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Math Help - Tricky Gauss Integral problem

  1. #1
    Senior Member bugatti79's Avatar
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    Tricky Gauss Integral problem

    Hi, I am having difficulty getting to the end with this one...see attached


    Thanks
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by bugatti79 View Post
    Hi, I am having difficulty getting to the end with this one...see attached


    Thanks
    Keep in mind that \displaystyle\int_{-\infty}^{\infty}e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}}. (*)

    I take it you mean the original problem is \displaystyle\int_{-\infty}^{\infty}\exp\left(-\left[\frac{1}{4a}+\frac{i\hbar t}{2m}\right]k^2+ikx\right)\,dk, because the 2m term appears out of nowhere in the second step.

    For the moment, let's focus on the exponent:

    -\left(\dfrac{1}{4a}+\dfrac{i\hbar t}{2m}\right)k^2+ixk.

    Rearranging, we get

    -\dfrac{m+2ai\hbar t}{4am}\left[k^2-\dfrac{4amix}{m+2ai\hbar t}k\right] (Verify)

    Complete the square to get

    -\dfrac{m+2ai\hbar t}{4am}\left[k-\dfrac{2amix}{m+2ai\hbar t}\right]^2-\dfrac{amx^2}{m+2ai\hbar t} (Verify)

    With this, we can now evaluate this integral instead:

    \displaystyle\int_{-\infty}^{\infty}\exp\left(-\frac{m+2ai\hbar t}{4am}\left[k-\frac{2amix}{m+2ai\hbar t}\right]^2-\frac{amx^2}{m+2ai\hbar t}\right)\,dk \displaystyle=\exp\left(-\frac{amx^2}{m+2ai\hbar t}\right)\int_{-\infty}^{\infty}\exp\left(-\frac{m+2ai\hbar t}{4am}\left[k-\frac{2amix}{m+2ai\hbar t}\right]^2\right)\,dk

    Now apply the substitution u=k-\dfrac{2amix}{m+2ai\hbar t}\implies \,du=\,dk.

    Thus, the integral becomes

    \displaystyle\exp\left(-\frac{amx^2}{m+2ai\hbar t}\right)\int_{-\infty}^{\infty}\exp\left(-\frac{m+2ai\hbar t}{4am}u^2\right)\,du.

    Now apply (*) to get

    \exp\left(-\dfrac{amx^2}{m+2ai\hbar t}\right)\sqrt{\dfrac{4am\pi}{m+2ai\hbar t}}=\dfrac{\exp\left(-\dfrac{ax^2}{1+2ai\hbar t/m}\right)}{\sqrt{\dfrac{m+2ai\hbar t}{4am\pi}}}=\dfrac{\exp\left(-\dfrac{ax^2}{1+2ai\hbar t/m}\right)}{\sqrt{\dfrac{1}{4a\pi}+\dfrac{i\hbar t}{2m\pi}}}

    which appears to be different from the answer you have supplied....so maybe I made a mistake somewhere??

    But this is the idea of how to approach this problem. Hope this helps!
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Keep in mind that \displaystyle\int_{-\infty}^{\infty}e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}}. (*)

    I take it you mean the original problem is \displaystyle\int_{-\infty}^{\infty}\exp\left(-\left[\frac{1}{4a}+\frac{i\hbar t}{2m}\right]k^2+ikx\right)\,dk, because the 2m term appears out of nowhere in the second step.

    For the moment, let's focus on the exponent:

    -\left(\dfrac{1}{4a}+\dfrac{i\hbar t}{2m}\right)k^2+ixk.

    Rearranging, we get

    -\dfrac{m+2ai\hbar t}{4am}\left[k^2-\dfrac{4amix}{m+2ai\hbar t}k\right] (Verify)

    Complete the square to get

    -\dfrac{m+2ai\hbar t}{4am}\left[k-\dfrac{2amix}{m+2ai\hbar t}\right]^2-\dfrac{amx^2}{m+2ai\hbar t} (Verify)

    With this, we can now evaluate this integral instead:

    \displaystyle\int_{-\infty}^{\infty}\exp\left(-\frac{m+2ai\hbar t}{4am}\left[k-\frac{2amix}{m+2ai\hbar t}\right]^2-\frac{amx^2}{m+2ai\hbar t}\right)\,dk \displaystyle=\exp\left(-\frac{amx^2}{m+2ai\hbar t}\right)\int_{-\infty}^{\infty}\exp\left(-\frac{m+2ai\hbar t}{4am}\left[k-\frac{2amix}{m+2ai\hbar t}\right]^2\right)\,dk

    Now apply the substitution u=k-\dfrac{2amix}{m+2ai\hbar t}\implies \,du=\,dk.

    Thus, the integral becomes

    \displaystyle\exp\left(-\frac{amx^2}{m+2ai\hbar t}\right)\int_{-\infty}^{\infty}\exp\left(-\frac{m+2ai\hbar t}{4am}u^2\right)\,du.

