0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ What do you mean by $P_k^m(x)$? The $m$-th derivative of the Legendre polynomial $P_k(x)$? $\endgroup$ – Jack D'Aurizio Jan 23 '15 at 13:59
  • $\begingroup$ Sorry, $P_k^m(x)$ are the associated Legendre Polynomials. $\endgroup$ – DaPhil Jan 23 '15 at 14:23
  • $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Jan 23 '15 at 14:45
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

The previous answer about using Gaunt's theorem won't hold unless some condition regarding the degrees are satisfied.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.