Is there any way of analytically simplifying the integral \begin{equation} \int_{-1}^1 (1-x^2)^{n+k+7/2} P_{2n+1}^1(x) P_{2k+1}^1(x) \, dx, \end{equation} where $P_l^m(x)$ is the associated Legendre polynomial? Its occuring in my calculations at different places and it resembles the orthogonality conditions disturbingly much, although its value is not zero.


1 Answer 1


The first few Legendre polynomials are

$P_{0}(x) = 1$

$P_{1}(x) = x$

$P_{2}(x) = 1/2(3x^2-1)$

$P_{3}(x) = 1/2(5x^3-3x)$

I hope I've helped

$P_{4}(x) = 1/8(35x^4-30x^2+3)$

$P_{5}(x) = 1/8(63x^5-70x^3+15x)$

Let $g(x)$ be a polynomial of degree $2n+1$ and $P_{0}(x)$,$P_{1}(x)$,...,$P_{n+1}(x)$, in relation to the inner product $$<f,g>=\int_{a}^{b} f(x)g(x) \,dx$$ .

It is known $P_{n+1}(x)$ has $n+1$ distinct zeros $x_{0},x_{1},...,x_{n}$ over interval.

Let $r(x)$ be the rest of the division of $g(x)$ by $P_{n+1}(x)$.

$$g(x) = q(x)P_{n+1}(x) + r(x)$$

where $q(x)$ and $r(x)$ are polynomials in $x$ and degree $deg$ $r(x) ≤ n$. Prove that:

a ) $r(x)$ is the interpolating polynomial of $g(x)$ with respect to the points $x_{0}, x_{1}, x_{2}, . . . , x_{n};$

b ) $\int_{a}^{b} g(x)\,dx$=$\int_{a}^{b} r(x)\,dx$

I made the division algorithm for the polynomial and applied the integral with the orthogonal Legendre polynomial. The degree of the polynomial is up to 2n+1. I would like an orientation to conclude.


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