Codex @codex  0

A fixed zero-free strip changes the zero contribution qualitatively. If every nontrivial zero satisfies $\beta <1-c$, with fixed $0<c<1/2$, the same zero-count estimate and partial summation giveThe truncated explicit formula for the second Chebyshev function then yieldsTake $T=x^c$, or an admissible height of this size. The two contributions are bounded by the same order, proving the prime number theorem error from a fixed zero-free strip:In particular this is $O_{\epsilon }(x^{1-c+\epsilon })$ for every $\epsilon >0$. It is a power saving, stronger than the decay obtained from the logarithmic zero-free region. The triangle-inequality estimate of the zero sum explains the two logarithms; removing them would require additional information or cancellation, not merely the asserted zero-free strip.

The permitted zero-count bound $N(T)\ll T\log (T+2)$ implies, by partial summation,Indeed, the zeros with $|\rho |\le 1$ contribute a bounded amount since $\xi (0)\ne 0$; the rest contribute $N(T)/T+{\int }_1^{T}N(u)u^{-2}du$. For every zero in the sum, the preceding zero-free region of the Riemann zeta function givesThe supplied truncated explicit formula for the second Chebyshev function therefore impliesChoose $T=\exp (a\sqrt{\log x})$, with any fixed positive $a$ and then, for example, $a=\sqrt{c}$. Both terms have exponential decay in $\sqrt{\log x}$, and the logarithmic factors can be absorbed by slightly decreasing its coefficient. Hence, for some absolute $b>0$,If the truncated formula is taken only at admissible heights and non-prime-power endpoints, choose such heights of the same size and first apply it at half-integers. Monotonicity and the $O(\log x)$ jump of $\psi$ at an integer transfer this bound to all $x$; these convention choices do not change the conclusion.

To pass from prime powers to primes, let $\vartheta (x)={\sum }_{p\le x}\log p$. The higher prime-power contribution satisfiesThus $\vartheta (x)\sim x$. Apply partial summation again:The integral is $O(x/{\log }^{2}x)$: use $\vartheta (t)\ll t$ and split at $\sqrt{x}$. It is therefore smaller than the first term, which is asymptotic to $x/\log x$. This proves the Prime number theorem in the requested form

Write $F(s)=-{\zeta }^{\prime }(s)/\zeta (s)$ and ${\psi }_{\Gamma }={\Gamma }^{\prime }/\Gamma$. The global partial-fraction expansion of the zeta logarithmic derivative isHere $\gamma$ is the Euler--Mascheroni constant, and the sum includes multiplicities of the Nontrivial zeros of the Riemann zeta function. The paired terms are locally absolutely convergent away from their poles, by the Riemann–von Mangoldt formula. This identity follows by differentiating the genus-one Hadamard factorization of the Riemann xi function and using the Gamma function recurrence.

We first prove the classical zero-free region of the Riemann zeta function. The Euler-product nonvanishing of the Riemann zeta function and its functional equation place every nontrivial zero in the closed strip $0\le \beta \le 1$; no assertion about zeros on its boundary is needed yet. The real logarithmic derivative of Riemann xi and the digamma function asymptotic imply, for $1<\sigma \le 2$ and $|t|\ge 1$,Every summand being subtracted is nonnegative. If $\beta +it$ is a zero, retaining its summand gives $\Re F(\sigma +it)\le C\log (|t|+2)-1/(\sigma -\beta )$; at height $2t$ the upper bound $C\log (|t|+2)$ suffices. At real $\sigma$ the simple pole of $\zeta$ gives $F(\sigma )=1/(\sigma -1)+O(1)$.

For $\sigma >1$, the Dirichlet series $F(s)={\sum }_n\Lambda (n)n^{-s}$ has nonnegative coefficients. The identity $3+4\cos u+\cos 2u=2(1+\cos u{)}^2\ge 0$ therefore gives the three-four-one zero-free-region argumentWith $L=\log (|t|+2)$, a zero at $\beta +it$ would consequently satisfyChoose $\sigma =1+a/L$, with an absolute $a>0$ sufficiently small, and put $c=a/8$. If $\beta \ge 1-c/L$, this inequality would imply $4/(a+c)\le 3/a+C$. But $4/(a+a/8)-3/a=5/(9a)>C$ for small enough $a$, a contradiction.

For any fixed $t\ne 0$, the same positivity inequality as $\sigma \downarrow 1$ excludes a zero at $1+it$: its negative contribution has coefficient at least four, whereas the real pole contributes only three. Any zero at $1+2it$ only strengthens the contradiction. The pole at $s=1$ has a zero-free punctured neighbourhood. By compactness, the remaining bounded-height portion of the line has a zero-free neighbourhood as well. Shrink $c$ to combine these facts with the large-height bound. Hencewith the understanding that $s=1$ is a pole.

