# Integral containing Associated Legendre Polynomials

I need to evaluated the following integral:

$\int_0^\pi \sin(x) \cos(x) P_l^m(\cos x) P_k^m(\cos x) \mathrm{d}x$

and I thought since a solution is known to a similar thing

$\int_0^\pi \sin(x) P_l^m(\cos x) P_k^m(\cos x) \mathrm{d}x = \frac{2(l+m)!}{(2l+1)(l-m)!}\delta_{l,k}$

maybe this is the case with an additional $\cos x$ as well.

• What do you mean by $P_k^m(x)$? The $m$-th derivative of the Legendre polynomial $P_k(x)$? – Jack D'Aurizio Jan 23 '15 at 13:59
• Sorry, $P_k^m(x)$ are the associated Legendre Polynomials. – DaPhil Jan 23 '15 at 14:23
• Ok. By the way, I think it would be nicer to include the definition $$P_\ell^{m}(x) = (-1)^m\ (1-x^2)^{m/2}\ \frac{d^m}{dx^m}\left(P_\ell(x)\right)$$ in your question to avoid counter-questions like mine. – Jack D'Aurizio Jan 23 '15 at 14:45

Since $$P_l^{m}(x) = (-1)^m\ (1-x^2)^{m/2}\ \frac{d^m}{dx^m}\left(P_l(x)\right)\tag{1}$$ the second integral equals: $$\int_{0}^{\pi/2}\cos(x)P_{l}^{m}(\sin x)P_{k}^{m}(\sin x)\,dx + \int_{0}^{\pi/2}\cos(x)P_{l}^{m}(-\sin x)P_{k}^{m}(-\sin x)\,dx$$ or: $$\int_{0}^{1}P_{l}^{m}(x) P_{k}^{m}(x)\,dx + \int_{0}^{1}P_{l}^{m}(-x) P_{k}^{m}(-x)\,dx=\int_{-1}^{1}P_l^m(x)P_k^m(x)\,dx$$ and $$\int_{-1}^{1}P_l^m(x) P_k^m(x)\,dx = \frac{2(l+m)!}{(2l+1)(l-m)!}\delta_{k,l}\tag{2}$$ follows from the orthogonality relation for the associated Legendre polynomials.
The first integral equals: $$\int_{-1}^{1}|x|\, P_{l}^{m}(x)\, P_{k}^{m}(x)\,dx\tag{3}$$ and it can be computed through Gaunt's formula once we expand $|x|$ in terms of Legendre polynomials.