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For $t\ge 2$ and $0<\eta \le 1/4$, suppose $|\zeta |\le M$, $M\ge 2$, on radius-$\eta$ discs centred at $1+\eta /4+it$ and $1+\eta /4+2it$. Every zero $\beta +it$ has the displayed gap. The reciprocal Euler product bounds both centre values below by a constant times $\eta$. The local logarithmic-derivative lemma gives errors $B/\eta$, $B=1+\log (M/\eta )$. The three-four-one zero-free-region argument then implies $4/(\sigma -\beta )-3/(\sigma -1)\ll B/\eta$. For a zero close to one, choose $\sigma =1+6(1-\beta )$, making the left side $1/(14(1-\beta ))$. Zeros farther away already satisfy the bound. Thus a logarithmic upper bound for zeta translates into a quantitative zero-free width.

Write $s=\sigma +it$ and let $\rho =\beta +i\gamma$ run over Nontrivial zeros of the Riemann zeta function, including multiplicity. The bounds involving $\log |t|$ below are for $|t|\ge 3$. At bounded heights their uniform versions use $\log (|t|+3)$; $\log |t|$ itself cannot be the right bound near zero or $|t|=1$.

Start with the Hadamard factorization of the Riemann xi function, an entire function of order one. Its logarithmic derivative has the convergent expansionDifferentiating $\xi (s)=\frac{1}{2}s(s-1){\pi }^{-s/2}\Gamma (s/2)\zeta (s)$ givesThe real sum of $1/\rho$ converges absolutely: $0\le \beta \le 1$ makes its terms $O({\gamma }^{-2})$ at large height, and the order-one Hadamard factorization ensures the required summability. Thus these terms and $\Re B$ form a fixed constant. The Stirling formula for the digamma function gives, uniformly for $1<\sigma \le 2$ and $|t|\ge 3$,This is the useful real form of the global partial-fraction expansion of the zeta logarithmic derivative. Every term subtracted is nonnegative because the Euler product excludes zeros with $\beta >1$. At the simple pole $s=1$, one also has $F(\sigma )=1/(\sigma -1)+O(1)$ as $\sigma \downarrow 1$.

The three-four-one zero-free-region argument now detects a zero too close to one. The Von Mangoldt function is nonnegative and the absolutely convergent Dirichlet series givesbecause the corresponding coefficient at $n$ is $\Lambda (n)n^{-\sigma }[3+4\cos (t\log n)+\cos (2t\log n)]$. If $\beta +it$ is a zero, retain its term in the expansion at height $t$ and discard the other nonpositive terms. The expansion at height $2t$ gives an upper bound even without retaining a particular zero. We obtain, for an absolute constant $C_0>0$,Choose $0<a<1$ so small that $5/(9a)>C_0$, put $b=a/8$, and take $\sigma =1+a/\log |t|$. If $1-\beta \le b/\log |t|$, the last inequality would requirea contradiction. Thereforefor some fixed $b>0$. This proves the required zero-free region of the Riemann zeta function. Nonvanishing on $\Re s=1$ at bounded nonzero heights, proved in Question 2, and the isolation of zeros allow a decrease of $b$ giving the uniform formulation with $\log (|t|+3)$ as well. Near $s=1$, apply this observation to the nonzero holomorphic function $(s-1)\zeta (s)$.

Next evaluate the real expansion at $\sigma =2$. The absolutely convergent Dirichlet series bounds $|{\zeta }^{\prime }/\zeta (2+it)|\le {\sum }_n\Lambda (n)n^{-2}$ uniformly in $t$, soSince $1\le 2-\beta \le 2$, each term on the left is at least $1/[4+(t-\gamma {)}^2]$. The smoothed zeta zero-count bound follows:Each zero satisfying $|t-\gamma |<1$ contributes more than $1/5$, so this also gives the local zero count for the Riemann zeta function:The same kernel sum is bounded at compact heights, since its tail converges uniformly there; the bounds with $\log (|t|+3)$ are consequently valid for all real $t$.

For completeness, the contour estimate used in Question 2 follows from these very formulas. Subtract the complex logarithmic derivative expansion at $2+it$ from that at $\sigma +it$, with $1-b/(2\log (T+3))\le \sigma \le 2$ and $|t|\le T$. Terms with $|t-\gamma |>1$ have differences $O((t-\gamma {)}^{-2})$, whose sum is $O(\log (|t|+3))$ by the kernel bound. There are $O(\log (|t|+3))$ remaining terms. After reducing $b$, the zero-free region of the Riemann zeta function separates every such zero horizontally from the half-width boundary by $\gg 1/\log (T+3)$: their heights satisfy $|\gamma |\le T+1$. Thus on the left edge, and on the horizontal edges at $t=\pm T$, each reciprocal is $O(\log (T+3))$. The Gamma function and pole terms satisfy the same required bound; at small heights on the left edge the pole distance is $\gg 1/\log (T+3)$. This proves $|{\zeta }^{\prime }/\zeta |\ll {\log }^2(T+3)$ on those edges and justifies the contour shift without choosing heights that dodge unknown zeros.

