Perron Inversion and the Explicit Formula

If $0<y<1$, close the segment to the right by a semicircle and let its radius tend to infinity; $y^s$ decays there and no pole is enclosed, so the limit is $0$. If $y>1$, close to the left instead; the enclosed simple pole at $0$ has residue $1$, so the limit is $1$.

If $y=1$, the integrand is $1\mathrm{/}s$, and direct parametrisation gives $(2\pi {)}^{-1}{\int }_{-T}^{T}c(c^2+t^2{)}^{-1}dt\to 1\mathrm{/}2$. These three cases prove the claim.

For $0<y<1$, move the finite segment rightward and then let the new real part tend to infinity; its two horizontal tails have modulus at most $y^c\mathrm{/}(T\mathrm{\mid }\log y\mathrm{\mid })$. A circular arc gives the independent bound $y^c$. The leftward contours give the same two bounds for $y>1$, after subtracting the residue $1$.

For $y=1$, integration of $ds\mathrm{/}s$ on the two omitted tails gives $\mathrm{\mid }I(1,T)-1\mathrm{/}2\mathrm{\mid }<c\mathrm{/}T$. Taking the smaller of the two preceding bounds proves the stated estimate.

For $F_N(s)={\sum }_{n\le N}{a}_n{n}^{-s}$, linearity and the kernel formula give $(2\pi i{)}^{-1}\int F_N(s)x^{s}ds\mathrm{/}s={\sum }_{n\le N}{a}_n\delta (x\mathrm{/}n)$.

Let first the height and then $N$ tend to infinity. The assumed domination permits both interchanges, while the right side tends exactly to the half-weighted sum defining $A^{\ast }(x)$.

Absolute convergence permits termwise integration on the finite segment; subtracting the defining starred sum leaves ${\sum }_n{a}_n(I(x\mathrm{/}n,T)-\delta (x\mathrm{/}n))$.

Apply the two branches of the truncated-kernel estimate termwise and the triangle inequality. The branch $n=x$ is precisely the displayed separate endpoint term.

The Hadamard product The Riemann xi function has its genus-one Hadamard product over the nontrivial zeros of zeta gives ${\xi }^{\mathrm{\prime }}\mathrm{/}\xi (s)=B+{\sum }_{\rho }(1\mathrm{/}(s-\rho )+1\mathrm{/}\rho )$. Write $\rho =\beta +i\gamma$, with $0<\beta <1$ by The only zeros of zeta on the nonpositive real axis are the negative even integers, and every other zero lies in the open critical strip. Since ${\sum }_{\rho }\mathrm{\mid }\rho {\mathrm{\mid }}^{-2}<\mathrm{\infty }$, the sum of $\mathrm{\Re }(1\mathrm{/}\rho )=\beta \mathrm{/}\mathrm{\mid }\rho {\mathrm{\mid }}^2$ converges. At $s_0=2+iT$, the xi identity, Stirling's formula and The logarithmic derivative of the zeta Dirichlet series is the Dirichlet series of the von Mangoldt function on Re s greater than 1 give ${\xi }^{\mathrm{\prime }}\mathrm{/}\xi (s_0)=O(\log T)$: the zeta term is bounded by ${\sum }_{n\ge 2}(\log n)n^{-2}$, and the Gamma term is $O(\log T)$. Differentiating Stirling here is justified by Cauchy's estimate for its analytic remainder on disks of radius proportional to $\mathrm{\mid }{s}_0\mathrm{\mid }$ in a larger sector. Taking real parts of the product formula, all variable summands are positive and ${\sum }_{\rho }\frac{1}{4+(T-\gamma {)}^2}\le {\sum }_{\rho }\frac{2-\beta }{(2-\beta {)}^2+(T-\gamma {)}^2}=O(\log T)\mathrm{.}$ In particular there are $O(\log T)$ zeros with $\mathrm{\mid }T-\gamma \mathrm{\mid }<1$, without using the present theorem or its unit-interval corollary.

