New generalizations of the Poisson kernel (Q1366412)

Recall that the Poisson kernel in two dimensions defined by \[ P(r,\theta) = \frac{1 - r^{2}}{1 - 2 r\cos\theta + r^{2}} \] has the property that \[ \frac{1}{2\pi} \int_{0}^{2\pi} P(r,\theta) d\theta = 1 \] for \( 0 \leq r < 1\) and \[ \frac{1}{2\pi} \int_{0}^{2\pi} P(r,\theta) d\theta = -1 \] for \(r>1\). \textit{A. E. Taylor}, in a short note [Am. Math. Mon. 57, 478-479 (1950; Zbl 0037.34704)], noticed that for \(0<r<1\), \[ F(r) = \frac{1}{2\pi} \int_{0}^{2\pi} P(r,\theta) d\theta \] satisfies the functional equation \[ F(r) = F(r^{2}). \] Since \(F(0) = 1,\) successive application leads to an alternative proof of the integral identity. The authors extend Taylor's technique to integrals with two and four parameters defined as follows: \[ Q(\theta;a,b) = \frac{1 - ab}{(1 - ae^{i\theta})(1 - be^{i\theta})}. \] \[ R(\theta;a,b,c,d) = \frac{L(a,b,c,d)}{(1 - ae^{i\theta})(1 - be^{-i\theta}) (1 - ce^{i\theta})(1 - de^{-i\theta})}, \] where \[ L(a,b,c,d) =\frac{(1-ab)(1-ad)(1-bc)(1-cd)}{1-abcd}, \] and \(a,b,c,d\) are complex numbers with modulus less than \(1.\) The integrals are evaluated by converting them to contour integrals on the unit circle in the complex plane and then calculating the residues at the poles of the integrand. It turns out that \[ \frac{1}{2\pi} \int_{0}^{2\pi} Q(\theta, a,b) d\theta = 1 \] and \[ \frac{1}{2\pi} \int_{0}^{2\pi} Q(\theta, a,b)^{2} d\theta = \frac{1+ab}{1-ab}, \] but a similar neat expression for \[ \frac{1}{2\pi} \int_{0}^{2\pi} Q(\theta, a,b)^{n} d\theta \] for \(n > 3 \) perhaps is not available. Hence it is stated as an open problem.