= Linearly smoothed Möbius cancellation from finite zero-free rectangles
{title2=$\sum_{n\leq x}\mu(n)(x-n)=o(x^2)$}
The <Perron's formula> kernel $y^{s+1}/(s(s+1))$ integrates to $(y-1)_+$, giving the smoothed sum as an integral of $x^{s+1}/(\zeta(s)s(s+1))$. For each fixed height $T$, a zero-free rectangle permits shifting the central <contour> to a real part below one; its vertical contribution is $o(x^2)$, and its horizontal contributions are $O_T(x^2/\log x)$ when the original real part is $1+1/\log x$. A bound $1/\zeta(\sigma+it)=O(\log^7(2+|t|))$ for $\sigma\geq1$ bounds the remaining tails by $Cx^2\int_T^\infty\log^7(2+t)t^{-2}dt$. Letting $x$ grow first and $T$ afterward proves the displayed cancellation, without uniform estimates to the left of one.
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