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Consider the $L$-series defined by $$L_{\alpha,\chi}(s) = \sum_{n\geq 1} \frac{e^{2\pi i \alpha \Omega(n)} \chi(n)}{n^s} = \prod_p \left(1 - \frac{e^{2\pi i \alpha} \chi(p)}{p^s}\right)^{-1}.$$ It should have an analytic continuation the left of $\Re s = 1$. Does it have any poles other than (possibly) $s=1$? Does it satisfy a functional equation? Or rather (since this is why I want a functional equation): what kind of growth estimates do we have on $L_{\alpha,\chi}(\sigma + i t)$ for fixed $\sigma < 1$ and variable $t$, $|t|\to \infty$?

(I'm sure this function is known and has most likely been studied - I just do not remember where I have ever seen it.)

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I dispute that this should be termed an "L-function", but anyway...

Taking logs for $\sigma>1$ you get

$$\log L_{\alpha,\chi}(s)=-\sum_p\sum_k{e(k\alpha)\chi(p^k)\over kp^{ks}}$$

So this is

$$\log L_{\alpha,\chi}(s)=-e(\alpha)\sum_p\sum_k{\chi(p^k)\over kp^{ks}} -\sum_p\sum_{k=1}^\infty{\chi(p^k)\over kp^{ks}}[e(k\alpha)-e(\alpha)]$$

and as the $k=1$ term in the latter double sum vanishes, this double sum, call it $G(s)$, is analytic for $\sigma>1/2$, and is bounded for $\sigma\ge \sigma_0$ there.

Exponentiating gives $$L_{\alpha,\chi}(s)=\exp(e(\alpha)\log L_{0,\chi}(s)+G(s))=L(s,\chi)^{e(\alpha)}\exp(G(s))$$

Thus: this gives an analytic continutation of ${L'_{\alpha,\chi}(s)\over L_{\alpha,\chi}(s)}$ to $\sigma>1/2$ away from zeros of $L(s,\chi)$, and similarly for $L_{\alpha,\chi}(s)$ though the zeros will now be branch points.

As for growth, you'd probably want to consider things in terms of the logarithm. For fixed $\sigma<1$, to get a fair-play asymptotic you'd likely have to assume that there are no zeros to the right of $\sigma$ of $L(s,\chi)$, or maybe additionally that $\sigma$ is not a limit point of real parts of zeros.

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  • $\begingroup$ Should we call it a zeta function instead? $\endgroup$
    – H A Helfgott
    Commented Feb 24, 2021 at 16:32
  • $\begingroup$ More seriously - this makes clear that it is necessary to work with a zero-free region, as $L_{\alpha,\chi}$ has essential singularities at any zeroes of $L(s,\chi)$ with $\Re s > 1/2$ (unless $\alpha$ is an integer or a half-integer). Thanks! $\endgroup$
    – H A Helfgott
    Commented Feb 24, 2021 at 16:36
  • $\begingroup$ There's a problem here: $\log L(s;\chi)$ is not a single-valued function -- its derivative $-\frac{L'(s;\chi)}{L(s;\chi)}$ does continue to a single-valued meromorphic function, but it does not integrate to zero around the poles. Thus the continuation above only makes sense for $s$ in a zero-free region. $\endgroup$
    – Lior Silberman
    Commented Feb 25, 2021 at 3:32
  • $\begingroup$ So I don't think your function continues to any region containing a circle around a simple zero, and in particular to take asymptotics on a vertical line $\sigma+it$ you need to assume there are no zeros on the line or to its right (there's no problem if the real parts converge to $\sigma$, since all you need is for the logarithm to be single-valued). $\endgroup$
    – Lior Silberman
    Commented Feb 25, 2021 at 3:38
  • $\begingroup$ Final comment: the logarithmic derivative of $L_{\alpha,\chi}$ will continue as a single-valued function to $\Re(s)>\frac12$ so estimates for that will follow from estimates on the logarithmic derivative of $L(s;\chi)$ and the boundedness of $G'(s)$ (which is also an absolutely convergent Dirichlet series in the region $\sigma>\frac12$) $\endgroup$
    – Lior Silberman
    Commented Feb 25, 2021 at 3:41
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This preprint by Kaczorowski and Perelli may contain the pieces of information you're looking for: https://arxiv.org/abs/1911.10497

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    $\begingroup$ No, the $\Omega(n)$ is missing. $\endgroup$
    – Stopple
    Commented Feb 24, 2021 at 18:37
  • $\begingroup$ Yes, good point. $\endgroup$
    – Sylvain JULIEN
    Commented Feb 24, 2021 at 19:39

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