Integration involving spherical harmonics

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The integral ∫∫dΩ V(Ω)Yml(Ω) is evaluated for a piecewise function V(Ω) defined over two hemispheres. The discussion highlights the challenge of applying the orthonormality properties of spherical harmonics due to the integration being split into two parts. It suggests rewriting V(Ω) in terms of spherical harmonics to utilize orthogonality, but the participant struggles with combining the integrals effectively. The integration process involves first handling the φ variable, leading to an integral that results in zero due to the properties of the spherical harmonics. The importance of correctly identifying the exponential term in the spherical harmonics is emphasized, as it significantly affects the outcome.
lordkelvin
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Homework Statement



Evaluate the integral ∫∫dΩ V(Ω)Yml(Ω) for V(Ω) = +V0 for 0<θ<π/2 ; -V0 for π/2<θ<π


Homework Equations


I was hoping to apply the orthonormality properties of the spherical harmonics but this is a little more difficult since the integral breaks into two integrals over half spheres instead of one integral over a full sphere.

I guess it may be useful to rewrite the V(Ω) as ±√(4π)Y00(Ω)V0 in order to invoke the orthogonality properties of the spherical harmonics when integrated on a sphere.


The Attempt at a Solution



Basically I've split the integral into two integrals over the upper and lower halves of the sphere. Now what? Surely I'm not to use the definition of the spherical harmonics; there must be a cleaner way to do this by invoking the properties of Ylm(omega). There is nothing I can say about ∫∫half of spheredΩYlm(Y00)* as far as I know. Is there some way to combine the two integrals again so that I can use the orthonormality property?
 
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Orthonormality won't really help here because you're not integrating the product of two spherical harmonics. The integral should be pretty straightforward though. Start by integrating φ out first. Then you'll be left with an integral involving a Legendre polynomial.
 
I get zero. I think this is so since Ylm(θ,\phi) = const* e^(i*\phi)*Plm(cos(\theta) and so the integral over phi yields zero.
 
It's e^{im\phi}, not e^{i\phi}. It makes a difference.
 
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