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Legendre Polynomials question

  1. Sep 17, 2012 #1
    I am doing a Laplace's equation in spherical coordinates and have come to a part of the problem that has the integral...

    ∫ P(sub L)*(x) * P(sub L')*(x) dx (-1<x<1)

    The answer to this integral is given by a Kronecker delta function (δ)...

    = 0 if L ≠ L'

    OR....
    = 2/(2L+1)*δ if L = L' (where δ = 1)

    I believe the reason why the integral is equal to zero when L ≠ L' is because of the orthogonality of Legendre polynomials, however i cannot figure out how the integral is equal 2/(2L+1). If L = L' then the integral would equal...

    ∫ [P(sub L)*(x)]^2 dx

    I tried to use a simple power rule of integration to solve the above integral, but i am afraid my method is flawed. Any suggestions?

    Pre-emptive thanks to whomever takes the time to help!!
     
  2. jcsd
  3. Sep 17, 2012 #2
    For clarity i should add that...

    P(sub L)(x)

    is a Legendre polynomial
     
  4. Sep 17, 2012 #3
    I think i have to use Rodrigues' formula
     
  5. Sep 18, 2012 #4

    vanhees71

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    2016 Award

    This is one possibility. Another is to use the generating function, which is used to define in the msot convenient way the Legendre polynomials.

    It's given by
    [tex]\Phi(r,u)=\frac{1}{\sqrt{1-2u r+r^2}}=\sum_{l=0}^{\infty} P_l(u) r^l.[/tex]
    From this you find
    [tex]P_l(u)=\frac{1}{l!} \left .\frac{\partial^l}{\partial r^l} \Phi(r,u) \right|_{r=0}.[/tex]
    This series expansion is obviously valid for [itex]|r|<1[/itex] if [itex]|u|<1[/itex].

    We know that the Legendre Polynomials are orthogonal to each other since they are solutions of the eigen-value equation
    [tex]\frac{\mathrm{d}}{\mathrm{d} u} \left [(1-u^2) \frac{\mathrm{d} P_l}{\mathrm{d} u} \right ]=-l(l+1) P_l,[/tex]
    where the differential operator is self-adjoint on [itex]L^2([-1,1])[/itex]. Thus we have
    [tex]\int_{-1}^{1} \mathrm{d} u P_l(u) P_{l'}(u)=N_l \delta_{ll'}.[/tex]
    To evaluate the normalization factor we take the following integral:
    [tex]I(r)=\int_{-1}^{1} \mathrm{d} u \Phi^2(r,u)=\int_{-1}^{1} \mathrm{d} u \frac{1}{1-2 u r +r^2}=-\left [\frac{1}{2r} \ln(1-2 u r+r^2) \right]_{u=-1}^{u=1}=\frac{1}{r} \ln \left (\frac{1+r}{1-r} \right ).[/tex]
    Expanding this into a power series in [itex]r[/itex] yields
    [tex]I(r)=\sum_{l=0}^{\infty} \frac{2}{2l+1} r^{2l}.[/tex]
    On the other hand from the definition of the Legendre Polynomials by the above given series expansion you get
    [tex]\Phi^2(r,u)=\sum_{l,l'=0}^{\infty} P_l(u) P_{l'}(u) r^{l+l'}.[/tex]
    Integrating over [itex]u[/itex] leads to
    [tex]I(r)=\sum_{l,l'=0}^{\infty} N_l \delta_{ll'} r^{2l}=\sum_{l=0}^{\infty} N_l r^{2l}.[/tex]
    Comparing the coefficients in both expansions of [itex]I(r)[/itex] yields
    [tex]N_l=\frac{2}{2l+1}.[/tex]
    QED.
     
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