The truncated <Perron formula> says that, for $c>1$ and $T\geq2$,
$$
\sum_{n\leq x}a_n
=\frac1{2\pi i}\int_{c-iT}^{c+iT}
\left(\sum_{n=1}^\infty\frac{a_n}{n^s}\right)\frac{x^s}{s}\,ds
+\text{a truncation error},
$$
where an endpoint has half weight and the error is controlled by
$$
\sum_n|a_n|\left(\frac xn\right)^c
\min\left(1,\frac1{T|\log(x/n)|}\right).
$$

Apply this with $a_n=\mu(n)$, $c=1+1/\log x$, and
$$
T=\exp(\sqrt{\log x}).
$$
Part (b) makes the integrand $x^s/(s\zeta(s))$. Move the contour to
$$
\sigma_0=1-\frac{c_2}{\log T}
=1-\frac{c_2}{\sqrt{\log x}},
$$
using a contour that stays inside the <Zero-free region of the Riemann zeta function> near small $|t|$. The estimates supplied in the question give $1/\zeta(s)\ll\log(|t|+3)$ on the new contour. Its vertical segment is therefore
$$
\ll x^{\sigma_0}(\log T)^2
\ll x\exp(-c_3\sqrt{\log x}),
$$
after decreasing $c_3$. The horizontal segments and the Perron truncation error are
$$
\ll\frac{x(\log x)^{O(1)}}T
\ll x\exp(-c_4\sqrt{\log x}).
$$
No residue is crossed because $1/\zeta(s)$ has a zero, rather than a pole, at $s=1$. Thus the <Mertens function> satisfies
$$
\boxed{\sum_{n\leq x}\mu(n)
\ll x\exp(-c\sqrt{\log x})}
$$
for some absolute $c>0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 3 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-150/3/d/solution