Wilson polynomials

Remarks

Since $(a + ix)_k (a - ix)_k = \prod_{j=0}^{k-1} \left( (a + j)^2 + x^2 \right)$, $W_n$ is a polynomial of degree $n$ in $x^2$. Example: $W_1(x^2) = (abc + abd + acd + bcd) - (a + b + c + d) x^2$. Check: $(a + b)(a + c)(a + d) - (a + b + c + d) a^2 = abc + abd + acd + bcd$, and the coefficient of $x^2$ is $-(a + b + c + d)$. This example is symmetric in $a, b, c, d$; in fact every $W_n$ is. Orthogonality: if $a, b, c, d$ have positive real parts and the nonreal ones come in complex conjugate pairs, then for $m \ne n$ $$\int_0^\infty W_m(x^2) W_n(x^2) \abs{\frac{\Gamma(a + ix) \Gamma(b + ix) \Gamma(c + ix) \Gamma(d + ix)}{\Gamma(2ix)}}^2 dx = 0,$$ where $\Gamma$ is the gamma function. The positivity hypothesis matters: if, say, $a < 0$, the integral alone no longer gives orthogonality, and the orthogonality measure needs finitely many extra point masses. Non-example: the polynomial $x$ is not a Wilson polynomial for any parameters, since every $W_n$ is a polynomial in $x^2$ and hence an even function of $x$.