Integral of Legendre polynomials 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. 
 A: 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.
