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Suppose $a$, $b$ two reals different from $0$, and consider the following infinite product: \begin{align} \prod_{n=1}^{\infty} \Bigg[ 1+ 2\ i \cdot \frac{n^2 + a\ n+b}{n^2+(n^2+a\ n+b)^2} \Bigg] \end{align}

where $i$ is the imaginary unit.

Question: is it possible to find a closed form for this product? Or at least with $a$, $b$ having some restrictions?

Infinite products involving terms like $1+\frac{1}{P(n)}$ where $P(n)$ is a polynomial of degree 2 can be sometimes solved with a closed form but I have not seen anything with polynomial of higher degree.

Lucenaposition
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1 Answers1

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From this post, one can find a reference to "On gamma quotients and infinite products" by M.Chamberland and A.Straub where we have the following:

\begin{equation} \prod_{k=0}^{\infty} \frac{(k+\alpha_1)...(k+\alpha_n)} {(k+\beta_1)...(k+\beta_n)} = \frac{\Gamma(\beta_1)...\Gamma(\beta_n)} {\Gamma(\alpha_1)...\Gamma(\alpha_n)} \end{equation}

if $\alpha_1+...+\alpha_n = \beta_1+...+\beta_n$.

Note that in the original question, the numerator $N$ and denominator $D$ of the product term are \begin{align} N &= n^2 + (n^2+a\ n+b)^2 + 2\ i\ (n^2+a\ n+b) \\ D &= n^2 + (n^2+a\ n+b)^2 \end{align} and so $N$ and $D$ have the same coefficient in $n^3$, i.e. the sum of the roots is the same for $N$ and $D$, which is exactly the requirement for applying the formula.

Since N and D are polynomials of degree 4, there exists exact formulas for computing their roots and then get a closed expression in terms of the gamma function for the infinite product.