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Choose real monic orthogonal polynomials $p_j$ of degree $j$, $j=0,1,\dots$, for $w(x)=x^a{e}^{-x}$ on $[0,\infty )$, and letExtend each ${\varphi }_j$ by zero to negative $x$. The resulting functions form an orthonormal set in $L^2(\mathbb{R},dx)$. Define the orthogonal polynomial projection kernelThe formula on the right is for nonnegative arguments; the kernel is zero when either argument is negative. With ${\Phi }_{ij}={\varphi }_{j-1}(x_i)$, the kernel evaluation matrix is $\Phi {\Phi }^T$. By the determinant multiplication identity and the preceding monic basis change,Thus initially ${\hat{c}}_m=c_m{\prod }_j{h}_j$.

Use the symmetric labelled joint probability density on the full orthant. Expanding the two copies of $\det \Phi$ into permutations and integrating all variables, orthonormality makes a term vanish unless the two permutations agree. Each of the $m!$ surviving terms integrates to $1$. HenceIf the eigenvalues are instead listed in increasing order on a single ordered chamber, that chamber's density is $m!$ times this symmetric density. We keep the full-orthant convention throughout.

An explicit choice is $p_j(x)=(-1{)}^{j}j!L_j^{(a)}(x)$ in terms of the Generalized Laguerre polynomials. To verify the normalization independently, their Rodrigues formula givesIntegrating by parts $j$ times proves orthogonality to every lower-degree polynomial. The boundary terms vanish at zero and infinity for $a\ge 0$. Taking the other factor to be $p_j$ and using $p_j^{(j)}=j!$ gives $h_j=j!\Gamma (a+j+1)$, where $\Gamma$ is the Gamma function. The Laguerre projection kernel is thereforefor $x,y\ge 0$.

Evaluate $R_n$ at $x=1$ using $(x^2-1{)}^n=(x-1{)}^n(x+1{)}^n$. In the Leibniz rule for the $n$th derivative, every term vanishes at $1$ except the term putting all $n$ derivatives on $(x-1{)}^n$. Therefore $R_n(1)=n!2^n$. Combining this with $P_n(1)=1$ gives the normalisation in the Rodrigues formula:For $n=0$, the formula gives ${\alpha }_0=1$, as required. As an independent check, the leading term of $R_n$ is $(2n)!x^n/n!$, so the resulting leading coefficient of $P_n$ is $(2n)!/(2^n(n!{)}^2)$, consistent with the Legendre polynomial recurrence relation.

The displayed Rodrigues formula gives the Legendre polynomialA four-node Gaussian rule is exact through degree $2\cdot 4-1=7$. Solving $q_4(x)=0$ first for $x^2$ gives the four nodesThe rule is not exact for every polynomial of degree eight: its quadrature value on $q_4^2$ is zero because every node is a root of $q_4$, whereas

Suppose $m<n$. The Rodrigues formula and $n$ integrations by parts givebecause $p_m^{(n)}=0$. Every boundary term vanishes: the exponential controls infinity, while the remaining power of $x$ has positive exponent at zero. Symmetry of the inner product gives orthogonality whenever $m\ne n$.

For the norm, monicity gives $p_n^{(n)}=n!$, so the same calculation yields