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The classical zero-free region of the Riemann zeta function and a truncated Perron formula givefor some absolute constant $c>0$. Take contour height $T=\exp (\sqrt{\log x})$ and move the Perron contour to $\Re s=1-c_1/\log T$.

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.

The truncated Perron formula says that if $F(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ converges absolutely for $\Re s>{\sigma }_a$, then for $c>{\sigma }_a$, $T\ge 2$, and $x$ not an integer,Take $a_n=\Lambda (n)$ and $c=1+1/\log x$. The logarithmic derivative identity gives $F(s)=-{\zeta }^{\prime }(s)/\zeta (s)$. Since $x-1/2$ is an integer, $|x-n|\ge 1/2$ for every integer $n$. In the range $x/2<n<2x$,and hence the contribution there isThe ranges $n\le x/2$ and $n\ge 2x$ are bounded by the same quantity using absolute convergence and $-{\zeta }^{\prime }(c)/\zeta (c)\ll \log x$. Therefore

For $c=1+1/\log x$, $1\le y\le x$ and $2\le T\le x$, apply the truncated Perron formula with Von Mangoldt function coefficients at both endpoints on the same vertical line. The difference kernel is $((x+y{)}^s-x^s)/s={\int }_x^{x+y}{u}^{s-1}du$, of modulus $O(y)$. The usual near-integer error bound contributes $O(x{\log }^{2}x/T)$. The logarithmic derivative on this line is finite because of the Euler product.

For $2\le T\le x$, the truncated Perron formula and a contour shift give the displayed form of the Riemann–von Mangoldt explicit formula. The pole at one contributes $x$ and the Nontrivial zeros of the Riemann zeta function contribute the finite sum. One can choose a separated height in $[T,T+1]$ and then restore $T$ using the local zero count for the Riemann zeta function. The error absorbs the half-weight discrepancy at a prime power. Outside this range of $T$, one needs the fuller error with the near-integer terms retained.

For $v\ne 1$, $K_T(v)$ differs from $1_{v>1}$ by $O(v^c\min (1,(T|\log v|{)}^{-1}))$. Close the contour to the left for $v>1$ and to the right for $v<1$; horizontal sides give the logarithmic denominator and the nearby transition is bounded. At $v=1$, direct integration gives $K_T(1)={\pi }^{-1}\arctan (T/c)=1/2+O(c/T)$. This explains the half-weight endpoint in the truncated Perron formula. The constants may be taken uniform for $1\le c\le 2$ and $T\ge 2$.