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I would like to know the asymptotic behaviour at large $n$ for $t\in\mathbb{R}$, $t\neq0$ of the following function: \begin{align*} A_n(t)&=\sum_{q=\frac{a}{b}\in \mathbb{Q}^+|\gcd(a,b)=1 \& ab=n}q^{it} \\ &=2^{\omega(n)}\prod_{i=1}^{\omega(n)}\cos(t\log p_i^{v_i}) \end{align*} where we have used the prime number decomposition of $q=\prod_{i}^{\omega(n)}p_i^{v_i}$ and $n=\prod_{i}^{\omega(n)}p_i^{|v_i|}$, and $\omega(q)=\omega(n)$ is their number of distinct prime factors.

On what can I rely to answer such question ? I can only notice: $$|A_n(t)|\leq A_n(0)=2^{\omega(n)}\sim 2^{\log\log n}, \quad n\gg 1$$ The aim is to find information about the convergence of the associated Dirichlet series $\sum_{n\geq 1}\frac{A_n(t)}{n^s}$ where $s\in \mathbb{R}$. The above remark leads to the absolute convergence for $s>1$, can we say better?

EDIT: As pointed out below, I got this expression writing $\frac{|\zeta(s+it)|^2}{\zeta(2s)}$ as a Dirichlet series.

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    $\begingroup$ Note that $2^{\log\log n} = (\log n)^{\log 2} \not\sim \log n$. Also note that $2^{\omega(n)}$ and $2^{\log\log n}$ are close only for most $n$, not all $n$; and even for typical $n$, the two functions are not really asymptotic to each other (their logarithms are, but that's not sufficient). $\endgroup$
    – Greg Martin
    Commented Apr 7, 2018 at 2:10

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Your series is never absolutely convergent in any half-plane of the form $\mathrm{Re}\,s>\delta$ with $\delta<1$ and there is even no convergence in the case $\mathrm{Re}\,s=1$. To prove this, let us notice that $A_n(t)$ is a multiplicative function of $n$, therefore we have

$$f(s):=\sum_{n=1}^{+\infty} A_n(t)n^{-s}=\prod_p f_p(s),$$

where

$$f_p(s)=1+\sum_{n=1}^{+\infty} \frac{2\cos(nt\log p)}{p^{ns}}=1+\frac{1}{p^{s+i t}- 1} +\frac{1}{p^{s-i t}- 1}$$

and the product is taken over all prime numbers.

Therefore, we have

$$f(s)=\frac{\zeta(s+it)\zeta(s-it)}{\zeta(2s)}.$$

Suppose now that the series converge absolutely for $\mathrm{Re}\,s=1$. Then the function $f(s)$ is bounded on this line, which is certainly not the case, as $f$ has a pole at $s=1+it$, because zeta is non-zero on the edge of critical strip and has a simple pole at $s=1$.

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  • $\begingroup$ Thank you for your answer. Your $f(s)$ was indeed my starting point, but I think I was not clear enough, I will add precisions. Actually, I am assuming $s,t\in \mathbb{R}$ and I would like to study the convergence of the Dirichlet series that yields $f(s,t)$ or the asymptotics of $A_n(t)$ for non zero $t$ (to avoid the pole at $s=1$). $\endgroup$
    – Alexandre
    Commented Apr 7, 2018 at 11:41
  • $\begingroup$ @Alexandre, the function $f(s)$ has a pole in the point $s=1+it$ for every $t \in \mathbb R$. I think one can also get an asymptotic formula for the partial sums of $A_n(t)$ and prove that there is no convergence in any point of the line $\mathrm{Re}\,s=1$. $\endgroup$
    – Alexander Kalmynin
    Commented Apr 7, 2018 at 12:11
  • $\begingroup$ I would be glad to have some insight on how to get such asymptotic formula. I think we should have at least the absolute convergence of the series for $\Im s =0$ and $\Re s >1$ $\forall t\in \mathbb{R}$, and hopefully the convergence for $\Im s = 0$ and $\forall \Re s$ for $t\neq 0$. Though I am not sure how to prove this. $\endgroup$
    – Alexandre
    Commented Apr 8, 2018 at 12:34

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