***Update**: I added $exp[i\theta\_k(s)]$ in the definition of $\eta^\*(s)$ to address some critical convergence issues. Thanks for the contributors who pointed to these issues.*


Prime numbers are denoted as $p\_1,p\_2,\dots$ with $p\_1=2$. The modulus of a complex number $s$ is denoted as $|s|$. Finally, $S$ denotes the right half of the critical strip, defined by $\frac{1}{2}<\Re(s)<1$.


Let us assume that one can find a product of the form


$$\eta^\*(s)=\prod\_{k=1}^\infty\frac{\exp[i\theta\_k(s)]\cdot\tau\_k(s)}{1-p\_k^{-s}},$$


converging in $S$, but not necessarily if $\Re(s)=\frac{1}{2}$, thus leaving open the possibility that it has zeroes or is undefined on the critical line. Let us further assume that


* $\theta\_k(s)=\arg(1-p\_k^{-s})$, with $\arg$ denoting the principal argument
* $\tau\_k(s)$ is a strictly positive real number
* $|\eta^\*(s)|>0$ if $s\in S$, thus no zero (in other words, the product converges if $s\in S$)
* $\eta^\*(s)$ is smooth enough


Would that imply that the Riemann Hypothesis (RH) is true? I guess the answer is *not necessarily*. I can not believe that the answer is yes, otherwise (barring some mistakes in my computations), I have found such a function, and I know that there is no way I could prove RH. So I am looking for an answer that explains why it does not necessarily imply that RH is true.


Note that if $\tau\_k(s)=1$, then $|\eta^\*|=|\zeta|$ is the modulus of the traditional zeta function.


Below is my function $\eta^\*$ satisfying all the requirements. The methodology to get there can be applied to other Dirichet $L$-functions. It is described in some details in my previous MO question [here](https://mathoverflow.net/questions/391178/truncated-euler-products-dirichlet-eta-function-and-convergence-issues) (not at all intended to prove RH), with the core idea explained in the "Update" section, just below the conclusions. The arguments are not very complicated. Instead, the approach (based on finite products ultimately converging) is somewhat unusual and involves some subtleties, and some luck in the sense that there are some rather unexpected simplifications taking place.


**About my re-scaled Riemann zeta product**


Let $s=\sigma + it$. It is defined using


$$\tau\_k(s)=\Big[1+\frac{2\cos(t\log p\_k)}{p\_k^\sigma + p\_k^{-\sigma}}\Big]^{-\frac{1}{2}}.$$


It results in


$$|\eta^\*(s)|^{-2} = \Big\{\prod\_{k=1}^\infty \Big(1+\frac{1}{p\_k^{2\sigma}}\Big)\Big\}
\cdot \Big\{\prod\_{k=1}^\infty \Big(1-\frac{4\cos^2(t\log p\_k)}{p\_k^{2\sigma}+p\_k^{-2\sigma}+2}\Big)\Big\}.$$


This simplifies to


$$|\eta^\*(s)|^{-2} = \frac{1}{\zeta(2\sigma)}
\cdot \prod\_{k=1}^\infty \Big[1-\Big(\frac{2\cos(t\log p\_k)}{p\_k^{\sigma}+p\_k^{-\sigma}}\Big)^2\Big] .$$


The above product converges if $s\in S$, but not always (if ever) when $\Re(s)=\frac{1}{2}$, and never if $\Re(s)<\frac{1}{2}$. And of course, due to the infinite product representation, $\eta^\*(s)$ can never vanish if $s\in S$. Again, details are available [here](https://mathoverflow.net/questions/391178/truncated-euler-products-dirichlet-eta-function-and-convergence-issues).


**Conclusions**


Just like $\eta$ (the Dirichlet eta function) is a scaled version of $\zeta$ to study its zeroes in $0<\Re(s)<1$, so is $\eta^\*$ in $\frac{1}{2}<\Re(s)<1$. We have $\eta(s)=(1-2^{1-s})\zeta(s)$ and $|\eta^\*(s)|=\tau(s)|\eta(s)|$ with $\tau(s)$ being the product of all $\tau\_k(s)$, properly defined if $\Re(s)>1$. Both functions $\eta$ and $\eta^\*$ are scaled analytic continuations. In particular, $\eta$ is an additive scaling in the sense that it is defined by a series (and thus easy to build but very hard to use to prove RH) while $\eta^\*$ is a multiplicative scaling directly defined by an infinite convergent product (thus hard to build but easy to gain insights about RH).



Scaled Riemann zeta function with no zero in the critical strip
The conditions presented are contradictory - for any $s=\sigma+it\in S$ with $t\neq 0$ it is impossible to choose positive real numbers $\tau\_k(s)$ such that $\prod\_{k=1}^\infty\frac{\tau\_k(s)}{1-p\_k^{-s}}$ converges. This is because the argument of the partial product cannot converge, and this is something that positive real $\tau\_k(s)$ can fix.


I will omit the details, but the idea is similar to that in my answer to your previous question. For some $n\in\mathbb N$ let $N=e^{2\pi n/t}$ and consider the product over primes in the interval $[e^{2n\pi/t+\pi/2t},e^{2n\pi/t+\pi/t}]$. By prime number theorem, there are $\gg N/\log N$ primes in this interval. For each such prime we have $\Im(p\_k^{-s})\gg N^{-\sigma}$, and so we deduce the the argument of $1-p\_k^{-s}$ is $\gg N^{-\sigma}$. The increase in the argument coming from all primes in this interval is $\gg N^{1-\sigma}/\log N$, which diverges. Hence the product can't converge.


Given the basic premise of the question is faulty, this voids the entirety of the rest of the question.


