The product $$ F(s)=\prod_{p}\frac1{(1-p^{-s})^p}, $$ converges for $\mathrm{Re}(s)>2$, when $p$ runs over all primes. Does it admit analytic continuation beyond the line $\mathrm{Re}(s)=2$? Any papers where it has been studied?
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$\begingroup$ Can you say something on where this equation comes from? The appearance of a $p$ in the exponent seems a bit weird to me, even if the answer apparently have no problem handling it at all. $\endgroup$– VincentCommented Aug 29, 2017 at 12:31
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2$\begingroup$ @Vincent: It comes from considering Ihara zeta functions attached to an arithmetic group and taking the product over these for varying primes. $\endgroup$– user1688Commented Aug 31, 2017 at 6:35
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2 Answers
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$$P(s) = \sum_p p^{-s}, \qquad \log F(s) = \sum_{p^k} \frac{p^{1-sk}}{k} = \sum_{k\ge 1} \frac{P(sk-1)}{k}$$
$P(s) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ and $P_N(s) = \sum_{n=N+1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ is analytic for $\Re(s) > \frac{1}{N+1}$ so that $$e^{N! P(s)} = e^{N! P_N(s)}\prod_{n=1}^{N-1} \zeta(ns)^{\mu(n) \frac{N! }{n}}$$ is meromorphic for $\Re(s) > \frac{1}{N+1}$ providing the analytic continuation of $P(s)$ :
$P(s)$ has a branch point at $\frac{\rho}{N}$ for each $N\ge 1$ and non-trivial zero $\rho$ of $\zeta$.
Therefore $P(s)$ has a natural boundary on $\Re(s) = 0$ and no analytic continuation exists beyond there.
For the same reason $F(s)^{N!}$ is meromorphic for $\Re(s) > 1+\frac{1}{N+1}$ and $\log F(s)$ has a branch point at $1+\frac{\rho}{N}$ for every $N\ge 1,\rho$ and hence a natural boundary on $\Re(s) = 1$ and no analytic continuation exists beyond there.
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1$\begingroup$ I think a branch point already prevents analytic continuation as a single valued function to a half-plane minus a discrete set of points of isolated singularities. So $P(s)$ already cannot be continued beyond the $\Re(s)=1$ line due to the logarithmic singularity of $\log\zeta(s)$ at $s=1$. $\endgroup$ Commented Jul 18, 2017 at 12:43
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3$\begingroup$ I agree though that $F(s)^{N!}$ is meromorphic for $\Re(s)>1+\frac{1}{N+1}$. This is a nice observation and extends what I said. $\endgroup$ Commented Jul 18, 2017 at 12:46
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$\begingroup$ @GHfromMO Hi, I'm trying to understand the theory behind the convergence of $\lim_{k \to \infty} \sum_{n=1}^\infty a_n n^{-s} 1_{gcd(n,N_k)=1}$ for arbitrary Dirichlet series with functional equation, with $ N_k = \prod_{p \le k}$. From that I obtain heuristics suggesting a proof of the RH would imply the analyticity of $\sum_p p^{-s} e^{2i \pi \xi p}$ for $\Re(s) > 1/2, \xi \in (0,\pi)$ (and hence GRH). Do you know this subject, and what it could be useful for compared to analyticity of $\sum_p p^{-s} \chi(p)$ ? $\endgroup$– reunsCommented Sep 4, 2017 at 18:33
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Let $\sigma:=\mathrm{Re}(s)$, and consider the principal branch of the logarithm. For $\sigma>3/2$ we have $$\begin{align*}p\log(1-p^{-s})-\log(1-p^{1-s})&=p\left(-p^{-s}+O(p^{-2\sigma})\right)-\left(-p^{1-s}+O(p^{2-2\sigma})\right)\\&=O(p^{2-2\sigma}),\end{align*}$$ hence the "Euler sum" $$H(s):=\sum_p\left\{p\log(1-p^{-s})-\log(1-p^{1-s})\right\},\qquad\sigma>3/2,$$ converges locally uniformly (and absolutely). This implies that the Euler product $$ G(s):=\exp(H(s))=\prod_p\frac{(1-p^{-s})^p}{1-p^{1-s}},\qquad\sigma>3/2,$$ defines a non-vanishing holomorphic function. In the original half-plane $\sigma>2$, we have $$ G(s)=\frac{\zeta(s-1)}{F(s)},\qquad\sigma>2,$$ hence $F(s)=\zeta(s-1)/G(s)$ extends to a meromorphic function in $\sigma>3/2$ with a simple pole at $s=2$ and no other pole there.
Regarding your second question, I am not aware of any papers where this function was studied. My argument above is rather standard though.
answered Jul 18, 2017 at 10:49
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1$\begingroup$ With the same argument you get the analytic continuation to $\Re(s) > 1$ and a natural boundary there $\endgroup$– reunsCommented Jul 18, 2017 at 11:29
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$\begingroup$ @reuns: I don't think so, because $H(s)$ diverges at $s=3/2$. Perhaps, with a more refined argument, one can get meromorphic continuation to $\Re(s)>1$, but even then it is unclear why $\Re(s)=1$ is a natural boundary. Feel free to add your more detailed analysis (in a separate post). $\endgroup$ Commented Jul 18, 2017 at 11:37
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