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Math Help - Normalizatoin Constant and a Gaussian Integral.

  1. #1
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    Normalizatoin Constant and a Gaussian Integral.

    Hello! I'm attempting to compute the normalization constant N for the following function f(x)=N e^{-(x-x_0)/2k^{2})}. This is generally done by evaluating \int |f(x)|^{2}dx=1, where the limits of integration are -inf to +inf.
    I can get this right up until the point where I need to make a u substitution.
    1=\int |N e^{-(x-x_0)/2k^{2})}|^{2}dx
    1=\int N^2 e^{-(x-x_0)^2/k^2} dx
    1=N^2 \int e^{-(x-x_0)^2/k^2} dx
    making the substitution u=x-x_0, du=dx
    1=N^2 \int e^{-u^2/k^2}du
    This integral is only slightly more clear to me. It looks like a Gaussian integral, but, how do I go about solving it? Any help is much appreciated.
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    Re: Normalizatoin Constant and a Gaussian Integral.

    I would actually make the subtitution

    u=\frac{x-x_{0}}{k}.

    That should make the integral a more standard integral.
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    Re: Normalizatoin Constant and a Gaussian Integral.

    So, that would yield: 1=N^{2} \int e^{-u^{2}}du=N^{2}\sqrt{\pi} ?
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    Re: Normalizatoin Constant and a Gaussian Integral.

    Quote Originally Posted by atomicpedals View Post
    So, that would yield: 1=N^{2} \int e^{-u^{2}}du=N^{2}\sqrt{\pi} ?
    I don't think you've quite got it yet. The substitution I suggested requires a change in the differential as well. The final answer had better have a 'k' in it somewhere.
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  5. #5
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    Re: Normalizatoin Constant and a Gaussian Integral.

    Woops! 1=N^{2}\frac{\sqrt{\pi}}{k} which then yeilds a normalization constant N=\pm\frac{\sqrt{k}}{\pi^{1/4}} .
    Last edited by atomicpedals; October 23rd 2011 at 10:49 AM.
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    Re: Normalizatoin Constant and a Gaussian Integral.

    Better, but still not quite there yet. Be careful as to numerator and denominator. Can you show your working, please?
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  7. #7
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    Re: Normalizatoin Constant and a Gaussian Integral.

    So my line of thinking was as follows:

    u=\frac{(x-x_{0})}{k} which I think yields du=\frac{1}{k}dx and so 1=N^{2}\int \frac{1}{k}e^{-u^{2}}du=1=N^{2}\frac{1}{k}\int e^{-u^{2}}du=N^{2}\frac{1}{k}\sqrt{\pi}
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  8. #8
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    Re: Normalizatoin Constant and a Gaussian Integral.

    Wait... I should have kdu=dx shouldn't I...
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  9. #9
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    Re: Normalizatoin Constant and a Gaussian Integral.

    Quote Originally Posted by atomicpedals View Post
    So my line of thinking was as follows:

    u=\frac{(x-x_{0})}{k} which I think yields du=\frac{1}{k}dx
    Fine so far.

    and so 1=N^{2}\int \frac{1}{k}e^{-u^{2}}du
    The problem here is that what you start with is a dx, not a du. Solve du = dx/k for dx. That's what you must plug in.

    =1=N^{2}\frac{1}{k}\int e^{-u^{2}}du=N^{2}\frac{1}{k}\sqrt{\pi}
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