Values where infinite products of primes and composites are equal Highly grateful for your help/steers on the following question (at the end):
Take the infinite product:
$$\displaystyle T(s) = \prod _{n=2}^{\infty } \left( \dfrac{{n}^{s}} {{n}^{s}-1}\right)$$
for $\Re(s) > 1$ it is equal to:
$$\displaystyle \prod _{primes}^{\infty } \left( \dfrac{{p}^{s}} {{p}^{s}-1}\right) * \prod _{composites}^{\infty } \left( \dfrac{{c}^{s}} {{c}^{s}-1}\right)$$
I.e. the Euler-product (equal to $\zeta(s)$) multiplied by its composite "equivalent" ( excluding 1 since that is a bit of a strange composite anyway).
Why my interest? I wanted to learn more about the composite infinite product (and see if it had a 'zeta' like version). Soon became clear to me that the only way to learn more about this product, is to concentrate on $T(s)$ and then divide it by $\zeta(s)$. 
I searched the web but there is hardly anything known about $T(s)$. F.i. Wolfram math only shows (formula 20) two different solutions (note: both need to be raised to $^{-1}$ to get $T(s)$ !) for odd and even integers and by reading through some arxiv math pre-prints the best I could find was a single, but still integer only formula that is:
$$\prod _{k=1}^{s-1}\Gamma  \left( 2- {{\rm e}} ^{{\frac {2 i \pi k}{s}}} \right), ( \Re(s) > 1, s \in \mathbb{N})$$
I then decided to explore ways to extend the domain for $s$ and derived the following formula:
$$\displaystyle \ln  \left( T\left( s \right)  \right) = \ln  \prod_{n=2}^{\infty } \left(   \left( -1+{n}^{-s} \right) ^{-1} \right) = \sum_{n=2}^{\infty } \ln  \left(  \left( 1-{n}^{-s} \right) ^{-1} \right)$$
$$\displaystyle = \sum_{m=1}^\infty \sum_{n=2}^{\infty } \frac{1}{mn^{ms}} =  \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n}$$
And this brings us to:
$$ T(s)={\rm e}^{\left( \displaystyle \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)}$$
Yep, there's always a $\zeta(s)$ hiding around the corner somewhere...
So, let's see what the plot looks like for $s>0$ ($T(s)$ diverges for $s<0$).
$T(s)=\displaystyle  {\rm e}^{\left( \displaystyle \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)} \text{   blue}$ 
$\displaystyle \zeta(s) = \prod _{primes}^{\infty } \left( \dfrac{{p}^{s}} {{p}^{s}-1}\right) \text{purple}$
$\displaystyle \frac{T(s)}{\zeta(s)} = \prod _{composites}^{\infty } \left( \dfrac{{c}^{s}} {{c}^{s}-1} \right) \text{   brownish}$
graph
For $s>1$ I could numerically solve the following equation:
$$\displaystyle \prod _{primes}^{\infty } \left( \dfrac{{p}^{s}} {{p}^{s}-1}\right) = \prod _{composites}^{\infty } \left( \dfrac{{c}^{s}} {{c}^{s}-1} \right)$$
giving this interesting number $s = 1.397737620...$ (there is only one for $\Re(s) > 1$ )
I obviously took a deep dive with this number on Google and Plouffe's inverter, but have not found anything 'beautiful' or related to other constants as yet.
Then the domain $0 < s < 1$. It is easy to see in the graph that $T(s)$, and therefore also $\dfrac{T(s)}{\zeta(s)}$, have 'trivial' poles for $s= \dfrac{1}{k}, k \in \mathbb{N}$ that are induced by the fact that for each $s= \dfrac{1}{k}$ there always is a $n s = 1$ that makes at least one term in the infinite sum equal to the pole $\zeta(1)$ (hence the whole sum turns into a pole).
But I'm actually mostly intrigued by what happens under the x-axis and especially where:
$$\zeta(s) = \dfrac{T(s)}{\zeta(s)}$$ or
$$|\zeta(s)| = {\rm e}^{\displaystyle \left(\frac12 \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)}$$.
