= Solution
{c}
Write $s=\sigma+it$; the source's rectangle notation $|\operatorname{Im}(t)|\leq T$ is interpreted as the intended height restriction $|t|\leq T$. We use the standard facts that $\zeta$ has no zeros on $\sigma\geq1$, has a <simple pole> at one, and is meromorphic with no other pole. The function $(s-1)\zeta(s)$ is <holomorphic function> and nonzero on the <compact set> $\{1+it:|t|\leq T\}$, including its removable value one at $s=1$. It remains nonzero in a neighborhood of this <compact set>. Together with nonvanishing to its right this gives some $0<\alpha_T<1$ such that the whole closed rectangle $\alpha_T\leq\sigma\leq2$, $|t|\leq T$ contains no zeta zero. The pole at one is harmless: $1/\zeta$ extends holomorphically there with value zero.
The linearly smoothed Perron kernel, for $y>0$ and $c>0$, is
$$
\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}
\frac{y^{s+1}}{s(s+1)}\,ds
=\begin{cases}y-1,&y>1,\\0,&0<y\leq1.\end{cases}
$$
For $y>1$ shift the <contour> left across the <simple poles> at zero and minus one, whose <residues> are $y$ and $-1$. For $y<1$ shift right, crossing no pole. At $y=1$ the integral is zero, also obtained by direct integration of $1/s-1/(s+1)$. The quadratic denominator gives absolute integrability on the original vertical line.
For $c>1$ use the absolutely convergent <Dirichlet series> $1/\zeta(s)=\sum\mu(n)n^{-s}$. Interchanging the sum and the <contour integral> is justified by $\sum|\mu(n)|n^{-c}<\infty$ and the integrable denominator. The term for $n$ is $n$ times the kernel at $y=x/n$, hence $(x-n)$ when $n<x$ and zero otherwise. Consequently
$$
\boxed{M_1(x):=\sum_{n\leq x}\mu(n)(x-n)
=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}
\frac{x^{s+1}}{\zeta(s)s(s+1)}\,ds.}
$$
Here is a <contour> argument that needs only the stated bound, with no unproved uniform estimate left of one. Take $x>e$, $c=1+1/\log x$, and first hold $T\geq2$ fixed. On the two tails $|t|>T$, the given bound and $x^{c+1}=ex^2$ give
$$
\left|\text{tail contribution}\right|
\leq Cx^2\int_T^\infty\frac{\log^7(2+t)}{t^2}\,dt.
$$
Shift the remaining segment to $\sigma=\alpha_T$. The integrand is <holomorphic function> inside the rectangle: the zero of $1/\zeta$ at one cancels the apparent zeta pole, and $0,-1$ lie outside. On the fixed left side its integral is $O_T(x^{1+\alpha_T})=o_T(x^2)$. On the two horizontal sides, $1/(\zeta(s)s(s+1))$ is bounded by some finite constant depending on $T$. Therefore their integrals are bounded by
$$
C_T\int_{\alpha_T}^{c}x^{\sigma+1}\,d\sigma
\leq C_T\frac{x^{c+1}}{\log x}
=o_T(x^2).
$$
It follows that
$$
\limsup_{x\to\infty}\frac{|M_1(x)|}{x^2}
\leq C\int_T^\infty\frac{\log^7(2+t)}{t^2}\,dt.
$$
Now let $T\to\infty$; the integrable tail tends to zero. Thus
$$
\boxed{\sum_{n\leq x}\mu(n)(x-n)=o(x^2).}
$$
The order of limits matters: first $x\to\infty$ for each fixed rectangle, then $T\to\infty$. This is <linearly smoothed Möbius cancellation from finite zero-free rectangles>.
\Image[/past-exam-of-the-analytic-number-theory-and-elliptic-curves-course-of-the-university-of-oxford/2009-mobius-contour.png]
{title=Shift a finite central Perron <contour> while bounding the two infinite tails on the right}
Back to article page
