Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am finding difficulty doing this integral involving Legendre polynomials. $$\int_{-1}^1 x^2 P_{n-1}(x)P_{n+1}(x)dx = \frac{2n(n+1)}{(2n-1)(2n+1)(2n+3)}$$ I have two strategies in my mind both of them have failed to produce results. One is that I could somehow use the orthogonality of Legendre polynomials after using Bonnet's recursion formula to get Legendre polynomials, to simplify the integrals, or I could use the Rodrigues formula by doing integration by parts. The first approach fails because of the $x^2$ in the integral. The second approach is not giving me integrated parts, where limits can be applied easily.How do I do this?

share|cite|improve this question
I guess the iterative formulas for Legendre polynomials would help in this case. – Patrick Li Oct 2 '12 at 11:20
up vote 1 down vote accepted

By Bonnet's formula, we have \[ (2n-1)xP_{n-1} = nP_n + (n-1)P_{n-2} \] and \[ (2n+3)xP_{n+1} = (n+2)P_{n+2} + (n+1)P_n \] so \begin{align*} \int_{-1}^1 x^2 P_{n-1}P_{n+1}\; dx &= \frac 1{(2n-1)(2n+3)}\int_{-1}^1 \bigl( nP_n + (n-1)P_{n-2}\bigr)\bigl((n+2)P_{n+2} + (n+1)P_n\bigr)\; dx\\ &= \frac 1{(2n-1)(2n+3)}\int_{-1}^1 n(n+1)P_n^2\; dx\\ &= \frac {2n(n+1)}{(2n-1)(2n+1)(2n+3)} \end{align*}

I have another denominator, I saw. But I will show also by example mine is right ;-): We have ($n=1$) that $P_0 = 1$, $P_2(x) = \frac 12(3x^2 - 1)$. It holds \begin{align*} \frac 12 \int_{-1}^1 (3x^4 - x^2) \; dx &= \frac 22\left(\frac 35 - \frac 13\right)\\ &= \frac 4{15}\\ &= \frac{2\cdot 1 \cdot (1+1)}{(2\cdot 1 - 1)(2 \cdot 1 \mathbin{{\color{red}+}} 1)(2\cdot 1 + 3)} \end{align*}

share|cite|improve this answer
Thanks for the answer. You are right, there was a typo in the answer in my question. SO the bonnets formula allows u to write the integrand as a product of legendre polynomial without the $x^2$, and hence can be integrated easily. Wonder why I didn't think of this. – user23238 Oct 2 '12 at 11:59

Recall the more general form of the orthogonality condition:

$$\int_{-1}^1 x^j P_k(x)\,\mathrm dx=0\qquad\text{if }j\neq k$$

Use this to simplify your integral to

$$2^{-n+1}\binom{2n-2}{n-1}\int_{-1}^1 x^{n+1} P_{n+1}(x)\,\mathrm dx$$

Now, there is the identity

$$\int_0^1 x^{n+2\rho}P_n(x)\,\mathrm dx=\frac{\left(2\rho+1\right)_n}{2^{n+1}\left(\rho+\tfrac12\right)_{n+1}}$$

Using this gives the result

$$\begin{align*} \int_{-1}^1 x^2 P_{n-1}(x)P_{n+1}(x)\mathrm dx&=2^{-n+1}\binom{2n-2}{n-1}\frac{(n+1)!}{2^{n+1}\left(\tfrac12\right)_{n+2}}\\ &=\frac{2n(n+1)}{(2n-1)(2n+1)(2n+3)} \end{align*}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.