If I have done my analysis correctly, this result would imply that there are an infinite number of values for $0 < s < 1$, where the (analytically continued) infinite products of primes and composites are equal (since $\zeta(s)$ remains negative between $0 < s < 1$ and there are an infinite number of poles separating the intersection points). And that would imply/reveal an infinite amount of tiny bits of information about how the primes 'grow like weed between the composites'.
Of course I checked $T(s)$ also for $s \in \mathbb{C}$, however, any graph I've produced sofar for $s=a+bi$ of $T(s)={\rm e}^{\left(\displaystyle \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)}$ did not reveal any non-trivial zeroes (nope, not even at $a=\frac12$...), although the curves do seem to be trending towards a large number of very chaotically distributed zeroes when $a \rightarrow 0$.
So, apologies for the relatively long intro to my question:
Since $\zeta(s)$ has been analytically continued throughout the entire complex domain, is it allowed to also analytically continue the division of $\dfrac{T(s)}{\zeta(s)}$ into the domain $s<1$? Or do the nominator and denominator each require an individual continuation and does the concept of division get 'lost in continuation'?
 A: I think that T is meromorphic on $\mathbb{C}$ just like $\zeta$, with a single pole at $s=0$. The ratio should be fine everywhere except at $s=1$, the negative integers, and the critical strip (or line, on the RH).
A: Just to elaborate a bit on my reaction to David Hansen's valid comment that I actually  should have explained in my original question ("my intended response was too large to fit in the margin" ;-) ).
My interest in the infinite product of composites is based on the following idea. 
Infinite products of the shape $\prod _{primes}^{\infty } \left( \dfrac{{p}^{s}} {{p}^{s}-1}\right)$ are only defined for $R_e >1$. The domain can be extended via analytical continuation as Riemann showed in his 1859 paper for the Euler product. He named the new function $\zeta(s)$ and proved it be valid for all $s \in \mathbb{C}$ with the only exception a pole at $s=1$. He found $\zeta(s)$ had zeroes (trivial ones at -2, -4, ... and non-trivial ones that appear to all be on the line $s = 0.5 + bi$). And via further transformations he also established a direct connection between the zeroes and the prime counting function $\pi(s)$. 
If we take the logarithmic version of the prime-counting function $\psi(s)$ (i.e. the sum of all prime powers less than x, weighted by a natural logarithm of the power e.g.:
$\psi(10) = 3 \log(2) + 2  \log(3) + 1  \log(5) + 1  \log(7)$
then the exact prime counting function is ($\rho_k$ is a non-trivial zero):
$\psi(x) = x - \log(2\pi) - \frac12 \log(1- \frac{1}{x^2})  -  \sum_{\rho} \dfrac{x^{\rho}}{\rho}$ 
Guess this is pretty standard stuff for the readers of this board and I also fully appreciate that the prime numbers are the atoms of the composites (what's in a name), but I wondered whether there could exist a Composite-counting-function that might be derived from $\prod _{composites}^{\infty } \left( \dfrac{{c}^{s}} {{c}^{s}-1}\right)$ in a similar fashion as Riemann did for the prime product. If so, one could use it as a sort of "detour" to approach the Riemann hypothesis from the other side. Let's just try a very small step backwards from the end result: 
$C(x)$ = number of composities < x.
$\psi(x) = x - C(x)$
$C(x) = (\log(2\pi) + \frac12 \log(1- \frac{1}{x^2})  +  \sum_{\rho} \dfrac{x^{\rho}}{\rho})$
A Composite-counting-function will therefore also be dependent on the non-trivial zeros. Since I couldn't find any way to obtain a zeta-like $C(x)$ function for the composite infinite product, I started exploring the route via $T(s)$ and got some success (I hope) by getting it expressed fully into $\zeta(s)$'s as:
$C(s) = \dfrac{e^{\sum_{n=1}^\infty \frac{\zeta(n s)-1}{n}}}{\zeta(s)}$
And before I even start dreaming about analytic continuation with contour integrals or further steps with Fourier/Mellin transforms, I'd been keen to understand whether the ratio can indeed be continued into $s \in \mathbb{C}$. If Charles' very encouraging response is indeed confirmed, then this would imply that the division $\dfrac{T(s)}{\zeta(s)}$ would induce a peak in $C(s)$ for each $s=\rho_k$. So, I'd need to count peaks rather than zeroes to compute the infinite sum of $\rho$'s in the Composite-counting-function.
