# 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.

• One way is to introduce $f_{n,k}(x) = (1-x^2)^{k/2} D^kP_n(x)$, and to prove by induction on $k$ that $f_{n,k}$ satisfy a certain 2nd order ODE. This makes them eigenfunctions of a differential operator for different eigenvalues, hence orthogonal.
– user147263
Jun 13 '14 at 5:39

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 $$I(k,n,n)=\frac{ (n+k)!}{(n-k)!}\frac{2}{2n+1}.$$