Expectation Value Homework: Integrating Gaussian Distribution

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Homework Help Overview

The discussion revolves around integrating the expression \(\int{x\cdot p(x)}\), where \(p(x)\) represents the Gaussian distribution. Participants are exploring the challenges associated with this integral, particularly focusing on the properties of the Gaussian function.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the difficulty of finding an analytical solution for \(\int e^{-x^2}dx\) and the implications of this for the original integral. There are attempts to clarify the use of substitutions and the validity of certain expressions related to the integral.

Discussion Status

The conversation includes some guidance regarding substitutions that could simplify the integral, although there is a lack of consensus on the correctness of specific expressions. Participants are actively questioning each other's reasoning and clarifying misunderstandings without reaching a definitive conclusion.

Contextual Notes

There is an ongoing examination of the assumptions related to the integrals being discussed, particularly regarding the limits of integration and the nature of the Gaussian function. Some participants express uncertainty about the application of certain mathematical rules in this context.

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

not

[tex]\frac{e^{-x^2}}{-2x}[/tex] + K ?
 
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:
No you can't
 
rock.freak667 said:
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?
 
HallsofIvy said:
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]
 
rock.freak667 said:
because [tex]\frac{d}{dx}(\frac{e^{-x^2}}{-2x}) = e^{-x^2}[/tex]

Why would you think that ?
 
dextercioby said:
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
ah oh my...stupid me...sorry...my bad
 

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