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Consider a multiplicative function $f(n)$ that we write in the form $f(n) = h(n)n^a$ for a certain $a>0$ and $h(n)$ a multiplicative function such that $h(n) \asymp 1$ (more precisely, we can take $h(p^k) = (1-p^{-2})^3$ for all $k>0$, even if in fact I am interested in this function with two different values of $h(p)$ and $h(p^2)$). I denote the Dirichlet series $D_f(s)$ associated to $f$.

My question is: what is known about the poles and residues of $D_f$?

I can write what I believe to be true, but maybe extra hypotheses are needed on the function $h(n)$. At least $D_f$ converges for $\Re(s)>a+1$ by the bound on $f(n)$. I think it has a pole at $s=a+1$, and meromorphic continuation up to $\Re(s)>a$. I am particularly interested in the residue of $D_f(s)$ at $s=a$. I guess it is $$R := \prod_p \left(1 - \frac 1p \right) \left( 1 + \frac{h(p)}{p} + \frac{h(p^2)}{p^2} + \cdots \right)$$

I would like to know if this is true, even up to adding some hypothesis, and why it is so. I thought of it as follows: we know the asymptotics $$\sum_{n<x} f(n) = \sum_{n<x} h(n)n^a \sim R \frac{x^{a+1}}{a+1}$$ and we know that the Dirichlet series is related by Mellin transform to this partial sum, more precisely $$D_f(s) = s \int_1^\infty x^{-s-1} \left(\sum_{n<x} f(n) \right) dx \sim \frac{Rs}{a+1} \int_1^\infty x^{a - s} dx \sim \frac{Rs}{(a+1)(a + 1 - s)} $$

Here the residue at $s=a+1$ is obviously $R$, but of course I cannot substitute that roughly the equivalent.

Wirdspan
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    What is your $h(p)$ exactly ? In most cases it won't have an analytic continuation beyond $\Re(s)>a+1$. Try with $h(p)=1+1/\log p$ to see an example where it is not meromorphic at $a+1$. – reuns Dec 11 '20 at 05:49
  • @reuns I have essentially the multiplicative function defined by $h(p^k) = (1-p^{-2})^3$ for all $k \geq 1$. I edited my post to make it clear, but I am also interested to what hypothesis we need to require to have a more general statement. – Wirdspan Dec 11 '20 at 05:56
  • To go beyond $-\sum_p \log(1-(1-p^{-2})^3 p^{-s}) = \sum_{p^k} \frac{p^{-s} (1-p^{-2})^{3k}}{k}=\sum_k\sum_{m=0}^{3k} {3k \choose m}\sum_p \frac{(-1)^{3k-m} p^{-sk-2m}}{k} $ $=\sum_k\sum_{m=0}^{3k} {3k \choose m} (-1)^{3k-m}\frac{1}{k}\sum_d \frac{\mu(d)}{d} \log \zeta(d(sk+2m))$ from which you know the (non-meromorphic due to many branch points) continuation and singularities of $\prod_p \frac1{1-(1-p^{-2})^3 p^{-s}}$. Those kind of things always have a natural boundary on the vertical line where the singularities accumulate – reuns Dec 11 '20 at 06:13
  • @reuns Thanks for the comment. But do you have an idea of the residue? – Wirdspan Dec 11 '20 at 08:06
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    @reuns Is there a way to make my heuristic argument valid? – Wirdspan Dec 11 '20 at 08:49
  • Which heuristic ? It is from the analytic properties of the Euler product that you'll prove that $\sum_{n\le x} h(n) \sim R x$ where $R=\prod_p \frac{1-p^{-1}}{1-(1-p^{-2})^3 p^{-1}}$. Concretely we need that $\sum_n (\sum_{d| n} \mu(d)h(d/n))n^{-s}$ converges absolutely on $\Re(s)=1$ and is non-zero at $1$. – reuns Dec 11 '20 at 09:06
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    @reuns In fact I don't really understand why you consider the Euler product $\prod_p \frac{1-p^{-s}}{1-h(p)p^{-s}}$. Why is there the $1-p^{-s}$ on top? And wouldn't $(1-h(p)p^{-s})^{-1}$ be for totally multiplicative functions?instead? – Wirdspan Dec 12 '20 at 06:07
  • @reuns Also, is the Euler product really converging because $h(p) = 1+O(p^{-2})$? In fact the log is not only $O(p^{-2-s})$ but there is also a second order term that in fact is $O(p^{-2s})$, no? That would give convergence only for $s>1/2$ – Wirdspan Dec 12 '20 at 06:14
  • @reuns In your residue computation, why is there a $\zeta(s)$ popping out? Is it a $\zeta(s)^{-1}$? – Wirdspan Dec 12 '20 at 06:18
  • And yes obviously it is $\sum_p \log(1-(1+O(p^{-2})p^{-s})+p^{-s}=\sum_p O(p^{-2-s})$ which is analytic for $\Re(s) > -1$ while $\sum_p \log(1-(1+O(p^{-2})p^{-s})-\log(1-p^{-s})=\sum_p O(p^{-2-s})+O(p^{-2s})$ is analytic for $\Re(s)>1/2$. So everything is clear to you now ? – reuns Dec 12 '20 at 06:24
  • @reuns Thanks for all these details. I still don't understand your original expression $\prod_p \frac{1-p^{-s}}{1-h(p)p^{-s}}$, where is it coming from? I only know the Euler product decomposition $\prod_p \sum_k h(p^k)p^{-ks}$, wouldn't these two match only for multiplicative functions? – Wirdspan Dec 12 '20 at 06:26
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    I thought you said $h(p^k)= (1-p^{-2})^{3k}$ but it doesn't change much with $h(p^k)=(1-p^{-2})^3$, you get $$\lim_{s\to 1} (s-1) \prod_p (1+(1-p^{-2})^3\frac{p^{-s}}{1-p^{-s}})=\lim_{s\to 1}\frac1{\zeta(s)}\prod_p (1+(1-p^{-2})^3\frac{p^{-s}}{1-p^{-s}})$$ $$=\prod_p (1-p^{-1})(1+(1-p^{-2})^3\frac{p^{-1}}{1-p^{-1}})$$ – reuns Dec 12 '20 at 06:34
  • @reuns Excellent, thank you very much for all these explanation. So this method is general, and we can conclude taking $s=1$ in the product provided it converges? – Wirdspan Dec 12 '20 at 06:40
  • To avoid making it longer, I wrote a question to justify the asymptotics I used in the heuristics: https://math.stackexchange.com/questions/3945323/property-on-average-orders-of-multiplicative-functions – Wirdspan Dec 12 '20 at 07:05

