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Normalization coefficient for Spherical Harmonics with m=l

  • #1
20
0

Homework Statement


Well it is not the problem itself that bothers me but the maths behind a part of it. As part of finding the coefficient I had to solve the integral of (Sin(x))^(2l+ 1). The solution given by the solution manual just pretty much jumps to the final answer http://i.imgur.com/hhoeLKE.png

Homework Equations


...

The Attempt at a Solution


Using substitution and the binomial theorem I was able to get a solution (of the integral part only, It would still need solving for the coefficient) and got this http://i.imgur.com/wnOUNIp.png

The problem is I dont see how to get from my answer to the one on the solution manual. I know they are equivalent because I checked numerically for different values of l.[/B]
 

Answers and Replies

  • #2
65
1
Hi,
They use a recurrence formula for sine integral:
[tex]\int sin^{2l+1} \left ( \theta \right )d\theta =-\frac{sin^{2l}\left ( \theta \right )cos\left ( \theta \right )}{2l+1}+\frac{2l}{2l+1}\int sin^{2l-1} \left ( \theta \right )d\theta[/tex]
And since
[tex]sin\left ( 0 \right )=sin\left ( \pi \right )=0[/tex]
You are left only with:
[tex]\int sin^{2l+1} \left ( \theta \right )d\theta =\frac{2l}{2l+1}\int sin^{2l-1} \left ( \theta \right )d\theta[/tex]
Finally they just sort of "calculate" this recurrently, and you are left only with the product of coefficients on the right side before the integral.
 

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