TITLE: Is there any significance to the product $\prod_\limits{n=1}^{\infty} \left(1+\frac{1}{n^x}\right)$?
QUESTION [5 upvotes]: Is there any significance to this product?
$$\prod\limits_{n=1}^{\infty} \left(1+\frac{1}{n^x}\right)$$

Basically taking the Riemann zeta function and trying to make it into a convergent product, because if you have a convergent sum then you can take the infinite product of 1+the sum and it will also converge.

REPLY [4 votes]: If $x$ is an integer $(x >1)$, there are nice formulae for the very first values of $x$
$$P_x=\prod\limits_{n=1}^{\infty} \left(1+\frac{1}{n^x}\right)$$
$$P_2=\frac{\sinh (\pi )}{\pi }$$
$$P_3=\frac{\cosh \left(\frac{\sqrt{3}  }{2}\pi\right)}{\pi }$$
$$P_4=\frac{\cosh \left(\sqrt{2} \pi \right)-\cos \left(\sqrt{2} \pi \right)}{2 \pi ^2}$$
$$P_6=\frac{\sinh (\pi ) \left(\cosh (\pi )-\cos \left(\sqrt{3} \pi \right)\right)}{2 \pi
   ^3}$$ It seems that, when $x$ is even, the formulae are quite "simple".
When $x$ is odd $(x>3)$, the result seems to be the reciprocal of products of gamma functions with complex arguments (the roots of unity). For example
$$P_5=\frac{1}{\Gamma \left(1-(-1)^{1/5}\right) \Gamma \left(1+(-1)^{2/5}\right) \Gamma
   \left(1-(-1)^{3/5}\right) \Gamma \left(1+(-1)^{4/5}\right)}$$