Define $\exp_x : \mathbb{N} \rightarrow \mathbb{C}$ by $\exp_x(d) = e^{ixd}$ for all $d \in \mathbb{N}$ and some $x \in \mathbb{R}$. I want to evaluate the Dirichlet convolution of the Mobius function $\mu$ with $\exp_x$, $$ f(x,n) := (\exp_x * \mu)(n) = \sum_{d | n} \mu\left(d\right) e^{ixn/d}. $$ Since $\exp_x$ is not multiplicative, it's not obvious to me what to do. Is there a trick for computing a sum like this?
Of course, an obvious thing to try is to Taylor expand $f$ about $x = 0$, which gives $$ f(x, n) = \left\lfloor \frac{1}{n} \right\rfloor + \sum_{k \geq 1}\frac{(ixn)^k}{k!}\sum_{d|n} \mu\left(\frac{n}{d}\right)d^{-k}. $$ Now the divisor sum is multiplicative, so it's not too hard to show that $$ \sum_{d|n} \mu\left(\frac{n}{d}\right)d^{-k} = \frac{1}{n^k}\prod_{p | n} \left(1-p^k\right) = \frac{(-1)^{\omega(n)} \text{rad}^k(n)J_k(n)}{n^{2k}}, $$ where $J_k(n)$ is the Jordan totient function, $\omega(n)$ is the number of distinct prime divisors $p$ of $n$, and $\text{rad}(n)$ is the radical of the integer $n$. Thus, $$ f(x, n) = \left\lfloor \frac{1}{n} \right\rfloor + \sum_{k \geq 1}\frac{(ixn)^k}{k!}\frac{(-1)^{\omega(n)} \text{rad}^k(n)J_k(n)}{n^{2k}}, $$ but I have no clue where to go from here (assuming a closed form even exists).