x}[/tex] + K ?
 
  • #4
782
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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?
 
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  • #5
rock.freak667
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No you can't
 
  • #6
HallsofIvy
Science Advisor
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Is [tex]\int e^{-x^2} dx [/tex]

not

[tex]\frac{e^{-x^2}}{-2x}[/tex] + K ?
No, it's not. Why in the world would you think it was?
 
  • #7
rock.freak667
Homework Helper
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No, it's not. Why in the world would you think it was?
because [tex]\frac{d}{dx}(\frac{e^{-x^2}}{-2x}) = e^{-x^2}[/tex]
 
  • #9
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Why would you think that ?
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).
 
  • #10
rock.freak667
Homework Helper
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ah oh my...stupid me...sorry...my bad
 

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