Show the Integration of Legendre Polynomials is 0 Show that for the Legendre polynomial $P_n$ with $n\neq0$,
$$\int_{-1}^{+1} P_n (x) dx =0$$
I put this polynomial in the Legendre equation then got stuck. Can you help me find out what to do next?
 A: By definition, the Legendre polynomial $P_n(x)$ is a polynomial solution of the Legendre ODE
$$\frac{d}{dx}\left[(1-x^2)\frac{d}{dx}P_n(x)\right] + n(n+1)P_n(x) = 0$$
Integrating both sides from $-1$ to $1$ and apply Fundamental theorem of calculus,
we have
$$\left[(1-x^2)\frac{d}{dx}P_n(x)\right]_{x=-1}^{1} + n(n+1)\int_{-1}^1 P_n(x) = 0$$
Because of the $1-x^2$ factor, the first term on LHS vanishes. This leaves us
with
$$n(n+1) \int_{-1}^1 P_n(x) = 0$$
When $n \ne 0$ , the coefficient $n(n+1)$ above $\ne 0$ and we are done.
A: You know that $$\int_{-1}^{1}P_{n}(x)P_{m}(x)dx=0$$
This is the orthogonality property. Now $P_{0}(x)=1$, and so you have the  following:
$$\int_{-1}^{1}P_{n}(x)dx=\int_{-1}^{1}1\times P_{n}(x)dx=\int_{-1}^{1}P_{0}(x)P_{n}(x)dx=0$$
This is true for $n\neq 0$ 
A: I would do this in two parts: first prove this for $n$ being odd then do the even case.
So assume $n$ is odd. From Rodridge's Formula we have 
$$P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} \left[(x^2-1)^n\right] $$
so if $n$ is odd $P_n(x)$ is a polynomial with all odd powers (justify this more). Thus $P_n(x)$ is an odd function and so by symmetry the symmetric integral about zero is zero. Therefore we have proven the result for all odd $n$.
More generally, this same argument can show that the Legendre polynomials have the following symmetry relation:
$$ P_n(-x) = (-1)^n P_n(x) $$
so the polynomials are odd if and only if $n$ is odd. Thus the previous argument won't work for $n$ being even. What we can do is use the following formula:
$$ (2n+1)P_n(x) = \frac{d}{dx}\left[ P_{n+1}(x) - P_{n-1}(x)\right] $$
(I just found that formula on the Wikipedia page, it might be in some book somewhere or have a name...) Thus take $n$ as even and integrate both sides. Notice that the right hand side gives
$$ \int_{-1}^{1} \frac{d}{dx}\left[ P_{n+1}(x) - P_{n-1}(x)\right] = \left[ P_{n+1}(x) - P_{n-1}(x)\right] _{-1}^{1}$$
Since $n+1$ and $n-1$ are odd, $P_{n+1}(1) = P_{n-1}(1) = 1$ and thus by the symmetry property $P_{n+1}(-1) = P_{n-1}(-1) = -1$. So 
$$ \left[ P_{n+1}(x) - P_{n-1}(x)\right] _{-1}^{1} = 0. $$
Notice this then means that the integral of $P_n$ is zero when $n$ is even.
Note that if this is for homework, you probably can't use all of these identities. But this gives a guideline for one way to get to the conclusion with details to fill in.