On the horizontal segment $s=\sigma +iT$, $1\mathrm{/}2\le \sigma \le 2$, subtract the product formula at $s_0$. For $\mathrm{\mid }T-\gamma \mathrm{\mid }\ge 1$, $\mid \frac{1}{s-\rho }-\frac{1}{s_0-\rho }\mid \le \frac{3\mathrm{/}2}{(T-\gamma {)}^2}\le \frac{15\mathrm{/}2}{4+(T-\gamma {)}^2}\mathrm{.}$ Thus step 1.1 bounds the nonlocal difference sum by $O(\log T)$, uniformly in $\sigma$. The local subtracted terms $1\mathrm{/}(s_0-\rho )$ also total $O(\log T)$. The integral of the imaginary part of each remaining local term $1\mathrm{/}(\sigma +iT-\rho )$ is the argument change of a horizontal segment missing zero, of absolute value at most $\pi$. Their number is $O(\log T)$. Including the reference value ${\xi }^{\mathrm{\prime }}\mathrm{/}\xi (s_0)$, the total argument change of $\xi$ on the top segment from $2+iT$ to $1\mathrm{/}2+iT$ is therefore $O(\log T)$.

Fix a height $t_0\in (0,2)$ not equal to any zero ordinate. On the vertical segment from $2+i{t}_0$ to $2+iT$, the factors $s$ and $s-1$ in $\xi (s)=\frac{1}{2}s(s-1){\pi }^{-s\mathrm{/}2}\mathrm{\Gamma }(s\mathrm{/}2)\zeta (s)$ have bounded argument changes. The zeta factor also has bounded argument change: $\mathrm{\mid }\zeta (2+it)-1\mathrm{\mid }\le {\sum }_{n\ge 2}{n}^{-2}<1$, so it remains in a fixed right half-plane. Stirling with a continuous logarithm in the right half-plane gives $\mathrm{\Im }\log \mathrm{\Gamma }(1+iT\mathrm{/}2)=\frac{T}{2}\log \frac{T}{2}-\frac{T}{2}+O(1)\mathrm{.}$ Consequently the argument change along this vertical segment is $\frac{T}{2}\log \frac{T}{2\pi }-\frac{T}{2}+O(1)\mathrm{.}$

By [L1], $\xi (s)=\xi (1-s)$, and conjugation symmetry gives $\xi (1-\bar{s})=\overline{\xi (s)}$. Apply the argument principle to the rectangle with real sides $-1,2$ and heights $t_0,T$. Its zeros are precisely the nontrivial zeta zeros in that height range. Reflection in $\mathrm{\Re }s=1\mathrm{/}2$ pairs the two vertical edges and the two halves of each horizontal edge, doubling the argument change on the right vertical edge followed by the right half of the top; the bottom contributes a constant independent of $T$. Thus $N(T)=\frac{1}{\pi }({\mathrm{\Delta }}_{2+i{t}_0\to 2+iT}\arg \xi +{\mathrm{\Delta }}_{2+iT\to 1\mathrm{/}2+iT}\arg \xi )+O(1)\mathrm{.}$ Steps 2.1 and 3.1 yield the claimed main term and $O(\log T)$ error.

If $T$ is a zero ordinate, take nonzero-ordinate heights decreasing to $T$. Discreteness of zeros in the bounded strip makes their counts eventually equal to the convention $0<\gamma \le T$, with full multiplicities. The preceding error constant is independent of the distance to zero ordinates, so passage to the limit preserves the estimate. Finally compact heights $2\le T\le T_0$ are covered by enlarging the constant.

For $T\ge 3$, the zeros with ordinates in $[T,T+1]$ are included in $N(T+2)-N(T-1)$. Subtract the Riemann--von Mangoldt formula at $T+2$ and $T-1$; its main term changes by $O(\log T)$ and its two errors have that size.

On the compact range $0\le T<3$, discreteness of the zeros gives a fixed finite bound, which is absorbed by enlarging the constant. The buffered count in step 1.1 already includes both endpoints, so the stated closed interval is covered.

