I'm tasked to solve the following integral: $$\int_0^{\pi/2}P^m_n(cos(\theta))P^m_l(cos(\theta))sin(\theta)d\theta$$
where $n,k\in Z^+$ and $m \in Z^{0+}$. Is it possible and wise to perform the expansion: $$P^m_n(cos(\theta))P^m_l(cos(\theta)) = \sum a_{xy}P_x^y(2cos(\theta)-1)$$ to leverage orthogonality of the shifted Legendre basis over $[0,\pi/2]$ to simplify the integration? Or perhaps another set of functions orthogonal on the interval? If this is the right train of thought, I'm currently stuck on how to obtain the expansion coefficients, $a_{xy}$. If the proposed approach further complicates the problem, I would greatly appreciate some insight on how to solve the above integral.
Thank you for the help!
Edit (21-01-02): I believe I have solved the integral directly using the closed form of the associated Legendre polynomials, $$ P^m_n(cos(\theta)) = (-1)^m2^nsin(\theta)^{m/2}\sum_{k=m}^n \frac{k!}{(k-m)!}cos(\theta)^{k-m} {n \choose k} {\frac{n+k-1}{2} \choose n}.$$
The resulting integral is then, $$2^{n+l}\sum_{k=m}^n\sum_{j=m}^l \frac{k!}{(k-m)!}{n \choose k} {\frac{n+k-1}{2} \choose n} \frac{j!}{(j-m)!}{l \choose j} {\frac{l+j-1}{2} \choose l}\int_0^{\pi/2}sin(\theta)^m cos(\theta)^{k+j-2m}sin(\theta)d\theta$$ Making the substitutions $x = cos(\theta)$, $sin(\theta) = (1-x^2)$, and $dx = -sin(\theta)d\theta$ we have $$\int_0^{1}(1-x^2)^{m}x^{k+j-2m}dx = \frac{\Gamma(m+1)\Gamma(\frac{1}{2}(k+j+1-2m))}{2\Gamma({\frac{1}{2}(k+j+3)})}$$ where $m \geq 0$ and $k+j \geq 2m$, which should be satisfied by the initial problem statement. Unfortunately the argument inside two of the Gamma functions may not be an integer (e.g., $k=1$, $j=1$, and $m=0$) so there seems to be no simplified form beyond this point..