P.S.
After I plotted the graph for $T(s)$ with $s=0.5 + xi$, the $C(s)$-peaks and the $\zeta(s)$-zeroes at $s=\rho_k$ do not fully balance out and keep hovering between 0 and 1.
A: i'st easy to see that:
$$\ln T(s)=\sum_{n=1}^{\infty}\frac{\zeta(ns)-1}{n}$$
using the integral definition of the zeta function, one can show that:
$$\ln T(s)=s\int_{0}^{\infty}\frac{E_{s}(x^{s})-1}{xe^{x}(e^{x}-1)}dx$$
where : 
$E_{\alpha}(z)$ is the mittag-leffler fuction . 
now, following Riemann's trick, here is what i did :
start with contour integral :
$$I(s)=-s\oint_{c}\frac{E_{s}((-x)^{s})-1}{xe^{x}(e^{x}-1)}dx$$ 
the contour is the usual Hankel contour. consider $I(-s)$ :
$$I(-s)=s\oint_{c}\frac{E_{-s}((-x)^{-s})-1}{xe^{x}(e^{x}-1)}dx=-s\oint_{c}\frac{E_{s}((-x)^{s})}{xe^{x}(e^{x}-1)}dx$$ 


*

*the Mittag-Leffler function admits the beautiful continuation :
$E_{\alpha}(z^{-1})=1-E_{-\alpha}(z)$ -


or 
$$I(s)-I(-s)=s\oint_{c}\frac{dx}{xe^{x}(e^{x}-1)}=s\oint_{c}(-x)^{-1}e^{-x}dx-s\oint_{c}\frac{(-x)^{-1}dx}{e^{x}-1}$$
now :$$\oint_{c}(-x)^{-1}e^{-x}dx=\frac{-2\pi i}{\Gamma(1)}=-2\pi i$$
and the second integral could be thought of as:
$$\oint_{c}\frac{(-x)^{-1}dx}{e^{x}-1}=\lim_{z\rightarrow 0}\oint_{c}\frac{(-x)^{z-1}dx}{e^{x}-1}=-2i\lim_{z\rightarrow 0}\sin(\pi z)\Gamma(z)\zeta(z)=i\pi$$
or : 
$$I(s)-I(-s)=-3\pi is$$
lets go back to the 1st integral, and expand the Mittag-leffler function :
$$I(s)=-s\oint_{c}\frac{E_{s}((-x)^{s})-1}{xe^{x}(e^{x}-1)}dx=-s\sum_{n=1}^{\infty}\frac{1}{\Gamma(1+ns)}\oint_{c}\frac{(-x)^{sk-1}dx}{e^{x}(e^{x}-1)}$$ $$=s\sum_{n=1}^{\infty}\frac{2i \sin(k\pi s)\Gamma(ks)}{\Gamma(1+ns)}\left(\zeta(ks)-1\right)=2i\sum_{n=1}^{\infty}\sin(k\pi s)\frac{\zeta(ks)-1}{k}$$ 
now the problem becomes finding a function of the variable s -lets call it $A(s)$- such that:
$$\sum_{n=1}^{\infty}\sin(k\pi s)\frac{\zeta(ks)-1}{k}=A(s)\sum_{n=1}^{\infty}\frac{\zeta(ks)-1}{k}$$ 
if we define :
$$k(s)=\sum_{n=1}^{\infty}\frac{\zeta(ks)-1}{k}$$
then :
$$A(s)k(s)-A(-s)k(-s)=-\frac{3}{2}\pi  s$$
and the problem becomes proving the existence of $A(s)$ for all s, and of course, finding it !!  