1 Answers1

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  • With $h(n)=\prod_{p| n} (1-p^{-2})^3$, $H(s)=\sum_n h(n)n^{-s}= \prod_p (1+(1-p^{-2})^3\frac{p^{-s}}{1-p^{-s}})$, $G(s)=\frac{H(s)}{\zeta(s)}=\sum_n g(n)n^{-s}$ you get that $$G(s)= \prod_p (1-p^{-s})(1+(1-p^{-2})^3\frac{p^{-s}}{1-p^{-s}})=\prod_p (1-p^{-2s}+(1-p^{-s})((1-p^{-2})^3\frac{p^{-s}}{1-p^{-s}}-p^{-s})) $$ from which $g(p)=0,g(p^k)=O(1)$ so that $\sum_{n\le x} |g(n)| = O(x^{1/2+\epsilon}),\sum_{n> x} g(n)/n = O(x^{-1/2+\epsilon})$ and

$$\sum_{n\le x}h(n)= \sum_{d\le x}g(d) \lfloor x/d\rfloor= \sum_{d\le x}x\frac{g(d)}{d}+\sum_{d\le x} O(g(d))=x (G(1)+O(x^{-1/2+\epsilon}))+O(x^{1/2+\epsilon})$$

  • Next $\log (1+(1-u)^3 \frac{v}{1-v})$ is analytic for $|u|<2r,|v|<2r$ (we don't really care of $r$ just that it is $\in (0,1/2)$) so that for $|u|\le r,|v|\le r$, $\log (1+(1-u)^3 \frac{v}{1-v})=\sum_{m,l} c_{m,l} u^l v^m=\sum_m c_{m,0} v^m+O(u)$ from some coefficients $c_{m,l}\in \Bbb{Q}$ and for $|p^{-s}|\le r$ $$\log (1+(1-p^{-2})^3\frac{p^{-s}}{1-p^{-s}})=\sum_{m\le M} c_{m,0}p^{-sm} +O(p^{-sM})+O(p^{-2})$$

    Thus for $\Re(s) > 1/M$ $$\log H(s)-\sum_{p\le r^{-M}}\log (1+(1-p^{-2})^3\frac{p^{-s}}{1-p^{-s}})= \sum_{m\le M}c_{m,0}\sum_{j\le M} \frac{\mu(j)}{j} \log \zeta(smj))+\sum_p O(p^{-sM})$$ $$=\sum_{d\le M} \log \zeta(sd)\sum_{m| d} c_{m,0} \frac{\mu(d/m)}{d/m}+\sum_p O(p^{-sM})$$

  • Since the $\sum_p O(p^{-sM})$ term is analytic for $\Re(s)>1/M$ you got the analytic continuation of $H(s)$ to $\Re(s)>1/M$ with some pole/zero/branch point at $1/d$ and $\rho/d$ whenever $\sum_{m| d} c_{m,0} \frac{\mu(d/m)}{d/m}$ is non-zero and/or not an integer, plus some trivial zeros coming from the Euler factors.

    It is supposedly obvious that $\sum_{m| d} c_{m,0} \frac{\mu(d/m)}{d/m}$ won't be zero for all $d$ large enough so that $H(s)$ has infinitely many poles/zeros/branch points accumulating near $\Re(s)=0$ (a natural boundary with no analytic continuation beyond).

reuns
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  • I understand the details about analytic continuations. However in the answer, do you compute the residue as in the original question? Isn't it reduced to the computations written in your last comment? – Wirdspan Dec 12 '20 at 10:25
  • The residue is found in the very first line, again $G(s)=H(s)/\zeta(s)$ is analytic for $\Re(s) > 1/2$ so $H(s)$ is meromorphic on $\Re(s)>1/2$ with only one simple pole at $s=1$ of residue $G(1)$. – reuns Dec 12 '20 at 10:55