Show the Integration of Legendre Polynomials is 0

Question

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?

Answer 1

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.

Answer 2

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.