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Insane Integral

  1. Mar 29, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img4.imageshack.us/img4/4224/landau.png [Broken]
    http://g.imageshack.us/img4/landau.png/1/ [Broken]

    Note: do = 2*pi*sin(x) dx

    Well, as you can see this is an extremely painful integral.

    2. Relevant equations

    3. The attempt at a solution

    I have tried u = cos(1/2x) resulting in:
    du = -sin(1/2 x) (1/2) dx
    sin(x) = 2*sin(x/2)*cos(x/2)

    (n^2 u - nu^2 - n + u)/(n^4 + 2n^2 - 4n^3 u + 4n^2 u^2 - 4nu) (by expanding both the top and bottom, but as you can see it's messy and useless, I'm also dropping the constant because I don't care about them right now).

    I have no idea what kind of substitution to use for these beast... It evaluates to pi*a^2 with the integration limits 0 to X_max..

    Anyone have any idea how to do this by hand? I am tempted to use Matlab, but I really am supposed to do it manually.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 30, 2009 #2


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    Homework Helper

    The substitution you used is fruitful and if you don't work out the parentheses you should have gotten the following integral. (note that as x goes from 0 to xmax u goes from 1 to 1/n)

    -2 a^2 n^2 \pi \int_1^{1/n} \frac{(n u-1)(n-u)}{(n^2+1-2 n u)^2}\,du

    Now note that [tex] \frac{d}{du}(n u-1)(n-u) = n^2+1-2 n u [/tex] then use integration by parts. After some algebra you should get the value [tex] a^2 \pi[/tex].
    Last edited: Mar 30, 2009
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