# Legendre Polynomial Integral

## Homework Statement

As part of a larger problem, I need to compute the following integral (over $-1<\theta<1$):

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

## Homework Equations

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

Also, solutions are known to the following integrals:

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

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

## 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.

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## Answers and Replies

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

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.

gabbagabbahey
Homework Helper
Gold Member

## Homework Statement

As part of a larger problem, I need to compute the following integral (over $-1<\theta<1$):

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

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?

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?

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.

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gabbagabbahey
Homework Helper
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I need to compute (in spherical)

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

I assume this integral is over all of space?

I assume this integral is over all of space?

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

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

gabbagabbahey
Homework Helper
Gold Member
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.

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 $\vec{x}$ on the positive z-axis.

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!

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

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.

gabbagabbahey
Homework Helper
Gold Member
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.

Got it! Thanks.