Let $\left \lfloor \cdot \right \rfloor$ be the floor function. Is there a way to express the function $A(x)$ given by : $$A(x)=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{2}} \left \lfloor x^{1/n}-1 \right \rfloor\;\;\;\;\; (x \geqslant 2)$$
in terms of the nontrivial zeros of the Riemann zeta function?
The motivation behind the question is that there is a somewhat similar situation with [the second Chebyshev function $\psi(x)$][1], where we have the formula: $$\log\left(\left \lfloor x \right \rfloor !\right )=\sum_{n=1}^{\infty}\psi\left(\frac{x}{n} \right )$$ and by Möbius inversion, we have : $$\psi(x)=\sum_{n=1}^{\infty}\mu(n)\log\left(\left \lfloor \frac{x}{n} \right \rfloor !\right )=x-\sum_{\rho}\frac{x^{\rho}}{\rho}-\log(2\pi)-\frac{1}{2}\log(1-x^{-2})$$ So, on the LHS we have a summation in terms of a trivial number-theoretic function - $\log\left \lfloor x \right \rfloor !$ - that terminates at a finite $n$, and on the RHS we have a Fourier series in terms of the nontrivial zeros of zeta.
My attempt:
We won't deal with the smooth part of $A(x)$, as it's quite easy to obtain. The oscillatory part is given by: $$f(x)=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{2}}\left \{ x^{1/n} \right \}$$ For convenience, we replace f(x) by the function $g(x)$: $$g(x)=f\left(e^{x}\right)-\frac{3}{\pi^{2}}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{2}}p\left(e^{x/n}\right)$$ where $p(\cdot)$ is the sawtooth function, which has the Fourier expansion : $$p\left(e^{x/n}\right)=-\sum_{m=1}^{\infty}\frac{1}{2\pi i m}\left(\exp(2\pi i me^{x/n} )-\exp(-2\pi i me^{x/n} )\right)$$$$=-\sum_{m=1}^{\infty}\frac{1}{2\pi i m}\left(\exp\left(2\pi i m\left(e^{x/n}-1\right)\right) -\exp\left(-2\pi i m\left(e^{x/n}-1\right)\right) \right)$$ Assuming we can reverse the order of the summation - i have no proof! - we have that : $$g(x)=\sum_{m=1}^{\infty}\frac{\phi(m,x)}{m}$$ where we define : $$\phi(m,x)=-\frac{1}{2\pi i }\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{2}}\left(\exp\left(2\pi i m\left(e^{x/n}-1\right)\right) -\exp\left(-2\pi i m\left(e^{x/n}-1\right)\right) \right)$$
Using the generating function of the [Touchard polynomials][2] $T_{k}(\cdot)$, we have : $$\frac{1}{2\pi i}\left(\exp\left(2\pi i m\left(e^{x/n}-1\right)\right) -\exp\left(-2\pi i m\left(e^{x/n}-1\right)\right) \right)=\frac{1}{\pi}\sum_{k=0}^{\infty}\frac{\tilde{T}_{k}(2\pi m)}{k!}\left(\frac{x}{n}\right)^{k}$$ Where we define : $$\tilde{T}_{k}(2\pi m)=\frac{1}{2i}\left(T_{k}(2\pi i m)-T_{k}(-2\pi i m)\right)=\sum_{l=0}^{\infty}\frac{(2\pi m)^{l}}{l!}l^{k}\sin\left(\frac{\pi l}{2}\right)$$ Thus : $$\phi(m,x)=-\frac{1}{\pi}\sum_{k=0}^{\infty}\frac{\tilde{T}_{k}(2\pi m)}{\zeta(k+2)k!}x^{k}=-\frac{1}{\pi}\sum_{k=0}^{\infty}\frac{(-1)^{k}\tilde{T}_{k}(2\pi m)}{\zeta(k+2)k!}(-x)^{k}$$ Making use of Ramanujan's master theorem, we have for some vertical strip in the complex $s$ plane : $$\int_{0}^{\infty}\phi(x,m)x^{s-1}dx=-\frac{1}{\pi}\frac{\Gamma(s)G_{m}(s)}{\zeta(2-s)}$$ where we define the Dirichlet-Taylor series : $$G_{m}(s)=(-1)^{-s}\sum_{l=1}^{\infty}\frac{(2\pi m)^{l}}{l^{s}l!}\sin\left(\frac{\pi l }{2}\right)$$ Now, by Millin inversion theorem, we have : $$\phi(m,x)=-\frac{1}{2\pi^{2} i }\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{\Gamma(s)G_{m}(s)}{\zeta(2-s)}x^{-s}ds$$ The function $G_{m}(s)$ is clearly holomorphic. and if one shifts the path of integration properly, the contributions to the resulting series come only from the nontrivial zeros of $\zeta(2-s)$
Is this line of thought legit? [1]: https://mathworld.wolfram.com/MangoldtFunction.html [2]: https://en.wikipedia.org/wiki/Touchard_polynomials