It can be shown that the divisor function $\sigma_s(k)=\sum_{d\vert k} d^s$ defined for $k\in\mathbb{Z}^+$ can be expressed as a Dirichlet series with the Ramanujan sums $c_n(k):=\sum\limits_{m\in(\mathbb{Z}/n\mathbb{Z})^\times}e^{2\pi i\frac{m}{n}k}$ defined for $k\in\mathbb{Z}$ and $n\in\mathbb{Z}^+$ with the following identity $$\boxed{\forall k\in\mathbb{Z}^+:\sigma_s(k)=\zeta(1-s)\sum_{n=1}^\infty\frac{c_n(k)}{n^{1-s}}}$$ It is easy to show that, since $m\mapsto-m$ is a bijection on $(\mathbb{Z}/n\mathbb{Z})^\times$, $c_n(-k)=c_n(k)$ as we only permute the invertible elements. This suggests by looking at the previous Dirichlet series that $\sigma_s(k)$ could be extended to $k\in\mathbb{Z}\setminus\{0\}$ as $\sigma_s(-k)=\sigma(k)$. This is quite intuitive as it means that $\sigma_s(k)=\sum_{d\vert k}d^s$ runs over the positive divisors $\forall 0\neq k\in\mathbb{Z}$. Notice as well that we could extend $\sigma_s(k)$ to $k=0$ by looking at the previous series as $$\sigma_s(0)=\zeta(1-s)\sum_{n=1}^\infty \frac{c_n(0)}{n^{1-s}}=\zeta(1-s)\sum_{n=1}^\infty \frac{\varphi(n)}{n^{1-s}}=\zeta(1-s)\frac{\zeta(-s)}{\zeta(1-s)}=\zeta(-s)$$ This is also intuitive noting that the positive divisors of $0$ are $\{1,2,3,...\}$ so one would informally guess $\sigma_s(0)=1^s+2^s+3^s+...=\zeta(-s)$. So the previous series let's us expand the domain of $\sigma_s(k)$ to $k\in\mathbb{Z}$. In fact, it can also be shown that this extension is also multiplicative in the sense that if $n_1$ and $n_2$ are coprime then $\sigma_s(n_1 n_2)=\sigma_s(n_1)\sigma_s(n_2)$.
But this is not the end of the story. Coprimality can also be considered in $\mathbb{Q}$ where we say that $x,y\in\mathbb{Q}$ are coprime $\iff \forall p\in\mathbb{P}:\nu_p(x)=0$ or $\nu_p(y)=0$ where $\nu_p$ is the rational $p$-adic valuation. Then, $\sigma_s(x)$ can be defined for all $x\in\mathbb{Q}$ by defining $\forall p\in\mathbb{P}$ prime and $\alpha\in\mathbb{Z}:\sigma_s(p^\alpha)=\frac{1-p^{s(1+\alpha)}}{1-p^s}$, $\sigma_s(0)=\zeta(-s)$ and allowing it to be multiplicative on $\mathbb{Q}$. As an example, $$\sigma_2\left(\frac{64}{63}\right)=\sigma_2\left(2^6\cdot3^{-2}\cdot7^{-1}\right)=\frac{1-2^{14}}{1-2^2}\cdot\frac{1-3^{-2}}{1-3^2}\cdot\frac{1-7^{0}}{1-7^2}=5461\cdot\left(-\frac{1}{9}\right)\cdot 0=0$$ In fact, if $\nu_p(x)=-1$ for some $p\in\mathbb{P}$, then $\sigma_s(x)=0$. My question then is
Q: Is there a natural extension of Ramanujan sums $c_n(x)$ to $x\in\mathbb{Q}$ such that the former Dirichlet series identity holds true for every $k\in\mathbb{Q}$? That is, such that the following identity holds $$\boxed{\forall x\in\mathbb{Q}:\sigma_s(x)=\zeta(1-s)\sum_{n=1}^\infty\frac{c_n(x)}{n^{1-s}}}$$ One guess could be $c_n(x):=\sum\limits_{m\in(\mathbb{Z}/n\mathbb{Z})^\times}e^{2\pi i\frac{m}{n}x}$ but I don't know if it actually works.
I won't consider the proof of existence as a complete answer since I would like to find an explicit extension of $c_n$ to $\mathbb{Q}$.
$\color{red}{\bf\text{Update:}}$ As noted on Wikipedia, Hölder proved the identity $$c_n(k)=\mu\left(\frac{n}{(n,k)}\right)\frac{\varphi(n)}{\varphi\left(\frac{n}{(n,k)}\right)}$$ where $(\cdot,\cdot):\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ is the gcd function. One can naturally extend the gcd to $\mathbb{Q}$ by allowing the property $\forall x,y,\lambda\in\mathbb{Q}:(\lambda x,\lambda y)=|\lambda|(x,y)$. This extension also respects the property that $\nu_p((x,y))=\min\{\nu_p(x),\nu_p(y)\}$. With this in mind, one can extend $c_{n}(x)$ to rational $x$ as above. For example, if $p$ is a prime, then $$c_n\left(\frac{1}{p}\right)=\mu\left(\frac{n}{(n,1/p)}\right)\frac{\varphi(n)}{\varphi\left(\frac{n}{(n,1/p)}\right)}=\mu\left(\frac{pn}{(pn,1)}\right)\frac{\varphi(n)}{\varphi\left(\frac{pn}{(pn,1)}\right)}$$ $$=\mu\left(pn\right)\frac{\varphi(n)}{\varphi\left(pn\right)}=\begin{cases}0 & \text{ if } p\ \vert\ n\\ -\frac{\mu(n)}{p-1} & \text{ if } p\ \not\vert\ n\end{cases}$$ It is easy to show that $\sum_{n=1}^\infty\frac{\mu(pn)}{n^s}=-\frac{p^s}{1-p^s}\frac{1}{\zeta(s)}$ which leads us to the following $$\sum_{n=1}^\infty\frac{c_n(1/p)}{n^s}= -\frac{1}{p-1}\sum_{n=1\\ p\not\vert n}^\infty\frac{\mu(n)}{n^s}=-\frac{1}{p-1}\left(\sum_{n=1}^\infty\frac{\mu(n)}{n^s}-\sum_{n=1}^\infty\frac{\mu(pn)}{p^sn^s}\right)$$ $$=-\frac{1}{p-1}\left(\frac{1}{\zeta(s)}+\frac{1}{1-p^s}\frac{1}{\zeta(s)}\right)=-\frac{1}{p-1}\left(1+\frac{1}{1-p^s}\right)\frac{1}{\zeta(s)}$$ But this doesn't work as it would lead to $$\sigma_s(p^{-1})=-\frac{1}{p-1}\left(1+\frac{1}{1-p^{1-s}}\right)\neq0$$ Moreover, if $\alpha\geq2$ is an integer, then $\mu\left(\frac{n}{(n,p^{-\alpha})}\right)=\mu(p^\alpha n)=0$ as $p^\alpha n$ is not square-free and, in turn, this would lead to $c_n(p^{-\alpha})=0\Rightarrow \sigma_s(p^{-\alpha})=0$. This is quite interesting but not what I wanted (nor what I expected). But maybe there's a better extension of the formula proven by Hölder that does the trick.