Understanding part of the proof that $\int_{x=-1}^1P_L(x)P_M(x)\,\rm{d} x =0\quad\text{(for}\,\, L\ne M)$

Question

Starting from Legendre's Differential Equation $$\begin{align*} (1-x^2)y^{\prime\prime}-2xy^{\prime}+L(L+1)y=0\tag{1} \end{align*}$$

In the text that follows $P_L(x)$, $P_M(x)$ represent general Legendre Polynomials.

We can rewrite $(1)$ in the form $$\dfrac{\rm{d}}{\rm{d}x}\Big((1-x^2)P_L\acute (x)\Big)+L(L+1)P_L(x)=0\tag{2}$$

Now after writing the same for $P_M(x)$ $$\dfrac{\rm{d}}{\rm{d}x}\Big((1-x^2)P_M\acute (x)\Big)+M(M+1)P_M(x)=0\tag{3}$$

and then we multiply $(2)$ by $P_M(x)$ to get $$\dfrac{\rm{d}}{\rm{d}x}\Big((1-x^2)P_L\acute (x)\Big)P_M(x)+L(L+1)P_L(x)P_M(x)=0\tag{4}$$ and $(3)$ by $P_L(x)$ to get $$\dfrac{\rm{d}}{\rm{d}x}\Big((1-x^2)P_M\acute(x)\Big)P_L(x)+M(M+1)P_L(x)P_M(x)=0\tag{5}$$

subtracting $(5)$ from $(4)$ gets us to

$$\color{red}{P_M(x)\dfrac{\rm{d}}{\rm{d}x}\Big((1-x^2)P_L\acute (x)\Big)-P_L(x)\dfrac{\rm{d}}{\rm{d}x}\Big((1-x^2)P_M\acute(x)\Big)}+\Big(L(L+1)-M(M+1)\Big)P_L(x)P_M(x)=0\tag{6}$$

To continue this proof I need to show that the first two terms of $(6)$ marked in $\color{red}{\rm{red}}$: $$\color{red}{P_M(x)\dfrac{\rm{d}}{\rm{d}x}\Big((1-x^2)P_L\acute (x)\Big)-P_L(x)\dfrac{\rm{d}}{\rm{d}x}\Big((1-x^2)P_M\acute(x)\Big)}$$ can be written as $$\color{blue}{\dfrac{\rm{d}}{\rm{d}x}\Big((1-x^2)(P_M(x)P_L\acute (x)-P_L(x)P_M\acute (x))\Big)}$$

Is anyone able to demonstrate how to reach the $\color{blue}{\rm{blue}}$ from the $\color{red}{\rm{red}}$ expression by showing the intermediate steps? Or some hints would be very nice. Thanks.

Answer 1

Use the normal add and subtract the same term technique, for example: $$ A(b C')' - C(b A')' = A(b C')' - C(b A')' + A'b C'-A'b C' = (A b C')' -(A' b C)' = (A b C' - A' b C)' = (b(A C' - C A'))' $$

Answer 2

This really isn't an answer to the question as stated but it shows why they are orthogonal.

Start with $-m(m+1)P_m(x) = \frac{d}{dx}((1-x^2)P_m'(x))$ multiply both sides by $P_l(x)$ and integrate over $x\in [-1,1]$, after an integration by parts we get

$$ (m)(m+1)\int_{-1}^1 P_m(x) P_l(x)\mathrm{d}x = \int_{-1}^1 P_l'(x) (1-x^2) P_m'(x)\, \mathrm{d} x $$

Similarly start with $-l(l+1)P_l(x) = \frac{d}{dx} ((1-x^2)P'_l(x))$ and multiply both sides by $P_m(x)$ and integrate over $x\in [-1,1]$, after we integrate this by parts we get

$$ l(l+1)\int_{-1}^1 P_l(x)P_m(x)\, \mathrm{d} x = \int_{-1}^1 P_m'(x) (1-x^2) P'_l(x)\, \mathrm{d}x $$

Comparing both of these equation's right hand sides we see that $$0=(m(m+1)-l(l+1)) \int_{-1}^1 P_l(x)P_m(x)\, \mathrm{d} x$$ so either the integral is zero or $m(m+1)=l(l+1)$, but the latter is only possible if $l=m$ so if $l\not = m$ we must have that the integral is zero.