One can find a few formulas for integrating combinations of spherical harmonics $Y_l^m(\theta,\phi)$ over the whole sphere. But I want to calculate the upper half sphere average of $\hat n \cdot \hat z = \cos(\theta)$ are there any nice result that yields the value of,
$$ \int_0^{2 \pi}\int_0^{\pi/2} Y_l^m(\theta,\phi)Y_l^{m}(\theta,\phi)^*\cos(\theta)\sin(\theta)\,d\theta d\phi $$
Note we deduce that the above is the same as to calculate the value of, $$ \int_0^1xP_l^m(x)P_l^{m}(x)\,dx, $$ with $P_l^m(x)$ the associated Legendre polynomials.
I did some search and found the following identities for the associate Legendre polynomials, $P_l^m$, $$ (l-m+1)P^m_{l+1} = (2l+1)x P_l^m - (l+m)P_{l-1}^m.\tag{1} $$ Also the derivative has some properties, $$ (x^2-1)(P_l^m)' = -(l+1)x P_l^m + (l-m+1)P_{l+1}^m \tag{2} $$ and $$ (x^2-1)(P_l^m)' = l x P_l^m - (l + m)P_{l-1}^m. \tag{3} $$ Let $$ X = \int_a^b x (P_l^m(x))^2\,dx, $$ $$ Y = \int_a^b P_{l-1}^m(x)P_l^m/(x)\,dx, $$ and $$ Z = \int_a^b P_{l+1}^m(x)P_l^m/(x)\,dx. $$
Then integrating $(1)$ yields, $$ X = \frac{l+m}{2l+1}Y + \frac{l-m+1}{2l+1}Z \tag{4} $$
Also partial integration of $X$ lead to $$ X = \frac{1}{2}\Big[x^2P_l^mP_l^m \Big ]_a^b - \frac{1}{2}\int x^2((P_l^m)^2)'\, dx. \tag{5} $$
Using $$ \frac{1}{2}\int_a^b((P_l^m)^2)'\, dx = \Big[P_l^mP_l^m \Big ]_a^b $$ in and adding and removing this quantity diveded by two lead to $$ X = \frac{1}{2}\Big[(x^2 - 1)P_l^mP_l^m \Big ]_a^b - \int (x^2-1)(P_l^m)'P_l^m\,dx. \tag{6} $$ Let $$ A = \frac{1}{2}\Big[(x^2 - 1)P_l^mP_l^m \Big ]_a^b, $$ and use $(3)$ we get. $$ X = A - l X + (l+m)Y. $$ Or $$ Y = \frac{l+1}{l+m}X - \frac{1}{l+m}A. \tag{7} $$ Similarly using $(2)$ we get,
$$ X = A + (l+1)X - (l-m+1)Z. $$ Solving for $Z$,
$$ Z = \frac{1}{l-m+1} A + \frac{l}{l-m+1}X \tag{9} $$ Finally trying $(8)$,$(9)$ in $(4)$ yields $$ X = \frac{l+m}{2l+1}\Big(\frac{l+1}{l+m}X - \frac{1}{l+m}A\Big ) + \frac{l-m+1}{2l+m}\Big ( \frac{1}{l-m+1} A + \frac{l}{l-m+1}X\Big ) $$ We can collect terms and get $$ X = \Big(\frac{l+1}{2l+1}+\frac{l}{2l+1}\Big) X + \Big (\frac{1}{2l+1} - \frac{1}{2l+1})A $$ which gives no extra information unfortunately. But reading about Legendre polynomials lead to the following equation, $$ P_{l+1}^l = (2l+1)xP_l^l. $$ So for $m=l$ we find $$ Z = (2l+1) X $$ Use this in $(9)$ lead to $$ (2l+1) X = \frac{l}{l-m+1}X + \frac{1}{l-m+1}A = X + A $$ And we can solve for $X$,
$$ X = \frac{1}{l+1}A $$
And for the integral region representing the half sphere we conclude at least for $m=l$,
$$ X = \frac{P_l^l(0)^2}{2(l+1)}. $$