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Legendre Polynomial Integral

  1. Nov 8, 2008 #1
    1. The problem statement, all variables and given/known data
    As part of a larger problem, I need to compute the following integral (over [itex]-1<\theta<1[/itex]):

    [tex]\int \sin \theta P_{l}(\cos \theta) d (\cos \theta)[/tex]

    2. Relevant equations

    [tex]\int P_{l}(x) P_{l'}(x) dx= \frac{2}{2l+1} \delta_{l',l}[/tex]

    Also, solutions are known to the following integrals:

    [tex]\int x P_{l}(x) P_{l'}(x) dx[/tex]

    [tex]\int x^{2} P_{l}(x) P_{l'}(x) dx[/tex]

    3. The attempt at a solution

    Trig identities for converting [itex]\sin \theta[/itex] into [itex]\cos \theta[/itex] don't seem to help, here - which is why I'm confused.

    I would presume to convert the integrand to a form that only has cosines. I would then use the fact that [itex]P_{0}(x)=1[/itex] with orthogonality (listed above) to get the explicit solution.

    How do I get rid of that [itex]\sin \theta[/itex]?!

    Alright, I just found another formula online that was not in my text:

    [tex]\int g(x) P_{l} (x) dx = \frac{(-1)^{l}}{2^{l} l!} \int (x^{2} - 1)^{l} \frac{d^{l}}{dx^{l}} g(x) dx[/tex]

    If I were to use this, I might first convert the sine using

    [tex]\sin \theta = (1-\cos ^{2} \theta)^{\frac{1}{2}}[/tex]

    and then make that my [itex]g(x)[/itex].

    I don't think this would really help me, though, since that would leave me with an awkward expression involving the l'th derivative inside an integral. So I'm still stuck.
    Last edited: Nov 8, 2008
  2. jcsd
  3. Nov 9, 2008 #2
    Look for the orthogonality of 'associated Legendre functions' in any mathematical physics textbook.
  4. Nov 9, 2008 #3
    Alright, the associated Legendre polynomials are given by:


    With [itex]x=\cos \theta '[/itex], the integral I'm having trouble with is given by:

    [tex]I=\int (1-x^{2})^{\frac{1}{2}}P_{l}(x)dx[/tex]

    The associated Legendre polynomials for m equal to 0 and 1 are:


    Since [itex]P_{1}(x)=x[/itex], we could write the integral as

    [tex]I=-\int P_{1}^{1}(x) P_{l}^{0}(x) dx[/tex]

    but then we can't use orthogonality, given for associated Legendre polynomials by

    [tex]\int P_{l}^{m}(x) P_{l'}^{m}(x) dx = \frac{2}{2l + 1} \frac{(l+m)!}{(l-m)!} \delta_{l',l}[/tex]

    I'm still working on it; I've got some ideas to play around with.
  5. Nov 9, 2008 #4


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    This integral seems a bit unusual to me....what was the larger problem that this came out of?...Are you sure you don't actually want to compute [tex]\int \sin \theta P_{l}(\cos \theta) d \theta[/tex] over [itex]-\pi<\theta<\pi[/itex] instead?
  6. Nov 9, 2008 #5
    I need to compute (in spherical)

    [tex]\int \frac{1}{|\vec{x} - \vec{x'}|} \sin \theta' \delta (\rho' - a) d^{3} x'[/tex]

    Originally, I thought to write out the difference between the two vectors in spherical, and do the integral (as you may have noticed by my question in the Calculus forum). After a bit of a struggle, I realized that I could just use an expansion in spherical harmonics.

    The problem has azimuthal symmetry, so the spherical harmonics can be reduced to Legendre polynomials. I am then left with (where I will write out the volume element explicitly):

    [tex]\int \int \int \sin \theta' \delta (\rho' - a) \frac{\rho_{<}^{l}}{\rho_{>}^{l+1}} P_{l}^{*} (\cos \theta') P_{l} (\cos \theta) \rho' ^{2} \sin \theta' d\rho' d\theta' d\phi'[/tex]

    And using [itex] \sin \theta' d\theta' = -d(\cos\theta')[/itex], I'm left trying to solve the angular integral mentioned above.
    Last edited: Nov 9, 2008
  7. Nov 9, 2008 #6


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    I assume this integral is over all of space?
  8. Nov 9, 2008 #7
    Yes, sorry. I've been lazy with the bounds of integration.

    Also, I forgot the prime on the rho. It's been fixed.
  9. Nov 9, 2008 #8


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    Are you trying to find the potential at a point [itex]\vec{x}[/itex] do to a spherical shell of mass (or charge) of radius [itex]a[/itex]?....If so, just orient your co-ordinate system so that [itex]\vec{x}[/itex] lies on the positive z-axis.
  10. Nov 9, 2008 #9
    I'm trying to find the vector potential everywhere for a charged rotating spherical shell of radius a. It would be nice if I could just concern myself with the z-axis, though!
  11. Nov 9, 2008 #10


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    Okay, and the shell is rotating about the z-axis?....Is it uniformly charged?
  12. Nov 9, 2008 #11
    Yes. Actually, I'm really looking at this now and thinking that the expression I came up for the charge density might be wrong. I think all of this discussion has suggested that the above integral is "too difficult/complicated to be correct" and that I might look at the charge density first. I'll take a little time and see if I can remedy the problem myself, and if not, I'll come back here. Thanks.
  13. Nov 9, 2008 #12


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    Yes, your expression for the current density looks very wrong to me....you need to first correct its magnitude, but also consider it's direction! (which varies over the surface).....If I were you, I'd divide the surface current up into a bunch of infinitesimally thin loops, find the vector potential due to one such loop and then integrate it over all the loops on the sphere's surface.
  14. Nov 10, 2008 #13
    Got it! Thanks.
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