Legendre polynomials and binomial series

[LANS]

Homework Statement

Where P_n(x) is the nth legendre polynomial, find f(n) such that
$$\int_{0}^{1} P_n(x)dx = f(n) {1/2 \choose k} + g(n)$$

Homework Equations

Legendre generating function:
$$(1 - 2xh - h^2)^{-1/2} = \sum_{n = 0}^{\infty} P_n(x)h^n$$

The Attempt at a Solution

I'm not sure if that g(n) term is necessary.

First I integrate both sides of the generating function on 0->1. I can then replace the (1+h^2)^1/2 term with a binomial series. I'm not sure how to cancel out the rest of the factors to solve for P_n. Any help would be appreciated, thanks.

 

[Ray Vickson]

Homework Statement

Where P_n(x) is the nth legendre polynomial, find f(n) such that
$$\int_{0}^{1} P_n(x)dx = f(n) {1/2 \choose k} + g(n)$$

Homework Equations

Legendre generating function:
$$(1 - 2xh - h^2)^{-1/2} = \sum_{n = 0}^{\infty} P_n(x)h^n$$

The Attempt at a Solution

I'm not sure if that g(n) term is necessary.

First I integrate both sides of the generating function on 0->1. I can then replace the (1+h^2)^1/2 term with a binomial series. I'm not sure how to cancel out the rest of the factors to solve for P_n. Any help would be appreciated, thanks.

The generating function is
$$\frac{1}{\sqrt{1-2xh + h^2}},$$
not what you wrote (see, eg., http://en.wikipedia.org/wiki/Legendre_polynomials ).

Anyway, why try to solve for P_n? When you integrate you just need the coefficient of h^n on both sides.

 

[LANS]

Sorry, I wrote the original question in a slightly confusing way.

Anyways, I solved it last night:
http://imgur.com/anpYO
http://imgur.com/lEjtc.jpg