Integral of a product of Legendre polynomials I would like to show that
$$
\int_{-1}^{1}P_{n}^{1}(x)P_{n'}^{0}(x)\frac{x}{\sqrt{1-x^{2}}}\,\mathrm dx
=
\begin{cases}
-\frac{2n}{2n+1},&n=n'>0\\
-2,&n>n'\text{ and } n-n' \text{ even}\\
0,&\text{otherwise.}
\end{cases}
$$
where $P_{n}^{m}(x)$ are the associated Legendre polynomials. One can easily see that this is the answer through inputting the following into Mathematica:
Table[Integrate[
   LegendreP[n, 1, x] LegendreP[n1, x] x/Sqrt[1 - x^2], {x, -1, 
    1}], {n, 0, 10}, {n1, 0, 10}] // MatrixForm

However, I'm at a loss at how to go about proving it. I can't find it in any of the usual sources (e.g. DLMF, Gradshteyn and Rhzhik etc.), perhaps because it is not quite an orthogonality relation. Please can you help me? Writing it as
$$
\int_{0}^{\pi}P_{n}^{1}(\cos(\theta))P_{n'}^{0}(\cos(\theta))\cos(\theta)d\theta
$$
might help, but I'm not sure. Thanks!
 A: By definition we have $2^nn!P_n^1(x)=-\sqrt{1-x^2}D^{n+1}(x^2-1)^n$ so $P_n^1(x)=-\sqrt{1-x^2}P_n'(x)$. Thus \begin{align}\int_{-1}^1\frac{xP_n^1(x)P_{n'}(x)}{\sqrt{1-x^2}}\,dx&=-\int_{-1}^1xP_{n'}(x)P_n'(x)\,dx\tag1\\&=-\left[xP_{n'}(x)P_n(x)\right]_{-1}^1+\int_{-1}^1(xP_{n'}(x))'P_n(x)\,dx\\&=\small-(1+(-1)^{n+n'})+\int_{-1}^1P_{n'}'(x)P_n(x)\,dx+\int_{-1}^1xP_{n'}'(x)P_n(x)\,dx\\&=-(1+(-1)^{n+n'})+\frac2{2n+1}\delta_{nn'}+\int_{-1}^1xP_{n'}'(x)P_n(x)\,dx.\tag2\end{align} Orthogonality means that $\int_{-1}^1P_n(x)f(x)\,dx=0$ whenever $n>\deg f$. In $(1)$, this means that $\deg P_{n'}\le\deg(xP_n')$ so that $n'\le n$. Hence, when $n'<n$, we must have $\delta_{nn'}=0$ and $\int_{-1}^1xP_{n'}'(x)P_n(x)\,dx=0$ so \begin{align}\boxed{\int_{-1}^1\frac{xP_n^1(x)P_{n'}(x)}{\sqrt{1-x^2}}\,dx=-(1+(-1)^{n+n'})=\begin{cases}-2&\quad\text{if}\quad n-n'\quad\text{is even}\\0&\quad\text{otherwise}\end{cases}}\end{align} When $n=n'$, we can equate $(1)$ and $(2)$ to give $2\int_{-1}^1xP_n(x)P_n'(x)\,dx=1+(-1)^{2n}-\frac2{2n+1}$ so $$\boxed{\int_{-1}^1\frac{xP_n^1(x)P_n(x)}{\sqrt{1-x^2}}\,dx=-\left(1-\frac1{2n+1}\right)=-\frac{2n}{2n+1}.}$$
A: Introduce the quantities
$$
A_{n,m}:=\int_{-1}^1\partial^{n+2}((x^2-1)^{n+1})\partial^{m}((x^2-1)^m)dx.
$$
Integration by parts yields
$$
A_{n,m}=-A_{m-1,n+1}+(1-(-1)^{n+m+1})2^{n+m+1}m!(n+1)!.
$$
Now, since clearly $A_{n,m}=0$ as long as $m>n$, this enables to compute all $A_{m,n}$'s.

Up to constants, we have
$$
P_n(x)=C_n\partial_x^n((x^2-1)^n),\qquad
P_n^1(x)=\frac{C_n^1}{\sqrt{x^2-1}}\partial_x^{n+1}((x^2-1)^n),
$$
thus the integral we are after is essentially
\begin{align}
&\qquad\int_{-1}^1 x\partial_x^{n+1}((x^2-1)^n)\partial_x^{m}((x^2-1)^m)dx
\\&=\int_{-1}^1 \left[\partial_x^{n+1}(x(x^2-1)^n)-(n+1)\partial_x^n((x^2-1)^n)\right]\partial_x^{m}((x^2-1)^m)dx
\\&=\int_{-1}^1 \partial_x^{n+1}(x(x^2-1)^n)\partial_x^{m}((x^2-1)^m)dx-(n+1)\int_{-1}^1 \partial_x^n((x^2-1)^n)\partial_x^{m}((x^2-1)^m)dx.
\end{align}
Now, the second integral in the last expression is computed by the orthogonality property (or, which is equivalent, directly by parts) and the first one is easily computed in terms of the $A_{n,m}$'s introduced above, because
$$
\partial_x^{n+1}(x(x^2-1)^n)
= \partial_x^{n+1}\left(\frac 1{2(n+1)}\partial_x(x^2-1)^{n+1}\right)
=\frac 1{2(n+1)}\partial_x^{n+2}((x^2-1)^{n+1}).
$$
