How to caculate this integral by Legendre Poly.

Question

How to caculate the integral

$$\int_{-1}^1(1-x^2)\mathrm{P}_k'(x)\mathrm{P}_l'(x)~\mathrm{d}x$$

Where $\mathrm{P}_l(x)$ is the $l$ - oeder Legendre Poly.

Answer 1

Integrating by parts we obtain $$ \int_{-1}^1(1-x^2)P_k'(x)P_l'(x)\,dx = \left.(1-x^2)P_k'(x)P_l(x)\right|_{-1}^1 -\int_{-1}^1\frac{d}{dx}[(1-x^2)P_k'(x)]P_l(x)\,dx. \tag{1} $$ The first term on the RHS of $(1)$ vanishes, and the second one can be simplified with the help of the differential equation satisfied by the Legendre polynomial $P_k(x)$, $$ (1-x^2)P_k''(x)-2xP_k'(x)+k(k+1)P_k(x)=0 $$ $$ \implies \frac{d}{dx}[(1-x^2)P_k'(x)]=-k(k+1)P_k(x). \tag{2} $$ Thus \begin{align} \int_{-1}^1(1-x^2)P_k'(x)P_l'(x)\,dx &= k(k+1)\int_{-1}^1P_k(x)P_l(x)\,dx \\ &= \frac{2k(k+1)}{2k+1}\,\delta_{kl}, \tag{3} \end{align} where the last step follows from the normalization of the Legendre polynomials.