Orthogonality of Legendre Polynomials

Question

I have a question: In many textbooks, it is said that:

$$\int_{-1}^1 x^kP_n (x) dx=0$$

How can this be deduced?

Answer 1

Let us generalise the fact and prove that: $$\int_{-1}^{1} f(x)P_n(x)\, dx = \frac{(-1)^n}{2^n n!}\int_{-1}^{1} f^{(n)}(x) (x^2-1)^n\, dx$$

Proof:

$$\begin{align} \int_{-1}^{1} f(x)P_n (x)\, dx = \frac{1}{2^n n!} \int_{-1}^{1} f(x) \frac{d^n(x^2-1)^n}{dx^n}\, dx\\ = \frac{1}{2^n n!}\left[ \left\{f(x) \frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n\right \}^{1}_{-1} - \int_{-1}^{1} f’(x)\frac{d^{n-1}}{dx^{n-1}} (x^2-1)^n\, dx \right]\\ =-\frac{1}{2^n n!} \int_{-1}^{1} f’(x) \frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n\, dx \end{align}$$ since the first term is zero at $x=\pm 1$. Integrating further $(n-1)$ times should give you the result.

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For $f(x) = x^m$, $$f^{(n)}(x) = \begin{cases} 0 & m<n \\ n! & m=n \end{cases}$$ and it thus follows that $$\int_{-1}^{1} x^m f(x)\, dx = \begin{cases} 0 & m< n \\ \frac{2^{n+1}(n!)^2}{(2n+1)!} & m=n \end{cases}$$

Proof:

$$\int_{-1}^{1} x^nP_n(x)\, dx = \frac{(-1)^n}{2^n n!} \int_{-1}^{1} n!(x^2-1)^n \, dx = \frac{2}{2^n} \int_{0}^{1} (1-x^2)^n \, dx $$ $$= \frac{2}{2^n} \int_{0}^{\frac{\pi}{2}} \cos^{2n+1}\, d\theta \text{ using } x = \sin \theta $$ $$= \frac{2^{n+1}(n!)^2}{(2n+1)!}$$

Answer 2

That is not true if $k\geq n$, but if we assume $k<n$, from the orthogonality relation $$\int_{-1}^{1}P_n(x)\,P_m(x)\,dx =\frac{2\,\delta(m,n)}{2n+1} $$ and the fact that $\deg P_n(x)=n$ we have $\int_{-1}^{1}x^k P_n(x)=0$, since $x^k$ can be clearly written as a linear combination of $P_0(x),P_1(x),\ldots,P_k(x)$.

What if $k\geq n$? In such a case, we may consider the generating function for Legendre polynomials, $$ \frac{1}{\sqrt{1-2xt+t^2}}=\sum_{n\geq 0} P_n(x) t^n $$ multiply both sides by $x^k$ and perform $\int_{-1}^{1}(\ldots)\,dx$ to state $$ \int_{-1}^{1}x^k\,P_n(x)\,dx =[t^n]\int_{-1}^{1}\frac{x^k}{\sqrt{1-2xt+t^2}}\,dx $$ or simply invoke Rodrigues' formula and integration by parts: $$ \int_{-1}^{1}x^k\,P_n(x)\,dx =\frac{1}{2^n n!}\int_{-1}^{1}\left(\frac{d^n}{dx^n}x^k\right)(1-x^2)^n\,dx $$ then compute the RHS through Euler's Beta function (of course if $k+n$ is odd the outcome is simply zero by parity). In particular we have $$ \int_{-1}^{1}x^n P_n(x)\,dx = \frac{2^n}{\left(n+\tfrac{1}{2}\right)\binom{2n}{n}}.$$