Put $s_0=2+it$. Logarithmic differentiation of the Hadamard product and subtraction at $s_0$ give $\frac{{\xi }^{\mathrm{\prime }}}{\xi }(s)-\frac{{\xi }^{\mathrm{\prime }}}{\xi }(s_0)={\sum }_{\rho }(\frac{1}{s-\rho }-\frac{1}{s_0-\rho })\mathrm{.}$ The constants and genus-one correction terms cancel; the difference series converges absolutely, since its terms are $O_s(\mathrm{\mid }\rho {\mathrm{\mid }}^{-2})$ for large $\mathrm{\mid }\rho \mathrm{\mid }$.

For $\mathrm{\mid }t-\mathrm{\Im }\rho \mathrm{\mid }\ge 1$, each difference has absolute value at most $3\mathrm{/}\mathrm{\mid }t-\mathrm{\Im }\rho {\mathrm{\mid }}^2$. By [L1] and [L2], grouping into the bands $k\le \mathrm{\mid }t-\mathrm{\Im }\rho \mathrm{\mid }<k+1$ bounds their total by $C{\sum }_{k\ge 1}\frac{\log (\mathrm{\mid }t\mathrm{\mid }+k+3)}{k^2}=O(\log (\mathrm{\mid }t\mathrm{\mid }+2))\mathrm{.}$ Here $\log (\mathrm{\mid }t\mathrm{\mid }+k+3)\le \log (\mathrm{\mid }t\mathrm{\mid }+2)+\log (k+3)$, and both resulting weighted series converge. For the remaining zeros, $\mathrm{\mid }{s}_0-\rho \mathrm{\mid }\ge 1$, so the sum of their subtracted terms is also $O(\log (\mathrm{\mid }t\mathrm{\mid }+2))$ by [L1] and [L2]. Thus $\frac{{\xi }^{\mathrm{\prime }}}{\xi }(s)={\sum }_{\mathrm{\mid }t-\mathrm{\Im }\rho \mathrm{\mid }<1}\frac{1}{s-\rho }+\frac{{\xi }^{\mathrm{\prime }}}{\xi }(s_0)+O(\log (\mathrm{\mid }t\mathrm{\mid }+2))\mathrm{.}$

By The Riemann xi function $\xi (s)=\frac{1}{2}s(s-1)\mathrm{\Lambda }(s)$ and the Gamma recurrence, $\xi (s)=(s-1){\pi }^{-s\mathrm{/}2}\mathrm{\Gamma }(1+s\mathrm{/}2)\zeta (s)\mathrm{.}$ The Gamma factor has argument with real part at least $1\mathrm{/}2$. Stirling's formula for Gamma, differentiated using Cauchy's estimate on disks of radius proportional to the argument's modulus in a slightly larger sector, gives ${\mathrm{\Gamma }}^{\mathrm{\prime }}(z)\mathrm{/}\mathrm{\Gamma }(z)=\mathrm{Log}z+O(1\mathrm{/}\mathrm{\mid }z\mathrm{\mid })$ there for large $\mathrm{\mid }z\mathrm{\mid }$; compact subsets of this half-plane supply the remaining bound. Also The logarithmic derivative of the zeta Dirichlet series is the Dirichlet series of the von Mangoldt function on Re s greater than 1 gives $\mathrm{\mid }{\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta (2+it)\mathrm{\mid }\le {\sum }_{n\ge 2}(\log n)n^{-2}<\mathrm{\infty }$. Hence ${\xi }^{\mathrm{\prime }}\mathrm{/}\xi (s_0)=O(\log (\mathrm{\mid }t\mathrm{\mid }+2))$. Substituting the displayed xi identity into step 2.1 leaves the pole term $-1\mathrm{/}(s-1)$ and a Gamma logarithmic derivative of size $O(\log (\mathrm{\mid }t\mathrm{\mid }+2))$, proving the stated uniform formula even at bounded ordinates.

