Riemann zeta function form for Dirichlet series of lcm$[m,n]^{-s}$

Question

Define $$f(k) = |\{(m, n) \in \mathbb{N}^2 : [m,n] = k\}|$$ and $$F(s) = \sum_{i=1}^\infty \frac{f(i)}{i^s}$$ where $s= \sigma + it \in \mathbb{C}.$ Write $F$ in Reimann zeta function form and determine its half-plane of absolute convergence.

I would like to study this problem. First, I notice that $$F(s) = \sum_{i=1}^\infty \frac{f(i)}{i^s} =\sum_{m = 1}^\infty \sum_{n=1}^\infty \frac{1}{[m, n]^{s}} = \sum_{m>1}\sum_{n=1}^\infty [m, n]^{-s} + \sum_{i=1}^\infty \frac{1}{i^s} = \sum_{m>1, n= 1,2,3...} [m, n]^{-s} + \zeta(s). $$ For $m = 2$, $$\sum_{n=1}^\infty \frac{1}{[m, n]^s} = \frac{1}{2^s} + \frac{1}{2^s} + \frac{1}{6^s} + ... = \frac{1}{2^s} + \sum_{k=1}^\infty \frac{1}{(2k)^s} + \sum_{k=3}^\infty \frac{1}{(2k)^s} = 2 \sum_{k=1}^\infty \frac{1}{(2k)^s} - \frac{1}{4}.$$ Not sure what is the Riemann zeta form of that. I also think that this is not a good idea to acheive the formula (if it has pattern, it might suggest the formula.)

Answer 1

We first claim the following:

Lemma. For any $n\in\mathbf Z_+$ it holds that $$f(n)=d(n^2),$$ where $d(\cdot)$ is the divisor function.

Proof. We prove it by induction on the number of factors of $k$. If $k$ is a power of a prime then the result is plainly true, so suppose that it holds when $k$ has $N$ distinct prime factors, and we will show that

$$f(k\cdot p^\gamma)=f(k)f(p^\gamma)$$

for any prime $p$ that does not divide $k$. Indeed, a pair $(a,b)$ satisfies that $[a,b]=kp^\gamma$ iff either $p^\gamma|a$, $p^j|b$ ($j=0,\dots,\gamma$) and $[a/p^\gamma,b/p^j]=k$ or $p^j|a$ ($j=0,\dots,\gamma-1$), $p^\gamma|b$ and $[a/p^j,b/p^\gamma]=k$. Since there are $f(k)\cdot (\gamma+1)$ possibilities in the former case, and $f(k)\cdot \gamma$ in the latter, we have that the number of possible pairs is

$$f(k)[\gamma+1+\gamma]=f(k)(2\gamma+1)=d(k^2)d(p^{2\gamma})=d(k^2p^{2\gamma})=f(kp^\gamma).$$

This proves the lemma. $\blacksquare$

From the lemma, we see then that

$$\sum_{n\geq1}\frac{f(n)}{n^s}=\sum_{n\geq1}\frac{d(n^2)}{n^s}.$$

Since $d(n^2)$ is a multiplicative function, we know that

$$\sum_{n\geq1}\frac{d(n^2)}{n^s}= \prod_{p}\left\{1+\frac{d(p^2)}{p^s}+\frac{d(p^4)}{p^{2s}}+\dots\right\} $$

where the product is extended to all the primes. This is,

$$\begin{align}\sum_{n\geq1}\frac{d(n^2)}{n^s} &=\prod_{p}\left\{1+\frac{3}{p^s}+\frac{5}{p^{2s}}+\dots\right\}\\ &=\prod_p\sum_{l=0}^\infty\frac{2l+1}{p^{sl}}\\ &=\prod_p \frac{p^s \left(p^s+1\right)}{\left(p^s-1\right)^2}\\ &=\prod_p\frac1{1-p^s}\prod_p(1+p^{-s})\prod_p\frac1{1-p^{-s}}\\ &=\zeta(s)\frac{\zeta(s)}{\zeta(2s)}\zeta(s)\\ &=\frac{\zeta(s)^3}{\zeta(2s)}. \end{align}$$

Hence

$$\sum_{n\geq1}\frac{f(n)}{n^s}=\frac{\zeta(s)^3}{\zeta(2s)},\qquad\text{whenever }\Re(s)>1.$$