I have to prove that the following integral
$$ I = \int_{-1}^{1}\sin(\theta)P_{l}^{m}(\cos \theta)P_{l'}^{m \pm 1}(\cos \theta)\,d(cos\theta)\tag{1}$$
Is non-zero for $l' = l \pm 1$ by making use of the relationship
$$ (2l+1)\sin(\theta)P_{l}^{m-1}(\cos\theta) = P_{l+1}^{m}(\cos\theta) + P_{l-1}^{m}(\cos\theta)\tag{2}$$
My method is to use the second equation to reduce the original equation to the form where the orthogonality relation can be used:
$$ \int_{-1}^{1}P_l^m(x) P_k^m(x)\,dx = \frac{2(l+m)!}{(2l+1)(l-m)!}\delta_{k,l}\tag{3}$$
But my problem is that equation 2 cannot be substituted directly into the integrand of equation 1 because of the different m values.
May someone provide an explanation or hint as to how I can go about solving this? Is there a certain formula for 'shifting' the m values or to make it compatible for substitution? Thank you.
Just plug in your equation (2) into the integral in (1): $$\int_{-1}^1\sin\theta\ P_l^m(\cos\theta)P_{l'}^{m-1}(\cos\theta)d(\cos\theta)=\int_{-1}^1P_l^m(\cos\theta)\frac{P_{l'+1}^m(\cos\theta)+P_{l'-1}^m(\cos\theta)}{2l+1}d(\cos\theta)$$ Now separate this into a sum of two integrals. Using (3), the first integral will be proportional to $\delta_{l,l'+1}$, the second will be proportional to $\delta_{l,l'-1}$. Therefore the integral is non-zero only for $l'=l\pm 1$.