A question (Problem $7.4$) in my textbook (Mathematical Methods in the Physical Sciences - 3rd Edition by Mary L. Boas P578) asks me to
Use $$\int_{x=-1}^{1}(P_L(x)\cdot\text{any polynomial of degree < L})\,\mathrm{d}x=0\tag{A}$$ to prove that $$\displaystyle\int_{x=-1}^{1}P_L(x)P_{L-1}\acute (x)\,\mathrm{d}x=0\tag{1}$$and gives the hint: $\color{#180}{\fbox{What is the degree of $P_{L−1}(x)$}}$?
Also, show that $$\displaystyle\int_{x=-1}^{1}P_L\acute(x)P_{L+1} (x)\,\mathrm{d}x=0\tag{2}$$
Where $P_L(x),P_{L-1}(x)$ represent any general Legendre Polynomial and $P_{L-1}\acute (x)$ is the derivative of the $(L-1)$th Legendre Polynomial.
I am genuinely stuck right from the beginning as equation $(\mathrm{A})$ doesn't say anything about whether derivatives can be used with it. In other words; if the question was asking me to "Show that $\displaystyle\int_{x=-1}^{1}P_L(x)P_{L-1} (x)\,\mathrm{d}x=0$", then I would understand completely and simply quote equation $(\mathrm{A})$
I simply don't understand how to even start and the hint in $\color{#180}{\mathrm{green}}$ is making no sense to me whatsoever.
Does anyone have any idea how to show $(1)$ and $(2)$ are equal to zero?
Many thanks.
If you let $q = P_{L-1}'$, then $q$ is a polynomial. Its degree is no greater than that of $P_{L-1}$ since differentiation can never increase the degree of a polynomial. However since $P_{L-1}$ is a polynomial of degree $L-1$, we have that $q$ has degree at most $L-1$. By $(\text{A})$, $(1)$ and $(2)$ follow.