integral with legendre generating functionby buffordboy23 Tags: function, generating, integral, legendre 

#1
Nov1908, 12:58 PM

P: 540

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
Nov1908, 01:18 PM

P: 588

[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. 



#3
Dec408, 09:15 PM

P: 540

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(1Ax\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[1Ax\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? 



#4
Dec408, 09:37 PM

HW Helper
P: 5,004

integral with legendre generating functionHint: if A>1, then 1/A<1 



#5
Dec408, 10:01 PM

P: 540

Here we go.
[tex] \int^{1}_{1} \frac{\left(1Ax\right)dx}{\left(A^{2}2Ax+1\right)^{1/2}} = \int^{1}_{1} \frac{\left(1/Ax\right)dx}{\left(12x/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/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)/AP_{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(1x^{2}\right)dx = \frac{4}{3A} [/tex] Thanks. 


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