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Gaussian Integral Help

  1. Sep 15, 2014 #1
    1. The problem statement, all variables and given/known data

    I need to evaluate the following integral: [tex]\sqrt{\frac{2}{\pi}}\frac{\sigma}{\hbar}\int\limits_{-\infty}^{\infty}p^2 e^{-32\sigma^2(p-p_0)^2/\hbar^2}\,dp[/tex]

    2. Relevant equations

    Integrals of the form: [tex] \int\limits_{-\infty}^{\infty}x^2e^{-ax^2}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{a^3}}[/tex], taken from a table

    3. The attempt at a solution

    Ok so, obviously the integral provided is not of the form of the one given in the table, however, it is close. My thought was to change the integral to be such that we let some new variable [itex] t = (p-p_0)^2[/itex] so then we would have the following new integral [tex]\sqrt{\frac{2}{\pi}}\frac{\sigma}{\hbar}\int\limits_{-\infty}^{\infty}(t+p_0)^2 e^{-32\sigma^2 t^2/\hbar^2}\,dt[/tex]. Then I would expand it so that there are three integral that need to be evaluated i.e. (by expanding the [itex](t + p_0)^2[/itex] term.

    My question is... is this mathematically sound (or allowable), I can't see a reason why this shouldn't work but I wanted to get some feedback on my proposed method. Thank you
     
  2. jcsd
  3. Sep 15, 2014 #2

    nrqed

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    What you wrote is not correct because with your change of variable, ##dt \neq dp##.
    But try instead ##t=p-p_0##, that will get you closer to what you want.
     
  4. Sep 15, 2014 #3
    Ok yes, I see that mistake now, so we have that [itex] t=p-p_0[/itex], then [itex]dp=dt[/itex] and so then [itex]p = t + p_0 \implies p^2 = p_0^2 + 2p_0 t + t^2[/itex] giving way to the following:
    [tex] \sqrt{\frac{2}{\pi}}\frac{\sigma}{\hbar}\left[\,\int\limits_{-\infty}^\infty p_0^2 e^{-32\sigma^2t^2/\hbar^2}dt + \int\limits_{-\infty}^\infty 2p_0 t e^{-32\sigma^2t^2/\hbar^2}dt + \int\limits_{-\infty}^\infty t^2 e^{-32\sigma^2t^2/\hbar^2}dt\right][/tex]. Which I believe is correct? Unless I made a algebra error, or am just missing something more subtle here.
     
  5. Sep 15, 2014 #4

    Ray Vickson

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    I disagree with nreqd; what you wrote was absolutely correct, at least as I see it on my screen (so assuming you did not 'edit out' an error). Expanding the square ##(t + p_0)^2## and integrating term-by-term is the standard way these things are done.
     
  6. Sep 15, 2014 #5
    I think my last reply and is the same thing I had written at first, just expanded out. I think I made an error in what I said [itex]t[/itex] should be. I wrote [itex]t=(p-p_0)^2[/itex], but what I wrote for the integral was correct, but my original statement that [itex]t=(p-p_0)^2[/itex] was not and hence he was right that [itex]dp\neq dt[/itex]. At least, that's what I gathered from the replies.
     
  7. Sep 15, 2014 #6

    Matterwave

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    Yes, this is what happened. You said ##t=(p-p_0)^2## in your OP but wrote the integral assuming ##t=p-p_0##.

    You are now ready to do each integral term by term. The first 2 terms should not be too tricky, but the third term might require a little bit of a trick.
     
  8. Sep 15, 2014 #7
    The first integral is just of the form [tex]\int\limits_{-\infty}^\infty e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}[/tex] the second integral is zero and the third should be [tex]\int\limits_{-\infty}^\infty x^2 e^{-ax^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{a^3}}[/tex], or at least I think so.
     
  9. Sep 15, 2014 #8

    Matterwave

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    Yes. Those formulas look alright to me.
     
  10. Sep 15, 2014 #9
    Perfect, thank you all for your help.
     
  11. Sep 15, 2014 #10

    Ray Vickson

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    In fact, if ##t = p \pm p_0## for constant ##p_0## then, indeed, ##dp = dt## is true. This is just standard change-of-variable stuff.
     
  12. Sep 15, 2014 #11

    nrqed

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    Yes but if you read carefully the first post, the OP wrote that he used ##t =(p-p_0)^2 ##. Then the expression written was incorrect. It's only after the fact that I realized that the integral was correct but that he actually had used ## t = p-p_0##, which is correct. So we are both right, it is just that I focused on the change of variable he wrote and did not look in details at the integral and you did the inverse.
     
  13. Sep 16, 2014 #12

    Ray Vickson

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    OK, I stand corrected.
     
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