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I would like to estimate the Dirichlet series of a multiplicative function. Consider the following:

$$\sum_{m \leqslant X} \frac{\mu^2(m)}{m^s}$$

When does it converges when $X$ grows? What is an equivalent for $s=1$?

More generally I have no method for doing so, obtaining an estimate of summatory functions or Dirichlet series knowing properties about the sum over divisors. Is there any suitable transform doing so?

If I generalize the settings a little, what can be said about the relation between such an estimate and the associated one on a more general number field? For instance, do we know something precise about:

$$\sum_{N\mathfrak{m} \leqslant X} \frac{\mu^2(\mathfrak{m})}{N\mathfrak{m}^s}$$

where $\mathfrak{m}$ runs along ideals of the integer ring of a number field $F/\mathbf{Q}$?

reuns
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Desiderius Severus
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  • @MarcoCantarini Yes indeed, I fixed the question. I wonder what can be said as asymptotic development when $s=1$ and $X$ goes to infinity? – Desiderius Severus Dec 18 '16 at 10:53
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    You can use the Abel's summation and the asymptotic of the number of squarefree numbers $$\sum_{n\leq x}\mu^{2}(n)=\frac{6}{\pi^{2}}x+O(\sqrt{x}).$$ – Marco Cantarini Dec 18 '16 at 10:55

2 Answers2

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Perron's inversion formula states that if $F(x) = \sum_{n \leq x} f(n)$ and $F(s) = \sum_{n = 1}^{\infty} f(n) n^{-s}$, then \[F(x) = \frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} F(s) x^s \, \frac{ds}{s}\] for $\sigma$ sufficiently large and $x > 1$ not an integer. We take $f(n) = \sum_{N(\mathfrak{a}) = n} \mu^2(\mathfrak{a}) N(\mathfrak{a})^{-\alpha}$. Then via multiplicativity, \[F(s) = \prod_{\mathfrak{p}} \sum_{k = 0}^{\infty} \frac{\mu^2(\mathfrak{p}^k)}{N(\mathfrak{p}^k)^{\alpha}} \frac{1}{k N(\mathfrak{p}^k)^s} = \prod_{\mathfrak{p}} \left(1 + \frac{1}{N(\mathfrak{p})^{\alpha + s}}\right) = \prod_{\mathfrak{p}} \frac{1 - N(\mathfrak{p})^{-2(\alpha + s)}}{1 - N(\mathfrak{p})^{-(\alpha + s)}} = \frac{\zeta_K(\alpha + s)}{\zeta_K(2(\alpha + s))},\] where $\zeta_K(s)$ denotes the Dedekind zeta function of the number field $K$. So \[\sum_{N(\mathfrak{a}) \leq x} \frac{\mu^2(\mathfrak{a})}{N(\mathfrak{a})^{\alpha}} = \frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \frac{\zeta_K(\alpha + s)}{\zeta_K(2(\alpha + s))} x^s \, \frac{ds}{s}\] for $\sigma > \max\{1 - \alpha,0\}$. The integrand has a pole at $s = 1 - \alpha$, at $s = 0$, and at $s = \frac{\rho}{2} - \alpha$ for any zero $\rho$ of $\zeta_K(s)$. By Cauchy's residue theorem, we may move the line of integration to the left, picking up residues along the way. These give the main terms for the left-hand side.

Suppose first that $\alpha \neq 1$. Then the residue of the integrand as $s = 1 - \alpha$ is \[\lim_{s \to 1} (s - 1) \zeta_K(s) \times \frac{x^{1 - \alpha}}{(1 - \alpha) \zeta_K(2)}.\] When $K = \mathbb{Q}$, the limit is $1$ and $\zeta(2) = \pi^2/6$. In general, the limit is given by the analytic class number formula, and $\zeta_K(2)$ may not have such a nice closed form.

Now suppose that $\alpha = 1$. Then the integrand has a double pole at $s = 0$, so this residue is a little more complicated to calculate. We write the integrand in the form \[\frac{\zeta_K(s + 1)}{\zeta_K(2s + 2)} \frac{e^{s \log x}}{s}\] and expand this as a Laurent series around $s = 0$. The main term (after writing $e^{s \log x} = 1 + s \log x + O_x(s^2)$) turns out to be \[\lim_{s \to 1} (s - 1) \zeta_K(s) \times \frac{\log x}{\zeta_K(2)}.\] There is also a constant term that is much more complicated (at least when $K$ is not $\mathbb{Q}$).

