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Homework Help: How does integrating this work?

  1. Apr 14, 2010 #1

    Consider the gaussian distribution:

    [tex] p(x) = Ae^(-\lambda (x-a)^2) [/tex]

    (a) use the equation, [tex] 1={\int_{-\infty}^{\infty}} p(x)dx [/tex]

    (b) find <x>, <x^2> and [tex]\sigma[/tex]


    a) if i take (x-a) to be u,

    [tex] 1=\int_{-\infty}^\infty Ae^(-\lambda(x-a)^2)dx[/tex]
    [tex] \int_{-\infty}^\infty Ae^(-\lambda(u)^2)dx [/tex] (not sure if this is right, when i substitute in u, is dx, supposed to be replaced by du?

    why? because of the integration limits? idk understand why please explain,

    - after i get told why that occurs,

    i need help integrating this
    [tex]{\int_{-\infty}^{\infty}} Ae^(-\lambda(u)^2)du [/tex]
  2. jcsd
  3. Apr 14, 2010 #2

    Char. Limit

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    Gold Member

    Assuming that a is a constant...

    Well, start with your replacement...


    then differentiate it with respect to x.


    And multiply both sides by dx (you can do this).


    So du=dx!
  4. Apr 15, 2010 #3
    my first question is how to solve for A and a

    I've been given that i need to figure out A first, before a

  5. Apr 15, 2010 #4

    Char. Limit

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    Gold Member

    Are you given a value for lambda? That would really help if you were trying to solve the integral...

    However, first thing I'd do, assuming A is constant, is factor it out of the integral. After all, the resulting integral is solvable, and would be even more solvable if we had a value lambda (and I hope it's a square number too).
  6. Apr 15, 2010 #5


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    Science Advisor

    By the way, with LaTex, put everything you want grouped together, such as an exponent, in { }.

    [ tex ]e^{-\lambda (x- a)^2} [ tex ]
    [tex]e^{-\lamda (x- a)^2}[/tex]
  7. Apr 15, 2010 #6
    Don't worry about integrating that one.

    Instead, try integrating
    [tex]{\int_{-\infty}^{\infty}} (u+a)Ae^{-\lambda(u)^2}du [/tex]
    [tex]{\int_{-\infty}^{\infty}} (u+a)^{2}Ae^{-\lambda(u)^2}du [/tex]
    If you integrate by parts (and you may need to do this more than once, to get rid of the "u+a" coefficients), you'll get (some expression)*(the integral you can't do, which is equal to 1)
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