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I am actually interested in the analytic continuation of $\phi_w(s)=\prod_p(1+w\cdot p^{-s})^{-1}$. Here $w$ is rational, or the imaginary unit multiplied by a rational. Consider for now that $w=1$. Of course,

$$\phi(s)=\sum_{k=1}^\infty \frac{\lambda(k)}{k^s}=\frac{\zeta(2s)}{\zeta(s)},$$

where $\lambda$ is the Louisville function, see here. It is known that $\lambda(k)=\pm 1$, being equal to $1$ half of the time, and $-1$ half of the time. The function $\phi(s)$ is analytic for $\Re(s)>1$. But what about $\Re(s)>\frac{1}{2}$?

Question: Does $\phi$ have an analytic continuation to $\Re(s)>\frac{1}{2}$? The closest reference I could find is this.

In the end, I am interested in the product $\rho(s)=\eta(s)\phi^*(s)$ where $\eta$ (the Dirichlet eta function) is the analytic continuation of $\zeta$, and $\phi^*$ is the analytic continuation of $\phi$, and thus $\rho$ is analytic. I'd like to say something about the zeros of $|\rho(s)|^2$ in $\frac{1}{2}<\Re(s)<1$, by looking at the zeroes of $\eta$ and $\phi^*$ separately, more specifically in the product expression. The Riemann hypothesis probably implies it has none, thus also my interest in a case like $w=i$ (the imaginary unit) where $\eta_w$ has known zeroes $s=(\log i)/(\log p_k)$ for any prime $p_k$ (see note at the bottom), but the product cancels them out:

$$(1-i p_k^{-s})(1+i p_k^{-s})=1+p_k^{-2s}.$$

Anyway, before even contemplating studying this problem, the first step is to get an analytic continuation of $\phi$, thus my question. Proving that the series for $\phi(s)$ convergences if (say) $\Re(s)>\frac{1}{2}$ would solve this issue. However I remember reading that proving this is as hard as proving RH (the Riemann hypothesis). If that is the case, there is no point for me to dig deeper into this.

Note:

The function $\eta_w(s)$ is the analytic continuation of $\prod_p (1-w\cdot p^{-s})^{-1}$.

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    $\begingroup$ $\zeta(2s)$ is analytic and nonzero for $\Re s>1/2$. In addition, $\zeta(s)$ is meromorphic for $\Re s>1/2$. Hence $\phi(s)$ is analytic for $\Re s>1/2$ if and only if $\zeta(s)$ is nonzero for $\Re s>1/2$. The last condition is the Riemann Hypothesis. I voted to close, because this question is not appropriate at this site (asking for the validity of a well-known conjecture). $\endgroup$
    – GH from MO
    Commented Apr 30, 2021 at 5:48
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    $\begingroup$ @GHf : We have $|\sum_{k>1} \lambda(k)/k|<\pi^2/6$, see mathoverflow.net/questions/198404/…. This suggests that $\phi(s)$ is defined at $s=1$. I am wondering if convergence of the series for $\phi(s)$ can be handled using the Dirichlet test (en.wikipedia.org/wiki/Dirichlet%27s_test). After all, convergence for $\eta(s)$ (where $\lambda(k)$ is replaced by $(-1)^{k+1}$ has been proved, using the Dirichlet test if I remember correctly, up to $\Re(s)>0$. $\endgroup$
    – Vincent Granville
    Commented Apr 30, 2021 at 12:33
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    $\begingroup$ $\rho(s)=(1-2^{1-s})\zeta(2s)$ on $\Re s >1$ hence everywhere, so not sure what is unclear about it beyond what is unclear about $\zeta$; same about $\phi(s)=\zeta(2s)/\zeta(s)$ everywhere (so it is meromorphic in the plane with poles at the non-trivial zeroes of $\zeta$) so not sure again what is unclear about that one either beyond what is unclear about $\zeta$ $\endgroup$
    – Conrad
    Commented Apr 30, 2021 at 14:17
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    $\begingroup$ In general, it is better to start mathematical research with a lesser known problem. Research and learning goes in small steps. I mean, you don't have to convince yourself that the RH is true, and it is pointless to try your hands at the RH without considerable research expertise. There are many problems about $\zeta(s)$ and related functions that are much easier than the RH, but still would be very valuable to make progress on. $\endgroup$
    – GH from MO
    Commented Apr 30, 2021 at 18:46
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    $\begingroup$ Statistics of patterns in consecutive values of $\lambda(k)$ is a difficult and active area of research. See arXiv:1509.01545 and arXiv:1904.05096, for example. $\endgroup$
    – GH from MO
    Commented Apr 30, 2021 at 21:41

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