Solving integrals of Legendre polynomial multiplied by a differential of another Legendre polynomial

Question

\begin{equation} \int_{-1}^1 P_{m}(x) \frac{d}{dx} P_{n}(x) \, dx =2, \end{equation}

for $m+r=n$ and $r$ is an odd number.

Does a proof of this exist? I am struggling to prove it.

Answer 1

Yes, this comes easily from the formula $\frac{x^2-1}k \frac{d}{dx} P_k = xP_k - P_{k-1}$ i.e. $$(2j+1)P_j = \frac{d}{dx}(P_{j+1} - P_{j-1})$$ i.e. $$\frac{d}{dx} P_{j+1} = (2j+1) P_j + (2(j-2)+1) P_{j-2} + \dots = \frac{2 P_j}{\|P_j\|_{L^2}^2} + \frac{2 P_{j-2}}{\|P_{j-2}\|_{L^2}^2} + \dots $$ since $ \|P_j\|_{L^2}^2 = \frac2{2j+1}$. Setting $n=j+1$ and the fact that the $P_k$s are $L^2$ orthogonal, the only term that survives is the $P_m$ term, giving $$ \int_{-1}^1 P_m(x) \frac{d}{dx} P_n(x) dx= \int_{-1}^1 P_m(x) \frac{2 P_m(x)}{\|P_m\|_{L^2}^2} dx = 2.$$