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Homework Help: Simple integral or not?

  1. Jan 29, 2005 #1
    This should be a fairly simple integral but I can't get it for some reason. Here's the problem:
    [tex]A\int_{-\infty}^{\infty} x e^{-\lambda(x-a)^2}dx[/tex]
    Now I know that
    [tex]\int_{-\infty}^{\infty} e^{-x^2}dx=\sqrt{\pi}[/tex]
    only for those limits.
    Okay so I do parts,
    When you evaluate the integral from dv to get v, you substitue say
    to make it like the second integral i put down, but you can't evaluate it between the limits of -infinity to infinity when doing parts right? And the indefinite integral of this form is not solvable as far as I know.

    BTW, This is for my QM class, finding the average/expectation value of x,

    Am I even going about this the right way? I don't know anymore. Please tell me where I screwed up and point me in the right direction. :cry:
  2. jcsd
  3. Jan 29, 2005 #2


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    Here's a hint :

    [tex]\frac{d}{dx}e^{-x^2} = -2xe^{-x^2}[/tex]

    Use a substitution to get the integrand to a form like the above. You'll still have a term that needs the error function, but you already know how to handle that.
  4. Jan 29, 2005 #3
    Awesome, I got it!
    Much thanks Curious :D
  5. Jan 29, 2005 #4


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    No problem, glad to help. :smile:
  6. Jan 29, 2005 #5

    I'm stuck again :grumpy:

    Okay, so I found <x> and now I have to find <x^2>, which gives,
    [tex]A\int_{-\infty}^{\infty} x^2 e^{-\lambda(x-a)^2}dx[/tex]
    I made the same substitution that I made last time, namely,
    which gives me a error function term that just equals [tex]\sqrt{\pi}[/tex], a term that goes to zero and another term that is,
    and I'm having trouble solving this now.

    Is there another substitution that I should have made initially? I couldn't find one. I swear this is the last time I'll need help :biggrin:. Any help is greatly appreciated.
  7. Jan 30, 2005 #6


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    You can solve the above integral using integration by parts.
  8. Jan 30, 2005 #7
    for an equation of form:


    do integration by parts with

    [tex] u=x [/tex]
  9. Jan 30, 2005 #8
    Geez how could I not see that? Thanks for spending your 300th post helping me out learningphysics! And thanks to you too FulhamFan3.
  10. Jan 30, 2005 #9


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    I hope your first integral was zero... :uhh: As u had to integrate the odd function on a symmetric interval wrt to the origin...


    P.S.The Cauchy principal value...
  11. Jan 30, 2005 #10
    I forget to mention that lambda and a are positive real constants (and A = sqrt(lambda/pi)), so the function is symmetric about x=a not the origin, and the first integral turns out to be just <x> = a, and the second one <x^2> = a^2 + 1/(2*lambda) (for anyone who's interested).
  12. Jan 30, 2005 #11


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    Yes,you're right,it's not symmetric wrt 0...I should have looked better...

    Good thing you finally pulled them through.

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