Take logarithmic derivatives of The Riemann zeta function satisfies the classical sine-gamma functional equation. The logarithmic derivative of the sine factor is $\frac{\pi }{2}\cot (\pi s\mathrm{/}2)$, which is uniformly bounded under the distance condition. Since $\mathrm{Re}(1-s)\ge 2$, the absolutely convergent series for $-{\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta$ in The logarithmic derivative of the zeta Dirichlet series is the Dirichlet series of the von Mangoldt function on Re s greater than 1 bounds ${\zeta }^{\mathrm{\prime }}(1-s)\mathrm{/}\zeta (1-s)$ uniformly.

Applying Stirling's formula for Gamma in the right half-plane to $1-s$ and differentiating its uniform analytic remainder by Cauchy's estimate on smaller sectorial disks gives ${\mathrm{\Gamma }}^{\mathrm{\prime }}(1-s)\mathrm{/}\mathrm{\Gamma }(1-s)=O(\log (\mathrm{\mid }s\mathrm{\mid }+2))$ for $\mathrm{Re}(1-s)\ge 2$. Combining the finitely many factor bounds proves the assertion.

A simple pole of zeta at $1$ makes $-{\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta$ have residue $1$; a zero $\rho$ of multiplicity $m$ makes it have residue $-m$. Multiplication by $x^s\mathrm{/}s$ gives the first two entries.

By [L1], the remaining zeros crossed on the nonpositive real axis occur at $-2k$. At $s=-2k$, the sine in [L2] has a simple zero, while all its other factors are finite and nonzero by [L3]; hence these zeros are simple. Their residues sum to ${\sum }_{k\ge 1}{x}^{-2k}\mathrm{/}(2k)=-\frac{1}{2}\log (1-x^{-2})$. Also by [L1], zeta is nonzero at $0$, so the pole of $1\mathrm{/}s$ gives the final entry.

Write $\mathrm{\Phi }=\tilde{\varphi }$. Integration by parts gives $\mathrm{\Phi }(s)=\frac{y^{s+1}-x^{s+1}}{(y-x)s(s+1)}\mathrm{.}$ Its only pole is $0$, with residue $1$; the apparent singularity at $-1$ is removable, with value $-\log (y\mathrm{/}x)\mathrm{/}(y-x)$. On each fixed vertical strip it is $O_{x,y}(\mathrm{\mid }\mathrm{\Im }s{\mathrm{\mid }}^{-2})$ at large height, with the constant also depending on the strip.

We compute the constant at zero. Put $\gamma ={\lim }_N(H_N-\log N)$ as in The Euler–Mascheroni constant and the harmonic asymptotic. The fractional-part formula in For $\mathrm{Re}s>0$, zeta admits the fractional-part integral formula with a simple residue-one pole at $1$ and ${\int }_1^N\frac{\{u\}}{u^2}du=\log N-H_N+1$ give $\zeta (1-s)=-1\mathrm{/}s+\gamma +O(s)$. Logarithmic differentiation of the locally uniform product in The Weierstrass product for reciprocal Gamma at $1$ gives $-\frac{{\mathrm{\Gamma }}^{\mathrm{\prime }}(1)}{\mathrm{\Gamma }(1)}=1+\gamma +{\sum }_{n\ge 1}(\frac{1}{n+1}-\frac{1}{n})=\gamma \mathrm{.}$ The same product at $1$ gives $\mathrm{\Gamma }(1)=1$, so $\mathrm{\Gamma }(1-s)=1+\gamma s+O(s^2)$. Finally The Riemann zeta function satisfies the classical sine-gamma functional equation and $2^s{\pi }^{s-1}\sin (\pi s\mathrm{/}2)=\frac{s}{2}(1+s\log (2\pi )+O(s^2))$ yield $\zeta (s)=-1\mathrm{/}2-s\log (2\pi )\mathrm{/}2+O(s^2)$: the two Euler constants cancel. Thus ${\zeta }^{\mathrm{\prime }}(0)\mathrm{/}\zeta (0)=\log (2\pi )$.

