Integral of spherical harmonics over sphere 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?
 A: 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
$$
\int_{S^{2}} Y_{l}^{m} \mathrm{~d} S^{2} = \sqrt{4 \pi} \delta_{\ell 0} \delta_{m 0}
= \left\{\begin{array}{ll}
\sqrt{4 \pi} & \text { if } l=0 \text { and } m=0 \\
0 & \text { otherwise }
\end{array}\right.
$$
A: The representation
$$
Y^m_l(\theta,\phi) = \sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}} P_l^m(\cos(\theta)) \exp(\text{i}m\phi)
$$
is instructive here. Clearly $Y^0_0(\theta,\phi)= (4\pi)^{-1/2}$, so the integral over the sphere is $\sqrt{4\pi}$. In all other cases, one can separate the integral into polar component ($\phi$) and zenithal component ($\theta$) and integrate separately. Since
$$
\int_0^{2\pi} \exp(\text{i}m\phi)\,d\phi = 0,\quad
\int_{0}^{\pi} P_l^m(\cos(\theta))\,d\theta = 0,
$$
(the latter being a propery of the associated Legendre polynomials)
one has
$$
\int_{S^2} Y^m_l(\theta,\phi) =
\begin{cases}
\sqrt{4\pi} \quad&\text{if } l = 0 \text{ and } m = 0,\\
0 \quad&\text{otherwise.}
\end{cases}
$$
