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Is this class of infinite products over primes well known?

Investigating $\dfrac{\zeta(s-1)}{\zeta(s)}$ I was able to generate this series for $\pi$:

$\displaystyle\dfrac{\pi^2}{15}=\prod_{p\in\text{ prime}}\dfrac{p^2}{p^2+1}$

It's fairly straightforward to obtain and it doesn't converge particularly quickly. For example by the fourth term it gives:

$\pi\approx\sqrt{2205/208}\approx3.25$

I can't find a reference to this product anywhere aside from the obvious, and interesting similarity to Euler's:

$\displaystyle\dfrac{\pi^2}{6}=\prod_{p\in\text{ prime}}\dfrac{p^2}{p^2-1}$

Equating their $\pi$s yields the interesting product of primes:

$\displaystyle\prod_p\frac{p^2-1}{p^2+1}=\frac{2}{5}$

This has some curious properties:

$\displaystyle\int_{-1}^1 \frac{p^2-1}{p^2+1} dp = 2 - \pi$

In fact if we subtract the diverging parts:

$\displaystyle\int_{0}^{\infty} \left(\frac{p^2-1}{p^2+1}-1\right) dp = - \pi$

Although I've not done so yet, I think an infinite number of such products over primes can be obtained by this method; all of which equal a rational number. I simply used the quotients between $\zeta(2), \zeta(3), \zeta(4)$ in such a way that $\zeta(3)$ cancelled out.

I was curious as a next step to have a look at what the sequence of such products looks like, and what is its limiting behaviour as $s\to\infty$ which might tell us something about exact forms for $\zeta(2n+1)$. But I wondered if I'm obviously following in somebody else's footsteps here.

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  • $\begingroup$ ... Do you see the difference between $\prod_{p}\dfrac{p^2}{p^2+1}$ and $\prod_p \frac{1}{1+p^{-2}} = \prod_p \left( 1+p^{-2} \right)^{-1}$ ? An infinite product is of the form $\prod_n (1+a_n)$ with $a_n \to 0$. $\endgroup$ – reuns Nov 2 '17 at 21:22
  • $\begingroup$ @reuns I see the importance that it can be expressed as $\prod_n (1+a_n)$ with $a_n \to 0$. Are you saying it's important not to express it in another way? I see the difference between the two products but I don't see the similarity or the relevance of $\frac{1}{1+p^{-2}}$. I see that $\frac{p^2}{p^2+1}=1+\frac{1}{p^2-1}$ which is of the required form. $\endgroup$ – samerivertwice Nov 2 '17 at 23:57
  • $\begingroup$ $\prod_n \frac{1}{1+a_n} = \frac{1}{\prod_n (1+a_n)}$. How do you check the convergence ? $\endgroup$ – reuns Nov 2 '17 at 23:58
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These are called Euler Products.

In particular, $$\frac{\pi^2}{15} = \frac{\zeta(4)}{\zeta(2)} = \prod_p \left( 1+p^{-2} \right)^{-1}.$$

See the examples here for more:

https://en.m.wikipedia.org/wiki/Euler_product

Edit: using the identities in that link, it can be seen that if the real part of $s$ is greater than 2 then we have $$\frac{\zeta(s-1)}{\zeta(s)} = \prod_p \frac{1-p^{-s}}{1-p^{1-s}}$$ and this is also equivalent to $$\sum_{n=1}^{\infty} \frac{\phi(n)}{n^s}.$$

I found the equivalency on page 231 of Apostol’s Introduction to Analytic Number Theory (pdf can be found here http://www.zuj.edu.jo/?wpdmdl=13005). There are a few identities there which are written as exercises or standard examples.

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  • $\begingroup$ Thanks for the reference. I was aware of Euler products but not the class of products over primes such as $5/2$ which is also given there! Which I hadn't seen elsewhere. Any hints as to how I might use my formula for $\dfrac{\zeta(s-1)}{\zeta(s)}$ to evaluate $\zeta(2n+1)$? $\endgroup$ – samerivertwice Nov 2 '17 at 12:42
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    $\begingroup$ @RobertFrost I know that finding the odd integer valued zeta functions is open. Given that the identities you mentioned are known and that the odd values of the Riemann zeta function are not, I imagine that there isn’t an easy way to go about this using those identities. I can think about this a little more though. $\endgroup$ – user328442 Nov 2 '17 at 12:48
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    $\begingroup$ @RobertFrost see my edit $\endgroup$ – user328442 Nov 2 '17 at 13:04

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