To justify inversion explicitly, let $J(z)=\frac{1}{2\pi i}{\int }_{\mathrm{\Re }s=2}\frac{z^s}{s(s+1)}ds(z>0)\mathrm{.}$ Closing a rectangle to the left for $z>1$, and to the right for $z<1$, gives $J(z)=1-z^{-1}$ and $J(z)=0$, respectively, by The residue theorem for a null-homologous cycle. Indeed, first let the height tend to infinity with the other vertical side fixed: the horizontal integrals are $O(T^{-2})$ times a fixed width. Then let that side tend to the appropriate infinity through half-integers; its integral is $O(z^{\sigma }\mathrm{/}\mathrm{\mid }\sigma \mathrm{\mid })$ and vanishes. At $z=1$, continuity of the absolutely convergent initial integral gives $J(1)=0$. Consequently $[yJ(y\mathrm{/}u)-xJ(x\mathrm{/}u)]\mathrm{/}(y-x)=\varphi (u)$ for every $u>0$, including $u=x,y$. Combining this with step 1.1 and The logarithmic derivative of the zeta Dirichlet series is the Dirichlet series of the von Mangoldt function on Re s greater than 1 gives ${\sum }_n\mathrm{\Lambda }(n)\varphi (n)=\frac{1}{2\pi i}{\int }_{\mathrm{\Re }s=2}-\frac{{\zeta }^{\mathrm{\prime }}}{\zeta }(s)\mathrm{\Phi }(s)ds\mathrm{.}$ The exchange of sum and integral is absolute, since ${\sum }_n\mathrm{\Lambda }(n)n^{-2}\le {\sum }_{n\ge 2}(\log n)n^{-2}<\mathrm{\infty }$ and $\mathrm{\Phi }(2+it)=O_{x,y}((1+\mathrm{\mid }t\mathrm{\mid }{)}^{-2})$.

By [L1], [L2], and $0<\mathrm{\Re }\rho <1$, step 1.1 gives absolute convergence of the zero sum: its bands at large $\mathrm{\mid }\mathrm{\Im }\rho \mathrm{\mid }\asymp j$ contribute $O_{x,y}(\log (j+2)\mathrm{/}{j}^2)$. The trivial-zero sum converges absolutely since $\mathrm{\mid }\mathrm{\Phi }(-2k)\mathrm{\mid }\le C_{x,y}{x}^{1-2k}\mathrm{/}{k}^2$. We also choose admissible heights explicitly. Let $M_j$ count zeros with ordinates in $[j-1,j+2]$, with multiplicity. It is $O(\log (j+2))$. Among the $2M_j+2$ equally spaced points of $[j,j+1]$, each such ordinate excludes at most one point at distance less than $1\mathrm{/}(4M_j+4)$. Choose the least remaining point $T_j$. Ordinates outside that larger interval are at distance at least $1$, so every zero ordinate is at distance at least $1\mathrm{/}(4M_j+4)$ from $T_j$. Conjugation gives the same separation at $-T_j$.

Fix an odd integer $R\ge 3$ and shift the integral of step 2.1 to $\mathrm{\Re }s=-R$, using heights $\pm T_j$ from step 2.2. On $-1\le \mathrm{\Re }s\le 2$, A local formula for the logarithmic derivative of zeta bounds ${\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta$ by $O({\log }^2{T}_j)$: there are $O(\log T_j)$ nearby zeros, each reciprocal is $O(\log T_j)$, and the pole term is bounded. On $-R\le \mathrm{\Re }s\le -1$, A left-half-plane bound for the logarithmic derivative of zeta gives $O_R(\log T_j)$. Thus each horizontal integral is $O_{R,x,y}({\log }^2{T}_j\mathrm{/}{T}_j^2)$ and tends to zero. By Residues in the von Mangoldt contour shift, $-{\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta$ has residues $1$ at $1$, $-m$ at a zero of multiplicity $m$, and $-1$ at each trivial zero. Multiplication by $\mathrm{\Phi }$ therefore gives residues $\mathrm{\Phi }(1)$, $-m\mathrm{\Phi }(\rho )$, $-\mathrm{\Phi }(-2k)$, and, by step 1.2, $-\log (2\pi )$ at $0$. There is no pole at $-1$. The residue theorem and absolute convergence in step 2.2 now express the initial integral as these residues for $-2k>-R$, the full nontrivial-zero sum, and the upward integral on $\mathrm{\Re }s=-R$.

