# Legendre Polynomial Integral

1. Nov 8, 2008

### Higgy

1. The problem statement, all variables and given/known data
As part of a larger problem, I need to compute the following integral (over $-1<\theta<1$):

$$\int \sin \theta P_{l}(\cos \theta) d (\cos \theta)$$

2. Relevant equations

$$\int P_{l}(x) P_{l'}(x) dx= \frac{2}{2l+1} \delta_{l',l}$$

Also, solutions are known to the following integrals:

$$\int x P_{l}(x) P_{l'}(x) dx$$

$$\int x^{2} P_{l}(x) P_{l'}(x) dx$$

3. The attempt at a solution

Trig identities for converting $\sin \theta$ into $\cos \theta$ don't seem to help, here - which is why I'm confused.

I would presume to convert the integrand to a form that only has cosines. I would then use the fact that $P_{0}(x)=1$ with orthogonality (listed above) to get the explicit solution.

How do I get rid of that $\sin \theta$?!

EDIT:
Alright, I just found another formula online that was not in my text:

$$\int g(x) P_{l} (x) dx = \frac{(-1)^{l}}{2^{l} l!} \int (x^{2} - 1)^{l} \frac{d^{l}}{dx^{l}} g(x) dx$$

If I were to use this, I might first convert the sine using

$$\sin \theta = (1-\cos ^{2} \theta)^{\frac{1}{2}}$$

and then make that my $g(x)$.

I don't think this would really help me, though, since that would leave me with an awkward expression involving the l'th derivative inside an integral. So I'm still stuck.

Last edited: Nov 8, 2008
2. Nov 9, 2008

### weejee

Look for the orthogonality of 'associated Legendre functions' in any mathematical physics textbook.

3. Nov 9, 2008

### Higgy

Alright, the associated Legendre polynomials are given by:

$$P_{l}^{m}(x)=(-1)^{m}(1-x^{2})^{\frac{m}{2}}\frac{d^{m}}{dx^{m}}P_{l}(x)$$

With $x=\cos \theta '$, the integral I'm having trouble with is given by:

$$I=\int (1-x^{2})^{\frac{1}{2}}P_{l}(x)dx$$

The associated Legendre polynomials for m equal to 0 and 1 are:

$$P_{l}^{0}(x)=P_{l}(x)$$
$$P_{l}^{1}(x)=-(1-x^{2})^{\frac{1}{2}}\frac{d}{dx}P_{l}(x)$$

Since $P_{1}(x)=x$, we could write the integral as

$$I=-\int P_{1}^{1}(x) P_{l}^{0}(x) dx$$

but then we can't use orthogonality, given for associated Legendre polynomials by

$$\int P_{l}^{m}(x) P_{l'}^{m}(x) dx = \frac{2}{2l + 1} \frac{(l+m)!}{(l-m)!} \delta_{l',l}$$

I'm still working on it; I've got some ideas to play around with.

4. Nov 9, 2008

### gabbagabbahey

This integral seems a bit unusual to me....what was the larger problem that this came out of?...Are you sure you don't actually want to compute $$\int \sin \theta P_{l}(\cos \theta) d \theta$$ over $-\pi<\theta<\pi$ instead?

5. Nov 9, 2008

### Higgy

I need to compute (in spherical)

$$\int \frac{1}{|\vec{x} - \vec{x'}|} \sin \theta' \delta (\rho' - a) d^{3} x'$$

Originally, I thought to write out the difference between the two vectors in spherical, and do the integral (as you may have noticed by my question in the Calculus forum). After a bit of a struggle, I realized that I could just use an expansion in spherical harmonics.

The problem has azimuthal symmetry, so the spherical harmonics can be reduced to Legendre polynomials. I am then left with (where I will write out the volume element explicitly):

$$\int \int \int \sin \theta' \delta (\rho' - a) \frac{\rho_{<}^{l}}{\rho_{>}^{l+1}} P_{l}^{*} (\cos \theta') P_{l} (\cos \theta) \rho' ^{2} \sin \theta' d\rho' d\theta' d\phi'$$

And using $\sin \theta' d\theta' = -d(\cos\theta')$, I'm left trying to solve the angular integral mentioned above.

Last edited: Nov 9, 2008
6. Nov 9, 2008

### gabbagabbahey

I assume this integral is over all of space?

7. Nov 9, 2008

### Higgy

Yes, sorry. I've been lazy with the bounds of integration.

Also, I forgot the prime on the rho. It's been fixed.

8. Nov 9, 2008

### gabbagabbahey

Are you trying to find the potential at a point $\vec{x}$ do to a spherical shell of mass (or charge) of radius $a$?....If so, just orient your co-ordinate system so that $\vec{x}$ lies on the positive z-axis.

9. Nov 9, 2008

### Higgy

I'm trying to find the vector potential everywhere for a charged rotating spherical shell of radius a. It would be nice if I could just concern myself with the z-axis, though!

10. Nov 9, 2008

### gabbagabbahey

Okay, and the shell is rotating about the z-axis?....Is it uniformly charged?

11. Nov 9, 2008

### Higgy

Yes. Actually, I'm really looking at this now and thinking that the expression I came up for the charge density might be wrong. I think all of this discussion has suggested that the above integral is "too difficult/complicated to be correct" and that I might look at the charge density first. I'll take a little time and see if I can remedy the problem myself, and if not, I'll come back here. Thanks.

12. Nov 9, 2008

### gabbagabbahey

Yes, your expression for the current density looks very wrong to me....you need to first correct its magnitude, but also consider it's direction! (which varies over the surface).....If I were you, I'd divide the surface current up into a bunch of infinitesimally thin loops, find the vector potential due to one such loop and then integrate it over all the loops on the sphere's surface.

13. Nov 10, 2008

### Higgy

Got it! Thanks.