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Integral with legendre generating function

  1. Nov 19, 2008 #1
    1. The problem statement, all variables and given/known data

    Use the Legendre generating function to show that for A > 1,

    [tex] \int^{\pi}_{0} \frac{\left(Acos\theta + 1\right)sin\thetad\theta}{\left(A^{2}+2Acos\theta+1\right)^{1/2}} = \frac{4}{3A} [/tex]

    2. Relevant equations

    The Legendre generating function

    [tex] \phi\left(-cos\theta,A\right) = \left(A^{2}+2Acos\theta+1\right)^{-1/2}} = \sum^{\infty}_{n=0}P_{n}\left(-cos\theta\right)A^{n}[/tex]

    3. The attempt at a solution

    Pretty much clueless, and the book makes no mention of anything else. I know that with these parameters the series should be convergent for |A| < 1, and that the interval now runs from [-pi, pi]. The only thing on the internet that I could find that deals with with integrals of Legendre functions is for orthogonality, which doesn't seem to apply here.
     
  2. jcsd
  3. Nov 19, 2008 #2
    Using the trick the integral becomes

    [tex]
    /int_0^\pi{\left(A\cos\theta+1\right)\sin\theta\sum_{n=0}^\infty{P_n(-\cos\theta)A^n}}
    [/tex]

    I suggest you are courageous and interchange the integration and summation (or maybe you can argue that this is allowed in our case?) then you expand [itex]A\cos\theta +1[/itex] in terms of Legendre functions. You should then be able to use some orthoganality property.
     
  4. Dec 4, 2008 #3
    I'm still stuck with this problem. Here is my work.

    Let [tex] x = -cos\theta \rightarrow d\theta = dx/sin\theta [/tex]. Substitution gives

    [tex] \int^{1}_{-1} \frac{\left(1-Ax\right)dx}{\left(A^{2}-2Ax+1\right)^{1/2}} = \int^{1}_{-1} \left(\sum^{\infty}_{k=0}P_{k}\left(x\right)A^{k}\right)\left[1-Ax\right]dx = \int^{1}_{-1} \left(\sum^{\infty}_{k=0}P_{k}\left(x\right)A^{k}\right)\left[P_{0}\left(x\right)-P_{1}\left(x\right)A\right]dx = \int^{1}_{-1} \left(P_{0}\left(x\right)^{2}-P_{1}\left(x\right)^{2}A^{2}\right)dx [/tex]

    where the last equation exploited orthogonality. Yet,

    [tex] \int^{1}_{-1} \left(P_{0}\left(x\right)^{2}-P_{1}\left(x\right)^{2}A^{2}\right)dx = 2 - \frac{2A^{2}}{3}[/tex]

    and is not the answer. What am I missing?

    I know that the legendre generating function requires |A|<1, but this problem says A>1. This implies that the series is divergent, but does that necessarily matter in this context?
     
  5. Dec 4, 2008 #4

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    For A>1, this is not the generating function.

    Hint: if A>1, then 1/A<1 :wink:
     
  6. Dec 4, 2008 #5
    Here we go.

    [tex]
    \int^{1}_{-1} \frac{\left(1-Ax\right)dx}{\left(A^{2}-2Ax+1\right)^{1/2}} = \int^{1}_{-1} \frac{\left(1/A-x\right)dx}{\left(1-2x/A+1/A^{2}\right)^{1/2}} =\int^{1}_{-1} \left(\sum^{\infty}_{k=0}P_{k}\left(x\right)A^{-k}\right)\left[1/A-x\right]dx = \int^{1}_{-1} \left(\sum^{\infty}_{k=0}P_{k}\left(x\right)A^{-k}\right)\left[P_{0}\left(x\right)/A-P_{1}\left(x\right)\right]dx [/tex]

    [tex] = \int^{1}_{-1} \left(\frac{P_{0}\left(x\right)^{2}}{A}-\frac{P_{1}\left(x\right)^{2}}{A}\right)dx =
    \frac{1}{A}\int^{1}_{-1} \left(1-x^{2}\right)dx =
    \frac{4}{3A}
    [/tex]

    Thanks.
     
    Last edited: Dec 4, 2008
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