Peter Humphries
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  • "moving the contour to the left" is not very clear. What contour, and what estimate of $\zeta_K(s)-\frac{C}{s-1}$ on the left of $Re(s) > 1$ ? – reuns Dec 18 '16 at 22:54
  • Oh, of course not. You need a zero-free region for $\zeta_K(s)$ and bounds for the growth of $\zeta_K(s)$ in this zero-free region, both of which are highly nontrivial and take a fair amount of effort to prove (they're in chapter 5 of Iwaniec and Kowalski). I was just sketching the standard analytic methods for this kind of problem, and showing where the leading order terms come from. – Peter Humphries Dec 18 '16 at 22:56
  • ?? No you don't need a zero-free region, the denominator is $\zeta_K(2s)$ whose analytic region $Re(s) > 1/2$ is trivial. As I sad what you need is an estimate for the growth rate of $\zeta_K(s)$ on $Re(s) > 1/2$ as $t \to \infty$, for example see there – reuns Dec 18 '16 at 23:00
  • Right, you just need to show that $\zeta_K(s)$ doesn't grow too fast in a strip of the form $\sigma > 1 - \frac{c}{\log t}$ (nothing needed about a zero-free region, sorry). Alternatively, you can just be lazy and use a Tauberian theorem; this won't give you a nice bound for the error term, but it means that you just need to know that $\zeta_K(s)$ is meromorphic on $\Re(s) \geq 1$ with only a simple pole at $s = 1$. – Peter Humphries Dec 18 '16 at 23:03
  • And optimally (I think) you can move the contour much more to $Re(s) = 1/2+\epsilon$ at least when $\mathbb{Q} = K$ – reuns Dec 18 '16 at 23:06
  • See what I can get by elementary methods – reuns Dec 19 '16 at 00:12
  • Yes right tks $A(x) =\lim_{s \to 1} \zeta(s) (\frac{1}{\zeta(2s)}-\frac{1}{\zeta(2)})+o(1)= 2\frac{-\zeta'(2)}{\zeta(2)^2}+o(1)$. And you are allowed to upvote ! – reuns Dec 19 '16 at 05:31
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We have that \[\sum_{n=1}^\infty \frac{\mu(n)^2}{n^s} = \frac{\zeta(s)}{\zeta(2s)} = \sum_{n = 1}^{\infty} \frac{1}{n^s} \sum_{d^2 \mid n} \mu(d)\] so that \[\sum_{n \leq x} \mu(n)^2 = \sum_{n \leq x} \sum_{d^2 \mid n} \mu(d) = \sum_{d^2 \leq x} \mu(d) \left\lfloor \frac{x}{d^2} \right\rfloor = x \sum_{d^2 \leq x} \frac{\mu(d)}{d^2} + \sum_{d^2 \leq x} \mu(d) \left(\left\lfloor \frac{x}{d^2} \right\rfloor - \frac{x}{d^2}\right)\] where \[\sum_{d^2 \leq x} \frac{\mu(d)}{d^2} = \sum_{d = 1}^{\infty} \frac{\mu(d)}{d^2} - \sum_{d^2 > x} \frac{\mu(d)}{d^2} = \frac{1}{\zeta(2)} + O\left(\frac{1}{\sqrt{x}}\right)\] and \[\sum_{d^2 \leq x} \mu(d) \left(\left\lfloor \frac{x}{d^2} \right\rfloor - \frac{x}{d^2}\right) = O\left(\sqrt{x}\right),\] i.e. \[\sum_{n \leq x} \mu(n)^2 = \frac{x}{\zeta(2)} + O\left(\sqrt{x}\right).\]

Now let \[B(x) = \sum_{n \leq x} \left(\mu(n)^2 - \frac{1}{\zeta(2)}\right) = O\left(\sqrt{x}\right), \qquad A(x) = \sum_{n \leq x} \frac{\mu(n)^2-\frac{1}{\zeta(2)}}{n}.\] By summation by parts, \[F(s) = \sum_{n=1}^\infty \frac{1}{n^s} \left(\mu(n)^2-\frac{1}{\zeta(2)}\right) = \lim_{x \to \infty} \left(\frac{B(x)}{x^s} - s \int_{1}^{x} \frac{B(t)}{t^s} \, \frac{dt}{t}\right)\] converges for $Re(s) > 1/2$, so taking $s = 1$ shows that $F(1) =\lim_{x \to \infty} A(x)$ and $A(x) = F(1)+o(1)$, and so \[\sum_{n \leq x} \frac{\mu(n)^2}{n} = A(x) + \frac{1}{\zeta(2)} \sum_{n \leq x} \frac{1}{n} = \frac{\log x+\gamma}{\zeta(2)} + F(1)+o(1).\]

reuns
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  • (and for $\zeta_K(s)$ it should work exactly the same) – reuns Dec 19 '16 at 05:32
  • The nice thing about this is that $F(s)$ converges absolutely, so you do truly get $\lim_{x \to \infty} A(x) = \lim_{s \to 1} F(s)$. – Peter Humphries Dec 19 '16 at 16:44
  • This won't work quite as nicely for number fields; the difficulty is the error term $\sum_{N(\mathfrak{a}) \leq x} 1 - C_K x$ isn't $O(1)$ unless $K = \mathbb{Q}$. (Here $C_K = \lim_{s \to 1} (s - 1) \zeta_K(s)$.) – Peter Humphries Dec 19 '16 at 16:48
  • @PeterHumphries you mean we need something like the https://en.wikipedia.org/wiki/Artin_L-function#The_Dedekind_conjecture for the general case ? – reuns Dec 19 '16 at 18:18
  • No, I mean that $\sum_{N(\mathfrak{a}) \leq x} 1 - C_K x$ provably isn't $O(1)$ except when $K = \mathbb{Q}$, so that the number field analogue of $\lfloor x \rfloor - x = {x} = O(1)$ doesn't hold. So you don't get such strong error terms for $\sum_{N(\mathfrak{a}) \leq x} \mu(\mathfrak{a})^2$. – Peter Humphries Dec 19 '16 at 18:27
  • @PeterHumphries Yes of course, but it is $\mathcal{O}(x^\epsilon)$ once we know this conjecture. – reuns Dec 20 '16 at 08:18
  • No, the error term is $\Omega(x^{\frac{1}{2} - \frac{1}{2n}})$, where $n = [K:\mathbb{Q}]$, and conjecturally this should be sharp. When $K=\mathbb{Q}(i)$, this is the Gauss circle problem. – Peter Humphries Dec 20 '16 at 19:36
  • @PeterHumphries tks I didn't know, so it is in some way similar to the Dirichlet divisor problem – reuns Dec 20 '16 at 20:27
  • sorry, but why $\sum_{d^2 > x} \frac{\mu(d)}{d^2} = O\left(\frac{1}{\sqrt{x}}\right)$ – L F May 12 '17 at 14:49
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    Because $\sum_{n > x} \frac{1}{n^2} = O(1/x)$ @LuisFelipe – reuns May 12 '17 at 14:56