I'm looking at Lebedev quadrature for the integration of functions of a sphere, where it says:
[…] integrate exactly all polynomials up to a given order. On the unit sphere, this is equivalent to integrating all spherical harmonics up to the same order.
I would like to check this, so I need the exact value of
$$
\int_{S^2} Y_l^m \,\text{d}S^2,
$$
i.e., the integral of the spherical harmonic over the 2-sphere.
Are those values known explicitly?
Best Answer
The accepted answer was not easy for me to see and it took me a while to do it in a more step-by-step manner:
The spherical harmonics are orthonormal by definition:
$$\int_{\theta=0}^{\pi} \int_{\varphi=0}^{2 \pi} Y_{\ell}^{m} Y_{\ell^{\prime}}^{m^{\prime} *} d \Omega=\delta_{\ell \ell^{\prime}} \delta_{m m^{\prime}}$$
where $d \Omega=\sin (\theta) d \varphi d \theta$ and $\delta$ is the Kronecker delta and is 1 if the indices are the same and 0 otherwise. We can now set $m^\prime = \ell^\prime =0$. If you insert this into the definition of the spherical harmonic, $Y_{l^\prime}^{m^\prime}(\theta, \phi)=\sqrt{\frac{2 l^\prime+1}{4 \pi} \frac{(l^\prime-m^\prime) !}{(l^\prime+m^\prime) !}} P_{l^\prime}^{m^\prime}(\cos (\theta)) \exp (\mathrm{i} m^\prime \phi)$ you can see that it yields $1/\sqrt{4 \pi}$. We substitute this back into the equation above to obtain
$$ \int_{\theta=0}^{\pi} \int_{\varphi=0}^{2 \pi} Y_{\ell}^{m} 1/\sqrt{4 \pi} d \Omega=\delta_{\ell 0} \delta_{m 0} $$
and multiply by $\sqrt{4 \pi}$ to see that the result to your desired integral is
$$ \int_{S^{2}} Y_{l}^{m} \mathrm{~d} S^{2} = \sqrt{4 \pi} \delta_{\ell 0} \delta_{m 0} = \left\{\begin{array}{ll} \sqrt{4 \pi} & \text { if } l=0 \text { and } m=0 \\ 0 & \text { otherwise } \end{array}\right. $$