Orthogonality of Legendre polynomials using specific properties I'm having significant issues with a problem and would appreciate any help at all with it. It is regarding proving the orthogonality of Legendre polynomials using a specific recursion formula and property of Legendre polynomials.
I have been given the following relation $$x P'_n(x) -P'_{n-1}(x) = n P_n (x) $$
And that $$ \int^{1}_{-1} f(x) P_n (x) dx = 0 $$ for any polynomial $f(x)$ of degree less than $n$.
I must then show that for $ n \geq  1$
$$ \int^{1}_{-1} P_n (x)^2 dx = \frac{1}{2n} \int^{1}_{-1} x \frac{d}{dx} (P_n (x)^2)dx $$
And then determine the value of the integral, that being $\frac{2}{2n+1}$. I must use the two stated facts, I'd be grateful for any help with this, thank you guys.
 A: $$\int\limits_{-1}^{1} P_n(x)^2 \;dx = $$
$$\int\limits_{-1}^{1} P_n(x)P_n(x) \;dx = $$
$$\frac{1}{2n}\int\limits_{-1}^{1} 2n P_n(x)P_n(x) \;dx = $$
Substituting your first relation in:
$$\frac{1}{2n}\int\limits_{-1}^{1} 2 \left[x P'_n(x) - P'_{n-1}(x)\right] P_n(x)\;dx = $$
$$\frac{1}{2n}\int\limits_{-1}^{1} 2 x P'_n(x) P_n(x) \; dx - \frac{1}{2n}\int\limits_{-1}^{1} 2 P'_{n-1}(x) P_n(x)\;dx = $$
Given your second property defining orthogonality (since after all $P'$ is of lesser degree than $P$) the entire second integral drops out.
$$\frac{1}{2n}\int\limits_{-1}^{1} x 2 P_n(x) P'_n(x) \; dx  = $$
Reverse of the chain rule:
$$\frac{1}{2n}\int\limits_{-1}^{1} x \frac{d}{dx} P^2_n(x) \; dx $$
QED
And for anyone who asks/wonders, or is inclined to criticize... I answered this question because it was asked, and this is a Q&A forum.  I shouldnt have to justify answering questions on a Q&A forum though, but this is the culture here.
A: I immediately thought of integration by parts. Here is my work:
By the product rule of derivative we have
$$ \frac{d}{dx} x\, P_n(x)^2 = P_n(x)^2 + x\, \frac{d}{dx}(P_n(x)^2)
\tag{1} $$
Note that Legendre polynomials have the property
$$ P_n(1) = 1,\quad \text{ and } \quad P_n(-1)=(-1)^n. \tag{2} $$
By integrating equation $(1)$ and using equation $(2)$ we get
$$ 2 = 1 - (-1) = \int_{-1}^1 P_n(x)^2 dx + \int_{-1}^1 
  x\, \frac{d}{dx}(P_n(x)^2) dx. \tag{3} $$
Define $\, a_n := \int_{-1}^1 P_n(x)^2 dx.\,$ Equation $(3)$
implies that
$$ 2 - a_n = \int_{-1}^1   x\, \frac{d}{dx}(P_n(x)^2)\,dx. \tag{4} $$
We are given the property that
$$ x P'_n(x) -P'_{n-1}(x) = n P_n (x). \tag{5} $$
Multiply this equation $(5)$ by $\,P_n(x)\,$ and integrate
to get
$$  \int_{-1}^1\!P_n(x)(xP'_n(x)\!-\!P'_{n-1}(x))dx
\!=\!\int_{-1}^1\!P_n(x)nP_n(x)dx. \tag{6} $$
We are also given the property that
$$ \int_{-1}^1 P_n(x)\,f(x)\,dx = 0 \tag{7} $$
for any polynomial $\,f(x)\,$ of degree less than $\,n.\,$
Now notice that
$$ x\, \frac{d}{dx}(P_n(x)^2) = 2\,x\,P_n(x)P'_n(x). \tag{8} $$
Combining equations $(6)$, $(7)$, and $(8)$ we get that
$$ \frac12(2-a_n) = n\,a_n. \tag{9}$$
Dividing this equation $(9)$ through by $\,n\,$ and using
the definition of $\,a_n\,$ and equation $(4)$ gives the result we wanted:
$$ \int_{-1}^1 P_n(x)^2 dx = \frac{2-a_n}{2n} = \frac1{2n}
\int_{-1}^1   x\, \frac{d}{dx}(P_n(x)^2)\,dx. \tag{10}$$
Also, solving for $\,a_n\,$ in equation $(9)$ gives
$$ a_n = \frac2{2n+1} \tag{11}$$
which was also requested.
NOTE: The property in equation $(2)$ was not given to
us. It would have to be proven using equations $(5)$
and $(7)$ which are given to us. It can be done, and
would take some effort, but I have not given the
steps required.