    Now apply (*) to get

    \exp\left(-\dfrac{amx^2}{m+2ai\hbar t}\right)\sqrt{\dfrac{4am\pi}{m+2ai\hbar t}}=\dfrac{\exp\left(-\dfrac{ax^2}{1+2ai\hbar t/m}\right)}{\sqrt{\dfrac{m+2ai\hbar t}{4am\pi}}}=\dfrac{\exp\left(-\dfrac{ax^2}{1+2ai\hbar t/m}\right)}{\sqrt{\dfrac{1}{4a\pi}+\dfrac{i\hbar t}{2m\pi}}}

    which appears to be different from the answer you have supplied....so maybe I made a mistake somewhere??

    But this is the idea of how to approach this problem. Hope this helps!
    Hi Chris,

    Thanks for your great help, I will have a look at it when I get the chance and will reply soon.

    Cheers!
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  4. #4
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Keep in mind that \displaystyle\int_{-\infty}^{\infty}e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}}. (*)

    I take it you mean the original problem is \displaystyle\int_{-\infty}^{\infty}\exp\left(-\left[\frac{1}{4a}+\frac{i\hbar t}{2m}\right]k^2+ikx\right)\,dk, because the 2m term appears out of nowhere in the second step.

    For the moment, let's focus on the exponent:

    -\left(\dfrac{1}{4a}+\dfrac{i\hbar t}{2m}\right)k^2+ixk.

    Rearranging, we get

    -\dfrac{m+2ai\hbar t}{4am}\left[k^2-\dfrac{4amix}{m+2ai\hbar t}k\right] (Verify)

    Complete the square to get

    -\dfrac{m+2ai\hbar t}{4am}\left[k-\dfrac{2amix}{m+2ai\hbar t}\right]^2-\dfrac{amx^2}{m+2ai\hbar t} (Verify)
    Yes your right, the 2m should be in there from the start.
    I can see the algebra but i dont understand the idea of completing the square. Why do we have to re-arrange it when we already have it the form -(ak^2 +bk)?....
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by bugatti79 View Post
    Yes your right, the 2m should be in there from the start.
    I can see the algebra but i dont understand the idea of completing the square. Why do we have to re-arrange it when we already have it the form -(ak^2 +bk)?....
    We have to rearrange it because completing the square only works when the coefficient of the k^2 term is 1.
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  6. #6
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by bugatti79 View Post
    Yes your right, the 2m should be in there from the start.
    I can see the algebra but i dont understand the idea of completing the square. Why do we have to re-arrange it when we already have it the form -(ak^2 +bk)?....
    You mean the coefficient of k has to be 1 and not k^ as below in order to get du=dk?

    -\dfrac{m+2ai\hbar t}{4am}\left[k-\dfrac{2amix}{m+2ai\hbar t}\right]^2-\dfrac{amx^2}{m+2ai\hbar t}
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by bugatti79 View Post
    You mean the coefficient of k has to be 1 and not k^ as below in order to get du=dk?

    -\dfrac{m+2ai\hbar t}{4am}\left[k-\dfrac{2amix}{m+2ai\hbar t}\right]^2-\dfrac{amx^2}{m+2ai\hbar t}
    As I said in my previous response, you need to have the coefficient of the k^2 term to be 1 in order to complete the square. Thats why I did the following:

    Quote Originally Posted by Chris L T521 View Post
    For the moment, let's focus on the exponent:

    -\left(\dfrac{1}{4a}+\dfrac{i\hbar t}{2m}\right)k^2+ixk.

    Rearranging, we get

    -\dfrac{m+2ai\hbar t}{4am}\left[k^2-\dfrac{4amix}{m+2ai\hbar t}k\right] (Verify)
    Now that its in the proper form, we complete the square -- i.e. k^2+bk=k^2+bk+\left(\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2 = \left(k+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2:

    Quote Originally Posted by Chris L T521 View Post
    Complete the square to get

    -\dfrac{m+2ai\hbar t}{4am}\left[k-\dfrac{2amix}{m+2ai\hbar t}\right]^2-\dfrac{amx^2}{m+2ai\hbar t} (Verify)
    So when you go back to the integral in question, the substitution becomes u=k-\dfrac{2amix}{m+2ai\hbar t}\implies\,du=\,dk

    Then the rest of the problem becomes algebra once you apply the formula for the Gaussian.

    Does this clarify things?
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  8. #8
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    As I said in my previous response, you need to have the coefficient of the k^2 term to be 1 in order to complete the square. Thats why I did the following:



    Now that its in the proper form, we complete the square -- i.e. k^2+bk=k^2+bk+\left(\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2 = \left(k+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2:



    So when you go back to the integral in question, the substitution becomes u=k-\dfrac{2amix}{m+2ai\hbar t}\implies\,du=\,dk

    Then the rest of the problem becomes algebra once you apply the formula for the Gaussian.

    Does this clarify things?
    aha, that is perfectly clear now. That is the first time I have seen the 'complete the square' used in an integral problem. Thanks alot Chris.
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