As part of a derivation for the question I asked here in Physics stackexchange, I am trying to calculate the following integral, but I am not sure how to proceed:
$\int_0^{2\pi}P_l^m(\cos\theta)P_{l-1}^m(\cos\theta) d\theta$,
where $P_l^m(\cos\theta)$ are the Associated Legendre Polynomials.
I believe that the answer is ultimately 0, but I am not quite sure of how reduce this integrable to a workable form, I have tried using recursion relations but have had no luck. I am sure that I am not trying to solve the integrand with an additional $\sin(\theta)$ factor, which would be easily solved using the change of variable $z = \cos(\theta)$ and applying the orthogonality of the Associated Legendre Polynomials. Does anyone have any advice?
Thanks.
Best Answer
We wish to evaluate: $$\int_0^{2\pi} P_l^m(\cos\theta)P_{l-1}^m(\cos\theta)\,\text{d}\theta $$ We use the transformation: $$x = \cos\theta\quad \frac{dx}{d\theta} = -\sin\theta \quad d\theta = \frac{-1}{\sqrt{1-x^2}}dx$$ to obtain$$\int_1^1 \frac{-P_l^m(x)P_{l-1}^m(x)}{\sqrt{1-x^2}}dx = 0$$
Since you're integrating from 1 to 1.
Weird? Yes I thought so too. Seeing as the integral comes from spherical harmoics, I assume you are messing up your limits of integration (should be from $0$ to $\pi$, since that is how the zenith angle is normally defined).
Then, I think you should be working with:
$$\int_0^{\pi} P_l^m(\cos\theta)P_{l-1}^m(\cos\theta)\,\text{d}\theta $$
which after the transformation becomes:
$$I = \int_{-1}^1 \frac{P_l^m(x)P_{l-1}^m(x)}{\sqrt{1-x^2}}dx$$
The Legendre Polynomials satisfy the relationship:
$$P_l^m(-x) = (-1)^{l+m}P_l^m(x)$$
This means that if $l+m$ is even (odd), the Legendre Polynomial is also even (odd). Here we have the two numbers: $l+m$ and $l+m-1$ One of these two has to be even and the other odd. So one of the polynomials is even and the other is odd. Since $\frac{1}{\sqrt{1-x^2}}$ is even, the integrand as a whole is odd. Thus the integral is also zero.
$$I = 0$$
Reference for parity relationship.