Closed form of a product of a ratio of polynomials

Question

I have been trying to approach finding a closed form of the following product$$\prod_{n=0}^\infty\frac{n^2+x_{1}n+x_{0}}{n^2+y_{1}n+y_{0}}$$

and in general$$\prod_{n=0}^\infty\frac{n^m+\sum_{k=0}^{m-1} x_{k}n^k}{n^{m}+\sum_{k=0}^{m-1} y_{k}n^k}$$

are there any techniques or methods that could help evaluate these?

Answer 1

The infinite product does converge when $x_1 = y_1$. In general, given any two polynomials $$\begin{align} A(t) &= \prod\limits_{i=1}^m (t + \alpha_i) = t^m + \alpha t^{m-1} + \cdots\\ B(t) &= \prod\limits_{i=1}^m (t + \beta_i) = t^m + \beta t^{m-1} + \cdots \end{align} $$ When $\alpha = \beta$, we can use the infinite product expansion of gamma function

$$\frac{1}{\Gamma(1+z)} = e^{\gamma z}\prod_{n=1}^\infty \left(1+\frac{z}{n}\right)e^{-\frac{z}{n}}$$ to deduce $$\prod_{n=1}^\infty \frac{A(n)}{B(n)} \stackrel{def}{=} \lim_{p\to\infty} \prod_{n=1}^p\frac{A(n)}{B(n)} = \prod_{i=1}^m \frac{\Gamma(1 + \beta_i)}{\Gamma(1+\alpha_i)}$$

When $\alpha \ne \beta$, the partial product depends on the upper limit $p$ like $O(p^{\alpha-\beta})$. This means the infinite product diverges to $\infty$ when $\Re \alpha > \Re \beta$ and $0$ when $\Re \alpha < \Re \beta$.

For the special case $m = 2$ and $x_1 = y_1 = q \in \mathbb{Z}$, the first infinite product becomes

$$\prod_{n=0}^\infty \frac{n^2 + x_1 n + x_0}{n^2 + y_1 n + y_0} = \frac{x_0}{y_0}\frac{\Gamma(1 + \frac{q}{2} + \nu)(\Gamma(1 + \frac{q}{2} - \nu)}{\Gamma(1 + \frac{q}{2} + \mu)\Gamma(1 + \frac{q}{2} - \mu)}\tag{*1}$$ where $\mu = \sqrt{x_0 - \frac{q^2}{4}}$ and $\nu = \sqrt{y_0 - \frac{q^2}{4}}$.

You can use following functional identifies of gamma function:

$$\Gamma(1+z) = z\Gamma(z)\quad\text{ and }\quad \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin \pi z}$$ to simplify $(*1)$ to a product of rational/sine/hyperbolic sine functions in $\mu$ and $\nu$.

Answer 2

To use simpler notations, consider $$A_p=\prod_{n=0}^p\frac{(n-a)(n-b)}{(n-c)(n-d)}=\frac{\prod_{n=0}^p (n-a)(n-b) } { \prod_{n=0}^p (n-c)(n-d)}=\frac{a\, b\,(1-a)_p \,(1-b)_p}{c\, d\, (1-c)_p\, (1-d)_p}$$ where appear Pochhammer symbols.

Using series expansions for infinitely large $p$ $$\log(A_p)=\log \left(\frac{1}{p}\right) (a+b-c-d)+\log \left(\frac{a b \Gamma (1-c) \Gamma (1-d)}{c d \Gamma (1-a) \Gamma (1-b)}\right)+$$ $$\frac{(a^2+b^2-a-b)-(c^2+d^2-c-d)}{2 p}+O\left(\frac{1}{p^2}\right)$$ This would converge if $a+b=c+d$ and then, if this is the case, $$A_\infty=\frac{a\, b \,\Gamma (1-c)\, \Gamma (1-d)}{c\, d\, \Gamma (1-a)\, \Gamma (1-b)}$$ Now, express $a,b,c,d$ as functions of $x_0,x_1,y_0,y_1$.