Legendre polynomials have explicit norms and weighted derivative orthogonality

Write $(x^2-1{)}^n=(x-1{)}^n(x+1{)}^n$. For every $j<n$, its $j$th derivative vanishes at both endpoints: by the product rule, every term differentiates each factor at most $j<n$ times and retains a positive power of the factor that vanishes at that endpoint. Thus $n$ integrations by parts in [F2] give, for every polynomial $q$ of degree less than $n$, ${\int }_{-1}^1{P}_{n}q=\frac{(-1{)}^n}{2^{n}n!}{\int }_{-1}^1(x^2-1{)}^n{q}^{(n)}=0,$ because $q^{(n)}=0$ by [F3]. Hence $P_n$ is orthogonal to every lower-degree polynomial.

The leading term of $(x^2-1{)}^n$ is $x^{2n}$; differentiating it $n$ times gives $(2n)!\mathrm{/}n!$ times $x^n$ by [F3]. All other terms have lower degree after differentiation. Therefore $P_n$ has degree $n$ and leading coefficient ${\kappa }_n=(2n)!\mathrm{/}(2^n(n!{)}^2)>0$. The distinct degrees imply the family $P_0,\dots ,P_N$ is linearly independent and spans every polynomial of degree at most $N$, by successive elimination of leading terms.

Since $P_n-{\kappa }_n{x}^n$ has degree less than $n$, step 1.1 gives ${\int }_{-1}^1{P}_n^2={\kappa }_n{\int }_{-1}^1{P}_n{x}^n\mathrm{.}$ Apply [F2] $n$ times to the Rodrigues formula against $x^n$. The endpoint terms vanish as in step 1.1 and $(x^n{)}^{(n)}=n!$, so ${\int }_{-1}^1{P}_n{x}^n=\frac{1}{2^n}{\int }_{-1}^1(1-x^2{)}^{n}dx\mathrm{.}$

Put $J_n={\int }_{-1}^1(1-x^2{)}^{n}dx$. Then $J_0=2$. For $n\ge 1$, the endpoint values of $x(1-x^2{)}^n$ are zero, so integrating its derivative gives $0=J_n-2n(J_{n-1}-J_n),J_n=\frac{2n}{2n+1}{J}_{n-1}\mathrm{.}$ Induction yields $J_n=2^{2n+1}(n!{)}^2\mathrm{/}(2n+1)!$. Substituting this and ${\kappa }_n$ from step 2.1 into step 3.1 gives $\int P_n^2=2\mathrm{/}(2n+1)$. Together with step 1.1 this proves the first orthogonality formula, including $n=0$.

When $n=0$, $P_0=1$ and $Q_0=-((1-x^2)P_0^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=0$, so $Q_0=0\cdot P_0$. Now take $n\ge 1$ and define $Q_n=-((1-x^2)P_n^{\mathrm{\prime }}{)}^{\mathrm{\prime }}$. Its leading term is $n(n+1){\kappa }_n{x}^n$, so $Q_n-n(n+1)P_n$ has degree less than $n$. For any polynomial $q$ of degree less than $n$, two integrations by parts give ${\int }_{-1}^1{Q}_{n}q={\int }_{-1}^1(1-x^2)P_n^{\mathrm{\prime }}{q}^{\mathrm{\prime }}=-{\int }_{-1}^1{P}_n((1-x^2)q^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=0,$ where both boundary terms vanish because of the factor $1-x^2$, and the last polynomial paired with $P_n$ has degree less than $n$, so step 1.1 applies. Therefore $Q_n-n(n+1)P_n$ is a lower-degree polynomial orthogonal to every lower-degree polynomial. Expand it in the basis from step 2.1; the first formula and the positive norms in step 4.1 force every coefficient to be zero. Thus $Q_n=n(n+1)P_n$.

One integration by parts and step 5.1 give ${\int }_{-1}^1(1-x^2)P_n^{\mathrm{\prime }}{P}_m^{\mathrm{\prime }}={\int }_{-1}^1{Q}_n{P}_m=n(n+1){\int }_{-1}^1{P}_n{P}_m\mathrm{.}$ The first formula, already proved in step 4.1 and step 1.1, yields the displayed weighted derivative identity for every $m,n\ge 0$.