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Integration of a gaussian function

  1. Aug 27, 2011 #1
    Hey, all. I need help computing a particular integral that's come up in my quantum mechanics homework (Griffith 1.3 for anyone who's interested). It involves integration of a gaussian function:

    [tex]\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {x}{e^{{-\lambda}{(x-a)^2}}} \, dx[/tex]

    Where [tex]\lambda[/tex] and [tex]a[/tex] are positive real constants.


    For the life of me I cannot figure out how to compute that integral... u-substitution doesn't seem to be an option, and integration by parts just makes things more confusing. Can someone please help me?
     
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  3. Aug 27, 2011 #2

    Ben Niehoff

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    You could do it easily if you had

    [tex]\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {(x-a)}{e^{{-\lambda}{(x-a)^2}}} \, dx[/tex]
    right? Because then it would be a simple substitution. So write

    [tex]\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {x}{e^{{-\lambda}{(x-a)^2}}} \, dx = \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {(x-a)}{e^{{-\lambda}{(x-a)^2}}} \, dx + \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {a}{e^{{-\lambda}{(x-a)^2}}} \, dx[/tex]
    and think about how to do the second piece. It should look familiar.
     
  4. Aug 27, 2011 #3
    I hadn't thought of that! That's a clever trick. Am I correct in thinking the second integral simplifies to:

    [tex]{\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {a}{e^{{-\lambda}{(x-a)^2}}} \, dx}={a}[/tex]

    And so the integral:


    [tex]\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {x}{e^{{-\lambda}{(x-a)^2}}} \, dx[/tex]

    is simply the first integral you wrote down (solvable by u-substitution) + a?
     
  5. Aug 27, 2011 #4

    gb7nash

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    How did you obtain a from that? I'm not saying that's wrong, but you need to show how you got it.
     
  6. Aug 27, 2011 #5
    I'm sorry! Here's the process I went through:

    [tex]{\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {a}{e^{{-\lambda}{(x-a)^2}}} \, dx}[/tex]

    Because a is a constant, it can be taken out of the integral:

    [tex]{{a}\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {e^{{-\lambda}{(x-a)^2}}} \, dx}[/tex]

    Then I performed a u-substitution:

    [tex]{u ={x-a}},{du = dx}[/tex]

    [tex]{{a}\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {e^{{-\lambda}{u^2}}} \, du}[/tex]

    I then used a second u-substitution:

    [tex]{v = {\sqrt{\lambda}u}}, {\frac{dv}{\sqrt{\lambda}} = du}[/tex]

    [tex]{{a}\sqrt{\frac{1}{\pi}}\int_{-\infty}^{+\infty} \! {e^{{-\lambda}{v^2}}} \, dv}[/tex]


    The integral is the well-known Gaussian integral:

    [tex]{\int_{-\infty}^{+\infty} \! {e^{{-\lambda}{v^2}}} \, dv}={\sqrt{\pi}}[/tex]

    And so I have:

    [tex]{{a}{\sqrt{\frac{1}{\pi}}}\sqrt{\pi}}={a}[/tex]
     
  7. Aug 28, 2011 #6
    Now I'm a bit confused, though... I came across an integral on Wikipedia's list of integrals for exponential functions, and they have this:

    [tex]\int_{-\infty}^{\infty} \! {x}{e^{-a{(x-b)^2}}} \ dx={b\sqrt{\frac{\pi}{a}}}[/tex]

    How does the integral we've broken into two parts become that?
     
  8. Aug 30, 2011 #7
    Is anyone able to help? This is seriously stumping me.
     
  9. Aug 30, 2011 #8

    Ray Vickson

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    You have already been given a complete solution, but it does require some final work on your part.


    RGV
     
    Last edited: Aug 30, 2011
  10. Aug 30, 2011 #9
    I'm sorry, but I'm having difficulty "seeing" this complete solution and how to work it.

    I understand what was said earlier, that:

    [tex]{\sqrt{\frac{\lambda}{\pi}} \int_{-\infty}^{+\infty} \! {x}{e^{-\lambda{(x-a)^2}}} \ dx} = {\sqrt{\frac{\lambda}{\pi}} \int_{-\infty}^{+\infty} \! {(x-a)}{e^{-\lambda{(x-a)^2}}} \ dx} + {\sqrt{\frac{\lambda}{\pi}} \int_{-\infty}^{+\infty} \! {a}{e^{-\lambda{(x-a)^2}}} \ dx}[/tex]

    It seems to me that the first integral is solvable by u-substitution:

    [tex]{u=(x-a)}, {du=dx}[/tex]
    [tex]{\sqrt{\frac{\lambda}{\pi}} \int_{-\infty}^{+\infty} \! {(x-a)}{e^{-\lambda{(x-a)^2}}} \ dx} = {\sqrt{\frac{\lambda}{\pi}} \int_{-\infty}^{+\infty} \! {u}{e^{-\lambda{u^2}}} \ dx}[/tex]

    The antiderivative of that integral is:
    [tex]\frac{-e^{-\lambda{u^2}}}{2\lambda}[/tex]

    Evaluating that antiderivative at the limits of negative and positive infinity seems to yield 0.

    Meanwhile, as I've shown about, the second integral can be solved by two round of u-substitution, yielding the constant a. Altogether, then:

    [tex]{\sqrt{\frac{\lambda}{\pi}} \int_{-\infty}^{+\infty} \! {x}{e^{-\lambda{(x-a)^2}}} \ dx}=a[/tex]

    However, that doesn't mesh with the definite integral identity I found online (shown in an earlier reply). Have I gone awry somewhere in my calculations, or is the identity I found incorrect?

    EDIT: Nevermind... I realized that in examining the integral identity I found in an integral table, I'd been neglecting to apply the constant in front of the integral. Once I did that, it all clicked and became equivalent to the answer I derived (with help) above. Thank you all so much. I apologize for being a bit dense toward the end.
     
    Last edited: Aug 31, 2011
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