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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.

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  • $\begingroup$ for what reason you DON'T wanna use $cos(\theta)=z$? $\endgroup$
    – tired
    May 4, 2015 at 8:08
  • $\begingroup$ Since I don't have a $\sin\theta$ factor in the integrand, if I do the $z=\cos\theta$ substitution I get $\frac{P_l^m(z)P_{l-1}^m(z)}{\sqrt{1-z^2}}$ $\endgroup$
    – Loonuh
    May 4, 2015 at 8:32

1 Answer 1

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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.

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  • $\begingroup$ Bravo. Bravo. Bravo. $\endgroup$
    – Loonuh
    Jul 14, 2015 at 5:09
  • $\begingroup$ The first derivation is wrong since $x=\cos \theta$ fails to be one-to-one on the domain of integration, note that the conclusion would imply $\int_0^{2\pi}f(\cos t)dt=0$ for any $f$. $\endgroup$ Jan 8, 2020 at 10:33

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