If $w$ follows a Polya-Gamma Distribution, denoted as $w\sim PG(b,0)$ with $b>0$ then
$$w\overset{D}{=}\frac{1}{2\pi^{2}}\sum_{k=1}^{\infty}\frac{g_{k}}{(k-1/2)^{2}},$$
where $g_{k}\sim\Gamma(b,1)$ mutually independent.
In the following paper, https://arxiv.org/abs/1205.0310, at page 4, equation (3) they derive that the Laplace Transformation of $w$ is the following
$$\mathbb{E}[e^{-wt}]=\prod_{i=1}^{t}(1+\frac{t}{2\pi^{2}(k-1/2)^{2}})^{-b}=\frac{1}{\cosh^{b}(\sqrt{t/2})},$$
where in the last equality they used the Weierstrass Factorization Theorem.
Does anyone know why the product appears in the Laplace Transformation, and overall how they derived the Laplace form??
I made an attempt which seems to be close to what they derived.
First, I define as $c_{k}=\frac{1}{2\pi^{2}(k-1/2)^{2}}$. Hence, the Laplace Transformation can be expressed as
$$\mathbb{E}[e^{-wt}]=\mathbb{E}[e^{-\sum_{k=1}^{\infty}c_{k}g_{k}}]$$ because $g_{k}$ are mutually independet I can write
$$=\mathbb{E}[e^{-c_{1}g_{1}}]\mathbb{E}[e^{-c_{2}g_{2}}]...\mathbb{E}[e^{-c_{k}g_{k}}]...=(1+c_{1}t)^{-b}(1+c_{2}t)^{-b}...(1+c_{k}t)^{-b}...=\prod_{i=1}^{\infty}(1+c_{i}t)^{-b}$$
So, should I assume that they just truncate the infinite product to only $t$ products, and if that's the case what's the reason behind that?