Orthogonality of associated Legendre polynomials Let $P_n(x)$ be the $n$-th degree Legendre polynomial. Let $k$ be a nonnegative integer less than or equal to both $n,m$. How to prove that
$$ \int_{-1}^1 (1-x^2)^k D^kP_n(x) D^kP_m(x)\,dx = \frac{2(n+k)!}{(n-k)!(2n+1)}\delta_{mn}, $$
where $D^k$ stands for derivative of order $k$, and $\delta_{mn}$ stands for Kronecker delta.
I think somehow we will use the orthogonality of $P_n, P_m$ and the norm of $P_n$, but I don't see how.
 A: Let $I(k,n,m)$ be the considered integral.
(1) Since $\deg P_n=n$, we see immediately that $I(k,n,m)=0$ if $k>n$ or $k>m$.
(2) Since $I(k,n,m)=I(k,m,n)$, we see that the remaining case to be considered is
$0\leq m\leq n$.
(3) Assume that $0\leq m\leq n$. Noting that $-1$ and $1$ are multiple zeros of $(x^2-1)^kD^kP_m$ of multiplicity $k$, we obtain by using k-successive integrations by parts
$$
I(k,n,m)=\int_{-1}^1P_n(x)D^k\left((x^2-1)^kD^{k}P_m(x)\right)dx
$$


*

*The polynomial $Q(x)=D^k\left((x^2-1)^kD^{k}P_m(x)\right)$ has degree $m$. So, if $m<n$ we obtain $I(k,n,m)=0$.

*The remaining case is when $m=n$. Here $\deg Q=n$. Recalling that the leading
coefficient of $P_n$ is $a_n=\frac{1}{2^n}\binom{2n}{n}$ we see that $P_n(x)=a_n(x^n+R_{n-1})$ with $\deg R_{n-1}<n$. Thus
$$
\eqalign{
Q(x)&=D^k\left((x^2-1)^kD^{k}P_n(x)\right)\cr
&=a_nD^k\left((x^2-1)^k\frac{n!}{(n-k)!}(x^{n-k}+R_{n-1}^{(k)})\right)\cr
&=a_n\frac{n!}{(n-k)!}D^k\left( x^{n+k}+S(x)\right)\quad\hbox{with $\deg S<n+k$}\cr
&=a_n\frac{n!}{(n-k)!} \frac{(n+k)}{n!}(x^n+T(x))\quad\hbox{with $\deg T<n$}\cr
&=a_n \frac{ (n+k)!}{(n-k)!} x^n+\hbox{terms of degree $<n$}\cr
&=\frac{ (n+k)!}{(n-k)!} P_n+\hbox{terms of degree $<n$}
}
$$
It follows using orthogonality
$$
I(k,n,n)=\frac{ (n+k)!}{(n-k)!} \int_{-1}^1 P_n^2(x)dx=
\frac{ (n+k)!}{(n-k)!}\frac{2}{2n+1}
$$
since $\Vert P_n\Vert^2=\frac{2}{2n+1}$.


As a conclusion, the only cases where $I(k,n,m)$ is different from  $0$ is when
$n=m$ and $0\leq k\leq n$, and in this case 
$$
I(k,n,n)=\frac{ (n+k)!}{(n-k)!}\frac{2}{2n+1}.
$$
