# Integral of Legendre polynomials

Is there any way of analytically simplifying the integral $$\int_{-1}^1 (1-x^2)^{n+k+7/2} P_{2n+1}^1(x) P_{2k+1}^1(x) \, dx,$$ 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.

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

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.