# Prove that $\displaystyle\int_{x=-1}^{1}P_L(x)P_{L-1}\acute (x)\,\mathrm{d}x=\int_{x=-1}^{1}P_L\acute(x)P_{L+1} (x)\,\mathrm{d}x=0$

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.

• Think of the polynomial $q(x) = P_{L-1}'(x)$. $P_{L-1}$ has degree $L-1$. What degree does $P_{L-1}'$ have? – Cameron Williams Apr 18 '16 at 1:51
• @Cameron Does it also have degree $L-1$? – BLAZE Apr 18 '16 at 1:52
• When you differentiate a (nonconstant) polynomial, its degree decreases by one, e.g. $x^n \Longrightarrow nx^{n-1}$. – Cameron Williams Apr 18 '16 at 1:53
• @Cameron So it has degree $L-2$? – BLAZE Apr 18 '16 at 2:01
• Yep. That is exactly right. – Cameron Williams Apr 18 '16 at 2:05

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.