On that last line, the distance to every trivial zero is at least $1$. Step 1.1 gives $\mathrm{\mid }\mathrm{\Phi }(-R+it)\mathrm{\mid }\le C_{x,y}\frac{x^{1-R}}{R^2+t^2}\mathrm{.}$ The left-half-plane bound therefore makes its integral at most $C_{x,y}{x}^{1-R}{\int }_{\mathbb{R}}\frac{\log (R+\mathrm{\mid }t\mathrm{\mid }+2)}{R^2+t^2}dt\le C_{x,y}^{\mathrm{\prime }}{x}^{1-R}\frac{\log (R+2)}{R}⟶0.$ Letting odd $R$ tend to infinity in step 3.1, using $x>1$ and the absolute convergence from step 2.2, proves the formula for every stated pair $1<x<y$.

Put $c=1+1\mathrm{/}\log (2x)$, so $1<c<2$ and $x^c=O(x)$. On $\mathrm{\Re }s=c$, The logarithmic derivative of the zeta Dirichlet series is the Dirichlet series of the von Mangoldt function on Re s greater than 1 gives $-{\zeta }^{\mathrm{\prime }}(s)\mathrm{/}\zeta (s)={\sum }_{n\ge 1}\mathrm{\Lambda }(n)n^{-s}$ absolutely. Apply A truncated Perron formula at height $H\ge 2$ to this series. Its starred sum is ${\psi }_0(x)$ by The half-weighted Chebyshev function, and its error is at most ${\sum }_{n\mathrm{\ne }x}\mathrm{\Lambda }(n)(x\mathrm{/}n{)}^c\min \{1,(H\mathrm{\mid }\log (x\mathrm{/}n)\mathrm{\mid }{)}^{-1}\}+O(\mathrm{\Lambda }(x)\mathrm{/}H)\mathrm{.}$ Here $\mathrm{\Lambda }(x)=0$ if $x$ is not an integer prime power.

The unit-interval zero count A unit-interval bound for zeta zeros bounds the number of zero ordinates in $[T-1,T+2]$ by $C\log (T+2)$. Removing intervals of radius $1\mathrm{/}(8C\log (T+2))$ about those ordinates leaves a point $H\in [T,T+1]$ at that distance or more from every zero ordinate; ordinates outside $[T-1,T+2]$ are farther away. Use this $H$ and its reflection $-H$ as the horizontal contour heights. By A local formula for the logarithmic derivative of zeta, for $-1\le \sigma \le c$ one has ${\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta (\sigma \pm iH)=O({\log }^2(H+2))$: at most $O(\log H)$ zeros enter the local sum and each denominator has modulus $\gg 1\mathrm{/}\log H$. The pole term is harmless since $H\ge 2$.

For $n\mathrm{\notin }(3x\mathrm{/}4,5x\mathrm{/}4)$, $\mathrm{\mid }\log (x\mathrm{/}n)\mathrm{\mid }$ is bounded below by an absolute constant. Since $\mathrm{\Lambda }(n)\le \log n$ and $c-1=1\mathrm{/}\log (2x)$, comparison with ${\int }_1^{\mathrm{\infty }}(\log u)u^{-c}du=(c-1{)}^{-2}$ gives ${\sum }_n\mathrm{\Lambda }(n)n^{-c}=O({\log }^2(2x))$. Thus these far terms contribute $O(x{\log }^2(2x)\mathrm{/}H)$.

