# Compute $\xi_p= \prod_{n=1}^{\infty} (1+\frac {1}{n^p})$

The main question I want to ask is inspired from this question

Find the value of $$\prod_{n=1}^{\infty} \left(1+\frac {1}{n^2}\right)$$

Now, I have solved this question easily using product representation of $$\displaystyle \frac {\sin x}{x}$$ and then substituting $$x=i\pi$$ in that representation to get $$\prod_{n=1}^{\infty} \left(1+\frac {1}{n^2}\right)=\frac {\sinh \pi}{\pi}$$

But this led me to investigate a more general infinite product systems as $$\displaystyle\xi_p= \prod_{n=1}^{\infty} \left(1+\frac {1}{n^p}\right)$$ where $$p\in N$$ and $$p\ge 2$$. But I don't know of any special function like that of $$\displaystyle \frac {\sin x}{x}$$ that could be used to calculate $$\xi_p$$, so I just started with finding values of $$\xi_2,\xi_3,\xi_4,\cdots$$, and what is quite surprising is that I could conjecture beautiful closed forms for odd and even $$p$$ (which I still think are the same). Here are the closed forms I could conjecture

$$\star$$For odd $$p$$ $$\displaystyle\xi_p=\frac {1}{\displaystyle \prod_{r=1}^{s-1} \Gamma\left(1+(-1)^r (-1)^{\frac rs}\right)}$$

$$\star$$While for even $$p=2k$$,$$k\ge 1$$ and $$k\in N$$ I got $$\displaystyle \xi_p=\alpha \frac {\displaystyle \prod_{r=1}^k \sin\left((-1)^{\frac {2r-1}{s}}\pi\right)}{\pi^k}$$ where $$\alpha$$ can take only the values $$+1,-1,i,-i$$ depending on some condition (which I was not able to find)

It did irk me that the closed form for even $$p$$ had product of sines instead of gamma function and noticed that using the properties of gamma function I can convert the RHS of the case of even $$p$$ to a similar form of RHS of odd $$p$$ but got stuck between at some point.

I would like to know whether or not my closed forms are correct and also some rigorous methods to find the closed forms for $$\xi_p$$.

I would also appreciate if someone could give a proof for the closed forms that he/she would have got for $$\xi_p$$

• There is a reason \displaystyle and the likes are discouraged in titles, but not everyone cares of course. – StubbornAtom Apr 8 at 16:18
• @StubbornAtom Happy ,buddy? – Rohan Shinde Apr 8 at 16:20
• Overjoyed, thank you. – StubbornAtom Apr 8 at 16:21
• $\frac{\pi^2}{\sin^2 \pi z}-\sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}$ is entire and bounded thus constant. The problem is that $\sin(z^{1/s}) \prod_{n=1}^\infty \frac{1}{1-\frac{z^2}{n^{2s}}}$ or $\sum_{n=-\infty}^\infty \frac{1}{(z-n)^s}$ (the completed Hurwitz zeta whose functional equation relates to $\sum_{n=-\infty}^\infty e^{2i \pi n z} n^{s-1}$) are not meromorphic, so the same argument fails to provide an expression in term of elementary functions. – reuns Apr 8 at 16:22
• The reciprocal of gamma function is entire and following infinite product expansion $$\frac{1}{\Gamma(z)} = z e^{\gamma z}\prod_{n=1}^\infty \left(1+ \frac{z}{n}\right) e^{-z/n}$$ valid for all complex $z$. From this, one can deduce $\xi_p = \frac{1}{\prod_{r=0}^{p-1}\Gamma\left(1 - (-1)^{\frac{2r+1}{p}}\right)}$ It is easy to verify this is equivalent to your formula for odd $p$. I haven't bothered to check the case for even $p$. – achille hui Apr 10 at 22:13