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Thread: Gaussian Integral

  1. #1
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    Gaussian Integral

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

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

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

    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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    Quote Originally Posted by synclastica_86 View Post
    I need to integrate:
    $\displaystyle 1=\int_{-\infty}^{\infty}Ae^{-\lambda(x-a)^2}dx$
    to find A. I tried 2 different methods but yielded different answers.

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

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

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

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

    Fixing this will reconcile the two answers you got.
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