 Solution

Let $A(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ converge absolutely at a real $c>0$, and take $T\ge \max (2,c)$. DefineThe truncated Perron kernel estimate isFor $v>1$ close the contour to the left, collecting the residue one at zero; for $v<1$ close it to the right, collecting no residue. On the horizontal sides, integrating $v^{\sigma }$ bounds the error by $O(v^c/(T|\log v|))$, and the remote vertical side tends to zero. Near $v=1$ use the bounded transition estimate instead, giving $O(v^c)$. These are the contours and bounds underlying the kernel formula. The constants are uniform for $1\le c\le 2$, the range needed below.

Absolute convergence permits termwise integration. The truncated Perron formula is consequentlyHere the primed sum has half weight when $x$ is an integer, and $a_x=0$ otherwise. To obtain the inclusive sum add $a_x/2$ at an integer. This endpoint convention avoids a false uniform assertion about the kernel at $v=1$.

Apply this with $a_n=\Lambda (n)$, $A=-{\zeta }^{\prime }/\zeta$, $c=1+1/\log x$ and $2\le T\le x$, for sufficiently large $x$. For $n\notin [x/2,2x]$, $|\log (x/n)|$ is bounded below, andusing $-{\zeta }^{\prime }(c)/\zeta (c)\ll (c-1{)}^{-1}$. In the central range $(x/n{)}^c\ll 1$, $\Lambda (n)\ll \log x$, and $|\log (x/n)|\asymp |n-x|/x$. Separate the nearest integers, then sum the harmonic tail over distances $j\ge 1$: its contribution is $O((x/T){\log }^{2}x)$. The possible endpoint weight and nearest terms, of size $O(\log x)$, are absorbed because $T\le x$. ThereforeFor $x+y$, use the same vertical line $c=1+1/\log x$; the error estimate remains valid since $x\le x+y\le 2x$. Subtract the two formulas. The identityhas absolute value $O(y)$ on that line, because $(2x{)}^{c-1}$ is bounded. Taking absolute values givesThis is the short-interval Perron bound for the second Chebyshev function.