Prove that Legendre polynomial satisfies $\int_{-1}^1 x P_n(x) P_m '(x) dx=0$

Question

I am aware that I have to use the equations

$(1-x^2)P_m''-2xP_m'+m(m+1)P_m=0$

and

$(1-x^2)P_n''-2xP_n'+n(n+1)P_n=0$

but I have been unable to do so.

I tried integrating by parts, like in this question: Prove that the Legendre polynomial satisfies $\int_{-1}^1(1-x^2) P_m' P_n' dx=0$

This gives me:

$\int_{-1}^1 x P_n P_m' dx= x P_n P_m \bigr |_{-1}^1-\int_{-1}^1P_nP_m dx-\int_{-1}^1 xP_mP_n'dx$

Again,

$\int_{-1}^1 x P_n' P_m dx= x P_n P_m \bigr |_{-1}^1-\int_{-1}^1P_nP_m dx-\int_{-1}^1 xP_m'P_ndx$

which is the same equation again.

This leads nowhere. Please help!

Answer 1

Hint

Use that $(1-x^2)P_m''-m(m+1)P_m=2xP_m^\prime$

to get $$\begin{aligned}2\int_{-1}^1 x P_n P_m' dx &= \int_{-1}^1 (1-x^2)P_nP_m^{\prime \prime}dx-m(m+1)\int_{-1}^1 P_mP_ndx\\ &=\int_{-1}^1 (1-x^2)P_nP_m^{\prime \prime}dx \end{aligned}$$

and use the previous question that you asked.