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Please help in integration of Associate Legendre function

  1. Jun 20, 2010 #1
    I don't understand why I solve the integration in two different ways and get two different answers!!

    To find:

    [tex]\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta [/tex]

    1) Solve in [itex] \theta[/itex]

    [tex] P_1(cos \theta) = cos \theta \;\Rightarrow \; P_1^1(cos \theta)= -sin \theta [/tex]

    [tex]\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int_0^{\pi}sin^2 \theta d \theta = -\frac{\pi}{2}[/tex]





    2) Let [tex]s=cos \theta \;\Rightarrow \; d\theta = \frac{ds}{-sin \theta} [/tex]


    [tex]\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int _1^{-1} P_1^1(s)ds[/tex]

    [tex]P_1(s)=s \;\Rightarrow P_1^1(s)=1[/tex]

    [tex]\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int _1^{-1} P_1^1(s)ds = -s|_1^{-1} = 2[/tex]

    You see the two method yield two answers!!! I know it should yield the same answer, the book show how to solve the problems in 2) form. Please tell me what did I do wrong.

    thanks

    Alan
     
  2. jcsd
  3. Jun 20, 2010 #2

    D H

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    Staff Emeritus
    Science Advisor

    Your error is here:
    You have the correct expression for the Legendre polynomial P1(s) but not for the associated Legendre function. The correct expression for the associated Legendre functions is given by

    [tex]P_l^m (x) = (-1)^m(1-x^2)^{m/2}\frac{d^m}{dx^m}P_l(x)[/tex]
     
  4. Jun 20, 2010 #3
    Thanks

    It must be too late last night. I somehow stuck with the idea that [itex] P_1^1(x) \hbox { is just derivative of } P_1(x)[/itex]! I should have gone to sleep instead of posting this!!!

    Thanks

    Alan
     
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