Normalization coefficient for Spherical Harmonics with m=l

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The discussion revolves around solving the integral of (Sin(x))^(2l+1) to find the normalization coefficient for spherical harmonics. The original poster struggles to connect their derived solution with the one provided in the solution manual. A user suggests using a recurrence formula for the sine integral, which simplifies the problem significantly. They explain that by applying this formula, the integral can be expressed in terms of previous integrals, leading to a straightforward calculation of coefficients. Ultimately, the discussion highlights the importance of recurrence relations in simplifying complex integrals in mathematical physics.
david13579
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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 don't 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]
 
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Hi,
They use a recurrence formula for sine integral:
\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
And since
sin\left ( 0 \right )=sin\left ( \pi \right )=0
You are left only with:
\int sin^{2l+1} \left ( \theta \right )d\theta =\frac{2l}{2l+1}\int sin^{2l-1} \left ( \theta \right )d\theta
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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