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Math Help - Gaussian Integral

  1. #1
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    Gaussian Integral

    I need to integrate:
    1=\int_{-\infty}^{\infty}Ae^{-\lambda(x-a)^2}dx
    to find A. I tried 2 different methods but yielded different answers.

    Using \int_{-\infty}^{\infty}e^{-u^2}du=\sqrt{\pi}:
    let u=\lambda(x-a) and du=\lambda dx
    then
    1=A\lambda^{-1}\int_{-\infty}^{\infty}e^{-u^2}du=\frac{A}{\lambda}\sqrt{\pi}
    thus A=\frac{\lambda}{\sqrt{\pi}}

    Attempt 2, using \int_{-\infty}^{\infty}Ae^{\frac{-(x-a)^2}{2c^2}}dx=Ac\sqrt{2\pi}
    Here A=\frac{1}{c\sqrt{2\pi}}; since  \frac{1}{2c^2}=\lambda, we have  c=\frac{1}{\sqrt{2\lambda}}
    Therefore A=\frac{1}{\frac{1}{\sqrt{2\lambda}}\sqrt{2\pi}}=\  sqrt{\frac{\lambda}{\pi}}

    I thought both formula should be valid as listed from wiki. What am I doing wrong?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Swampscott, MA
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    Quote Originally Posted by synclastica_86 View Post
    I need to integrate:
    1=\int_{-\infty}^{\infty}Ae^{-\lambda(x-a)^2}dx
    to find A. I tried 2 different methods but yielded different answers.

    Using \int_{-\infty}^{\infty}e^{-u^2}du=\sqrt{\pi}:
    let u=\lambda(x-a) and du=\lambda dx Here's your problem.
    then
    1=A\lambda^{-1}\int_{-\infty}^{\infty}e^{-u^2}du=\frac{A}{\lambda}\sqrt{\pi}
    thus A=\frac{\lambda}{\sqrt{\pi}}

    Attempt 2, using \int_{-\infty}^{\infty}Ae^{\frac{-(x-a)^2}{2c^2}}dx=Ac\sqrt{2\pi}
    Here A=\frac{1}{c\sqrt{2\pi}}; since  \frac{1}{2c^2}=\lambda, we have  c=\frac{1}{\sqrt{2\lambda}}
    Therefore A=\frac{1}{\frac{1}{\sqrt{2\lambda}}\sqrt{2\pi}}=\  sqrt{\frac{\lambda}{\pi}}

    I thought both formula should be valid as listed from wiki. What am I doing wrong?
    You said: "let u=\lambda(x-a) and du=\lambda dx"

    In this case though, -u^2 = -\underbrace{\lambda^2}_{bad}(x-a)^2. You want u=\sqrt{\lambda}(x-a) and du=\sqrt{\lambda}\,dx.

    Fixing this will reconcile the two answers you got.
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