# Perron formula and sum of $\mu$

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.

• see this proof of the PNT (with error term) for L-functions from Iwaniec Kowalski analytic number theory's book. (in the case of $\zeta(s)$ there is no exceptional zero). It should work more or less the same with $\frac{1}{\zeta(s)}$ instead of $\frac{\zeta'(s)}{\zeta(s)}$. – reuns Nov 15 '16 at 21:15
• Errrr ... is there a more understandable way to do it. I suppose that the information I know should be enough since it is in the same section as $$\pi(x) = li(x) + O(xe^{-c\sqrt{\log x}}).$$ Moreover, I do not know the $L-$function, and the $\otimes$ (tensor product ?)symbol. – Both Htob Nov 16 '16 at 1:24
• Here $q=1, d=1, \Lambda_f(n) = \Lambda(n), L(s,f) = \zeta(s)$. On the right page, it shows how the error term in $\sum_{n < x} \Lambda(n) = x+ \mathcal{O}(x e^{-c \sqrt{\log x}})$ comes from $\int_Z |\hat{\phi}(s) \frac{\zeta'(s)}{\zeta(s)}\ |d|s|$ where $\frac{\zeta'(s)}{\zeta(s)} =\mathcal{O}(\log^2 |Im(s)|)$ and $Z$ is the boundary of the zero-free region and $\hat{\phi}(s)$ is the Mellin transform of a smoothed approximation of $1_{[0,x]}$, so nothing really complicated here ! – reuns Nov 16 '16 at 1:31
• See also dms.umontreal.ca/~koukoulo/documents/notes/primes.pdf page 69 for the same proof (in the special case of $\zeta'(s)/\zeta(s)$) with more details – reuns Nov 16 '16 at 1:35