The truncated von Mangoldt explicit formula

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.