For odd positive integers $U$, shift the Perron segment from $c\pm iH$ to $-U\pm iH$. The horizontal parts with $-1\le \sigma \le c$ contribute $O\text{\hspace{0.17em}⁣}\left(\frac{{\log }^2(H+2)}{H}{\int }_{-1}^c{x}^{\sigma }d\sigma \right)=O\text{\hspace{0.17em}⁣}\left(\frac{x{\log }^2(H+2)}{H\log x}\right)\mathrm{.}$ For $\sigma \le -1$, the line is at distance at least $H$ from every negative even integer, so A left-half-plane bound for the logarithmic derivative of zeta bounds the logarithmic derivative by $O(\log (\mathrm{\mid }\sigma \mathrm{\mid }+H+2))$. These horizontal tails are $O((1+\log (H+2))\mathrm{/}(Hx\log x))$ after integrating $x^{\sigma }$ over $(-\mathrm{\infty },-1]$. Along $\mathrm{\Re }s=-U$, the same lemma with distance at least $1$ from the negative even integers gives ${\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta =O(\log (U+H+2))$; the vertical integral is $O(H\log (U+H+2)\mathrm{/}(U{x}^U))$ and tends to zero as $U\to \mathrm{\infty }$. All three bounds are absorbed by $O(x{\log }^2(xH)\mathrm{/}H)$.

In the middle interval, $(x\mathrm{/}n{)}^c=O(1)$ and $\mathrm{\mid }\log (x\mathrm{/}n)\mathrm{\mid }\gg \mathrm{\mid }n-x\mathrm{\mid }\mathrm{/}x$. On each side of $x$, single out the nearest prime power $q\mathrm{\ne }x$, if one lies in that middle interval. Its contribution is $O((\log x)\min \{1,x\mathrm{/}(H\mathrm{\mid }x-q\mathrm{\mid })\})$, at most the term involving $⟨x⟩$ in the statement. For all other integers on that side, put $m=\mathrm{\mid }n-q\mathrm{\mid }\ge 1$; because $q$ is the nearest contributing prime power, $\mathrm{\mid }n-x\mathrm{\mid }\ge m$, so their total is at most $O\text{\hspace{0.17em}⁣}\left(\frac{x\log (2x)}{H}{\sum }_{1\le m\le 2x}\frac{1}{m}\right)=O\text{\hspace{0.17em}⁣}\left(\frac{x{\log }^2(2x)}{H}\right)\mathrm{.}$ The endpoint $n=x$ in step 1.1 is $O(\log (2x)\mathrm{/}H)$. Consequently the Perron integral $P_H(x)$ satisfies $P_H(x)={\psi }_0(x)+O\text{\hspace{0.17em}⁣}(\frac{x{\log }^2(2x)}{H}+(\log x)\min \{1,\frac{x}{H⟨x⟩}\})\mathrm{.}$

The residue ledger Residues in the von Mangoldt contour shift now gives, after $U\to \mathrm{\infty }$, $P_H(x)=x-{\sum }_{\mathrm{\mid }\mathrm{\Im }\rho \mathrm{\mid }<H}\frac{x^{\rho }}{\rho }-\frac{{\zeta }^{\mathrm{\prime }}(0)}{\zeta (0)}-\frac{1}{2}\log (1-x^{-2})+O\text{\hspace{0.17em}⁣}\left(\frac{x{\log }^2(xH)}{H}\right)\mathrm{.}$ Its negative-even residue series converges absolutely since $x>1$. Combining this with step 3.1 proves the formula at height $H$.

The only zeros gained between the strict cutoffs $T$ and $H$ have $T\le \mathrm{\mid }\mathrm{\Im }\rho \mathrm{\mid }<H\le T+1$. There are $O(\log (T+2))$ of them, with multiplicity, and $0<\mathrm{\Re }\rho <1$ by The only zeros of zeta on the nonpositive real axis are the negative even integers, and every other zero lies in the open critical strip, so each contributes at most $x\mathrm{/}T$. Their total $O(x\log (T+2)\mathrm{/}T)$ is absorbed by $O(x{\log }^2(xT)\mathrm{/}T)$. Since $T\le H\le T+1$, the two errors in step 3.1 at height $H$ are bounded by the displayed errors at height $T$ up to absolute constants. This proves the statement, including when $T$ itself is a zero ordinate.