\begin{align} \prod_{n=1}^{\infty} \left(1+ (\frac{2\pi n}{\beta})^{-2} \right)^{-1} = \frac{\beta}{2 \sinh(\frac{\beta}{2})} \end{align} I'd like to prove the following products. Can you give me some explicit calculation for this? I try to expand all the term and make exponential, but eventually i failed.
Above products is convergent product (not zeta-function regularized products).
From the perspective of Basel problem
\begin{align} &\sin(x) = x- \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \\ & \frac{\sin(x)}{x} = 1- \frac{x^2}{3!} + \frac{x^4}{5!} + \cdots \end{align} Euler argue that the solution of $\frac{\sin(x)}{x}$ is $\pm n\pi$ for $ n=1,2,3...$ so \begin{align} \frac{\sin(x)}{x} &= (1-\frac{x}{\pi})(1+\frac{x}{\pi})(1-\frac{x}{2\pi})(1+\frac{x}{2\pi})\cdots \\ & = (1- \frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2}) \cdots = \prod_{n=1}^{\infty}(1- (\frac{x}{n\pi})^2) \end{align}
using \begin{align} & \sinh(x) = -i \sin(ix) \end{align}
\begin{align} \frac{\sinh(x)}{x}=\frac{\sin(ix)}{ix}= \prod_{n=1}^{\infty} (1- (\frac{ix}{n\pi})^2) =\prod_{n=1}^{\infty}(1+ (\frac{x}{n\pi})^2) \end{align}