Integral with Legendre Polynomial $\int x^2(P_l(x))^2dx$

Question

I have the integral where $P_l$ is the $l$th legendre polynomial. $$ \int^1_{-1}x^2\left(P_l(x)\right)^2dx $$

I think the way to do this is by integration by parts but I am not sure how to start. I can compute this numerically but I want a general expression in terms of $l$.

Answer 1

Recall that $$\int_{-1}^1P_mP_ndx=\frac{2}{2n+1}\delta_{mn}.$$Following @gt6989b's tip, for $l\ge 1$ your integral is $$\frac{1}{(2l+1)^2}\int_{-1}^1(lP_{l-1}+(l+1)P_{l+1})^2dx=\frac{2}{(2l+1)^2}\left(\frac{l^2}{2l-1}+\frac{(l+1)^2}{2l+3}\right).$$In the case $l=0$, this would be $\frac23$. Indeed, $P_0=1$ so the integral is $\int_{-1}^1x^2dx=\frac23$ as required.