Half sphere weighted average of spherical harmonics 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.
 A: 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 $m=l$,
$$
X = \frac{P_l^l(0)^2}{2(l+1)}.
$$
