Find the value of $\prod_{k=1}^{\infty} \big(1+\frac{1}{k^s}\big)$ I am attempting to find values to the family of products given by
$$p(s)=\prod_{k=1}^{\infty} \Bigg(1+\frac{1}{k^s}\Bigg)$$
where $s\in \mathbb{C}$ and the real part of $s$ is greater than $1$. In particular the values of $p(2)$ and $p(3)$ are
$$p(2)=\frac{\sinh{\pi}}{\pi}$$
$$p(3)=\frac{\cosh{\Big(\frac{\sqrt{3}}{2}\pi\Big)}}{\pi}$$
by using Wolfram: Alpha.
Firstly, I would like to know if there is a method which allows $p(s)$ to be written in a closed form for any $s\in \mathbb{C}$. Otherwise, can someone explain how to find the values of $p(2)$ and $p(3)$ shown above?
 A: For integers at least:
Following in the lines of this answer, we can write
$$\prod_{k=1}^{N-1} \frac{P(k)}{k^s},$$
where $P(x)=(x-\alpha_1)\cdots(x-\alpha_s)$, as
$$\prod_{j=1}^s \frac{\Gamma(N-\alpha_j)}{\Gamma(N)\Gamma(1-\alpha_j)}.$$
Then, using the lemma also described in that answer that
$$\lim_{N\to\infty} \prod_{i=1}^n \frac{\Gamma(N+\alpha_i)}{\Gamma(N+\beta_i)}=1$$
if $\sum_{i=1}^N \alpha_i=\sum_{i=1}^N \beta_i$, the terms involving $N$ cancel out in the limit and we get
$$\prod_{j=1}^s \Gamma(1-\alpha_j)^{-1}.$$
Using that $\Gamma(x)=(x-1)\Gamma(x-1)$, we can write this as
$$(-1)^{s+1}\prod_{j=1}^s \Gamma(-\alpha_j)^{-1}.$$
This doesn't seem to have a nice closed form (or at least not one WolframAlpha can find for $s=5$), but we can evaluate it for $s=2$ as follows, by noting that
$$\Gamma(i)\Gamma(-i)=i\Gamma(i)\Gamma(1-i)=\frac{i\pi}{\sin(i\pi)}=\frac{\pi}{\sinh \pi},$$
which gives your result. For $s=3$, the product becomes the reciprocal of
\begin{align*}
\prod_{\omega^3=1}\Gamma(\omega)
&=\Gamma(\omega)\Gamma(\overline{\omega})\\
&=|\Gamma(\omega)|^2\\
&=\frac{|\Gamma(1+\omega)|^2}{|\omega|^2}\\
&=\left|\Gamma\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)\right|^2\\
&=\frac{\pi}{\cosh\left(\frac{\pi\sqrt{3}}{2}\right)}.
\end{align*}
