integral with legendre generating function


by buffordboy23
Tags: function, generating, integral, legendre
buffordboy23
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#1
Nov19-08, 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.
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Pere Callahan
Pere Callahan is offline
#2
Nov19-08, 01:18 PM
P: 588
Quote Quote by buffordboy23 View Post
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]
.
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.
buffordboy23
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#3
Dec4-08, 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(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?

gabbagabbahey
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#4
Dec4-08, 09:37 PM
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P: 5,004

integral with legendre generating function


Quote Quote by buffordboy23 View Post
Use the Legendre generating function to show that for A > 1

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]

.... I know that with these parameters the series should be convergent for |A| < 1
For A>1, this is not the generating function.

Hint: if A>1, then 1/A<1
buffordboy23
buffordboy23 is offline
#5
Dec4-08, 10:01 PM
P: 540
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.


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