 Solution

Write $s=\sigma +it$; the source's rectangle notation $|\mathrm{Im}(t)|\le T$ is interpreted as the intended height restriction $|t|\le T$. We use the standard facts that $\zeta$ has no zeros on $\sigma \ge 1$, 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 $\{1+it:|t|\le T\}$, 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 ${\alpha }_T\le \sigma \le 2$, $|t|\le T$ 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 $c>0$, isFor $y>1$ 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

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 $T\ge 2$ fixed. On the two tails $|t|>T$, the given bound and $x^{c+1}=e{x}^2$ giveShift 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 byIt follows thatNow let $T\to \infty$; the integrable tail tends to zero. ThusThe 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.