Let $F(x) = \sum_{n \leq x} \mu(n)$. I want to show that $$F(x) = O(xe^{-c\sqrt{\log x}})$$ where $c$ is a positive constant.
I see some example of $\pi(x) = li(x) + O(xe^{-c\sqrt{\log x}})$ where $li(x) = \int_1^x \frac{dt}{\log t}$ using Perron formula
Perron: Let $G(s) = \sum_{n = 1}^\infty g(n)n^{-s}, K(x) = \sum_{n \leq x} g(x).$ Define
$$K^*(x) = K(x) (x \not\in \mathbb{N}), K^*(x)= K(x) - 1/2 (x \in \mathbb{N}).$$ Then $$\big|K^*(x) - \frac{1}{2\pi i}\int_{c - iT}^{c + iT}G(s)x^ss^{-1}ds \big| \leq x^c\sum_{n=1}^\infty \frac{|g(n)|}{n^c} \min\big(1, (T |\log(x/n)|)^{-1}\big) + \frac{c}{T\pi}|g(x)|^*.$$
Sol. Let $x \in \mathbb{R} \setminus \mathbb{N}.$ Set $f(n) = \mu(n).$ Use Perron $$|F(x) - \frac{1}{2\pi i}\int_{c - iT}^{c + iT}(\sum_{n=1}^\infty \mu(n)n^{-s}) x^ss^{-1}ds| \leq x^c\sum_{n=1}^\infty \frac{1}{n^c} \min\big(1, (T |\log(x/n)|)^{-1}\big) + \frac{c}{T\pi}.$$
In case of $\pi(x)$, a Theorem concerning zero free region of $\zeta$ is used
Theorem : For some $\alpha > 0$, $$\zeta(\sigma + it) \neq 0$$ for $$\sigma \geq 1 - \frac{\alpha}{\log(|t|+2)}.$$ I also have a result: Suppose that $\zeta(s)$ is in zero free region, then $$|\zeta(s)|^{-1} = O_\alpha(\log (|t| + 2))$$ for $\sigma \geq 1 - \frac{\alpha/2}{\log(|t| + 2)}.$
Since $\sum \frac{\mu(n)}{n} = \frac{1}{\zeta}$, I think I might need to us this two theorem.
Any help please ?