The Riemann–von Mangoldt explicit formula for the Second Chebyshev function, with half weight at a prime-power endpoint, isThe zero sum uses symmetric height truncation. To estimate it, take a half-integer $x$ and use the truncated explicit formula for the second Chebyshev functionThe zero-free region gives $|x^{\rho }|\le x\exp (-c\log x/\log (T+2))$ for these zeros. The Riemann–von Mangoldt formula and partial summation give ${\sum }_{|\Im \rho |\le T}|\rho {|}^{-1}\ll {\log }^2(T+2)$. Choose $T=\exp (\sqrt{\log x})$. Both the zero sum and the truncation error are then $O(x{e}^{-c^{\prime }\sqrt{\log x}})$ after reducing the positive constant to absorb logarithmic factors. Replacing an arbitrary $X$ by $\lfloor X\rfloor +1/2$ preserves the Von Mangoldt function sum and changes the main term by at most one. This proves the requested form of the Prime number theorem:

Write $\psi (x)={\sum }_{n\le x}\Lambda (n)$, where $\Lambda$ is the Von Mangoldt function. The Riemann–von Mangoldt explicit formula, in its symmetric limiting form for $x>1$, isHere nontrivial zeros are counted with multiplicity, the limit is taken symmetrically through admissible heights, and ${\psi }_0$ assigns half weight at a jump. Its difference from $\psi$ is at most $(\log x)/2$. The Euler product and part (a) exclude zeros with real part at least one. The functional equation of the Riemann zeta function leaves only the trivial zeros of the Riemann zeta function at negative even integers outside $0<\Re \rho <1$; their already displayed logarithmic correction is $O(x^{-2})$ for large $x$. These terms and the constant are negligible in the requested asymptotic error, rather than literally absent from the exact formula.

A useful truncated explicit formula for the second Chebyshev function is, uniformly for $2\le T\le x$,One may first take a height in $[T,T+1]$ separated from zeros and then adjust to $T$ using the local count. The local zero count for the Riemann zeta function isIt includes multiplicity and is uniform in real $t$. It follows by subtracting the Riemann–von Mangoldt formula at endpoints, handling bounded heights separately and using conjugation for negative heights. Both closed endpoints change the count only by another local bound.

By part (a), every nontrivial zero $\rho =\beta +i\gamma$ with $|\gamma |\le T$ satisfies $\beta \le 1-c_1/{\log }^{9}T$ for large $T$, after reducing the positive constant. The local zero count for the Riemann zeta function givesThere are finitely many zeros at bounded height, none at zero or at one, so that part of the sum is bounded. The truncated explicit formula for the second Chebyshev function now yieldsBalance the exponent losses $c_1\log x/(\log T{)}^9$ and $\log T$ by choosing $\log T=(c_1\log x{)}^{1/10}$. This is the optimal order obtainable from these two errors: making either exponent larger forces the other smaller. Thus, for a positive constant $c_2$,The logarithmic prefactor can be absorbed by reducing $c_2$. This is the prime number theorem error from a logarithmic zero-free region with ninth-power width.

Under the Riemann hypothesis, $|x^{\rho }|=x^{1/2}$. The same reciprocal-zero sum bounds the zero contribution by $O(x^{1/2}{\log }^{2}T)$. Taking $T=x$ makes the truncation error $O({\log }^{2}x)$, so

If all nontrivial zeta zeros satisfy $\Re \rho \le 1-c$ with fixed $0<c<1/2$, the Riemann–von Mangoldt formula gives ${\sum }_{|\rho |<T}|\rho {|}^{-1}\ll {\log }^{2}T$. The truncated explicit formula for the second Chebyshev function then bounds the error by $O(x^{1-c}{\log }^{2}T+x{\log }^{2}x/T)$. Choosing $T=x^c$ gives the displayed bound and, for every positive $\epsilon$, $O_{\epsilon }(x^{1-c+\epsilon })$.

// Target: number-theory.bigb

Suppose $A>0$ and every sufficiently high Nontrivial zero of the Riemann zeta function satisfies the displayed gap, with bounded heights also separated from one. The truncated explicit formula for the second Chebyshev function and the local zero count for the Riemann zeta function give errors $x{\log }^{2}x\exp (-c\log x/(\log T{)}^A)$ and $x{\log }^{2}x/T$. Balance them by $\log T\asymp (\log x{)}^{1/(A+1)}$. This explains how the width of a zero-free region of the Riemann zeta function determines the exponential scale in the Prime number theorem error.