How to evaluate the infinite product $\prod_{n\geq1}\prod_{k\geq1}e^{\frac{(-1)^{n+k}}{n^2k^2+3nk^2+3n^2k+2k^2+2n^2+9kn+6k+6n+4}}$

Question

So I was bored, and decided to do some infinite products and sums for fun. After a while, I came up with this:$$\prod_{n\geq1}\prod_{k\geq1}e^{\frac{(-1)^{n+k}}{n^2k^2+3nk^2+3n^2k+2k^2+2n^2+9kn+6k+6n+4}}$$which I thought that I might be able to evaluate. Here is my attempt at doing so:

Here's the $L$_$A$$T$_$E$$X$ code for the double infinite product so no confusion arises: $$\prod_{n\geq1}\prod_{k\geq1}e^{\frac{(-1)^{n+k}}{n^2k^2+3nk^2+3n^2k+2k^2+2n^2+9kn+6k+6n+4}}$$

$$\begin{align}\prod_{n\geq1}\prod_{k\geq1}e^{\frac{(-1)^{n+k}}{n^2k^2+3nk^2+3n^2k+2k^2+2n^2+9kn+6k+6n+4}}\implies&\prod_{n\geq1}\prod_{k\geq1}e^{\frac{(-1)^n(-1)^k}{(n^2+3n+2)(k^2+3k+2)}}\\\implies&e^{\left(\sum_{n\geq1}\frac{(-1)^n}{(n+1)(n+2)}\right)\left(\sum_{k\geq1}\frac{(-1)^k}{(k+1)(k+2)}\right)}\\\implies&e^{(-1)(-1)\left(\int_0^1\frac{u-u^2}{1+u}du\right)^2}\\=&e^{\left(\int_0^1\frac{u-u^2}{1+u}du\right)^2}\end{align}$$Evaluating the integral and then squaring the result gets us$$\begin{align}e^{\frac32-2\ln(2)}=&e^{\frac32\ln(e)-\ln(4)}\\=&e^{\ln\left(\frac{e^{1.5}}4\right)}\\=&\dfrac{e^\frac32}4\\\approx&1.1204223\end{align}$$

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My question

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Did I evaluate the infinite products correctly, or what could I do to evaluate the products correctly/does it diverge to $\pm\infty$?

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To clear up some confusion:

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Here's how I determined the "any constant $a$ to the power of a sum from $c=n$ to $d$ of $b_n$ is the product from $c=n$ to $d$ of $a^{b_n}$":

Say we have a number, let's just choose $e$ because why not. Now, say we have $e$ raised to the sum from $a=n$ to $b$ of $c_n$, right? Now, because of the laws of exponents, (that being $a^{b+c}=a^ba^c$) we have$$e^{\sum_{a=n}^bc_n}\gets\prod_{a=n}^be^{c_n}$$

Answer 1

Everything is fine (and well done) up to

$$e^{\left(\int_0^1\frac{u-u^2}{1+u}du\right)^2}=e^{\left(\frac{3}{2}-\log (4)\right)^2 }=\frac 1{64}e^{\frac{9}{4}+\log ^2(4)}$$

Side note

If you write $$P(a)=\prod_{n\geq1}\prod_{k\geq1}e^{\frac{(-1)^n(-1)^k}{(n^2+an+b)\,(k^2+ak+b)}}\qquad \text{with} \qquad a^2-4 b=1$$ for even values of $a$ $$P(a)=\exp\big(\left(\pi - \alpha_a \right)^2\big)$$ where the $\alpha_a$ are $$\left\{\frac{10}{3},\frac{46}{15},\frac{334}{105},\frac{982}{315} ,\frac{10942}{3465},\frac{140986}{45045}\right\}$$ The numerators and denominators form respectively sequences $A166107$ and $A025547$ in $OEIS$.