Put $F(s)=-{\zeta }^{\prime }(s)/\zeta (s)$. The Euler product gives $F(s)={\sum }_{n\ge 1}\Lambda (n)n^{-s}$ for $\Re s>1$. The elementary nonnegative polynomialtherefore gives the three-four-one zero-free-region argument:The coefficients $\Lambda (n)$ are nonnegative; applying the identity with $u=t\log n$ justifies the inequality term by term.

Take real parts in the global partial-fraction expansion of the zeta logarithmic derivative. For $1<\sigma \le 2$, the digamma function bounds yieldAll terms in the zero sum are nonnegative, because the Euler product and completed functional equation locate the nontrivial zeros in the closed critical strip. The constant terms are bounded: in particular ${\sum }_{\rho }\Re (1/\rho )={\sum }_{\rho }\beta /|\rho {|}^2$ converges. Near the real pole, $F(\sigma )=1/(\sigma -1)+O(1)$.

First exclude zeros on the line one. If $1+it$, $t\ne 0$, were a zero of multiplicity $m\ge 1$, the selected zero term would giveInserted into the positivity inequality, this yields $0\le (3-4m)/(\sigma -1)+O_t(1)$, impossible as $\sigma \downarrow 1$. At $t=0$, zeta has a pole, with a punctured neighbourhood containing no zeros. Thus the line one has no zeros.

Now let $\rho =\beta +i\gamma$ with $|\gamma |\ge 1$ and put $L=\log (|\gamma |+2)$. Select this zero in the formula at $t=\gamma$ and discard every other nonnegative zero kernel. The pole terms at heights $\gamma ,2\gamma$ are bounded. ConsequentlyChoose a fixed positive $a$ small enough that $C_{1}a\le 1/2$ and $1+a/L\le 2$, and set $\sigma =1+a/L$, $h=1-\beta$. Rearranging givesThis is the required logarithmic gap at large height.

For $|\gamma |\le 1$, zeros with $\beta \le 1/2$ already have a fixed gap. The remaining compact region contains finitely many zeros, none on the line one; the pole at one cannot be an accumulation point of zeros. Hence those zeros also have a fixed positive distance from that line. Reduce the absolute constant to cover this bounded-height region and to make the preceding weak inequality strict. The zero-free region of the Riemann zeta function is thereforeThis proof has established boundary nonvanishing as well as the quantitative gap, rather than assuming the former implicitly.

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:

Absolute convergence and the fundamental theorem of arithmetic give the Euler productFor a finite set of primes, expand the geometric factors: their product sums $n^{-s}$ over integers whose prime factors lie in that set. Let the finite sets increase through all primes. Absolute convergence permits passage to the limit and recovers the full Dirichlet series. Moreover ${\sum }_{p,m\ge 1}|p^{-ms}|/m<\infty$, so the logarithm converges and the product has no zeros there.

The same absolutely convergent logarithm, and $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2\ge 0$, giveThis is the product version of the three-four-one zero-free-region argument. It also proves there are no zeros on $\Re s=1$: if $\zeta (1+i{t}_0)=0$ for $t_0\ne 0$, its factor has order at least four as $\sigma \downarrow 1$, while the real pole contributes only order minus three and the $2t_0$ factor remains bounded. The displayed left side would tend to zero, a contradiction. At $t_0=0$ there is a pole, not a zero.

For large $|t|$, put $L=\log (|t|+2)$. The Hardy-Littlewood approximation to the Riemann zeta function at $x\asymp |t|$ gives $|\zeta (\sigma +it)|\ll L$ for $1-2/L\le \sigma \le 3$; its finite sum is bounded by ${\sum }_{n\le x}{n}^{-1+2/L}\ll L$ and the integral term is bounded. The Cauchy estimate for derivatives on circles of radius comparable to $1/L$ consequently gives $|{\zeta }^{\prime }(\sigma +it)|\ll L^2$ for $1-L^{-9}\le \sigma \le 2$.

Take ${\sigma }_0=1+a{L}^{-9}$ with a small fixed $a>0$. The product inequality, $\zeta ({\sigma }_0)\ll L^9/a$, and $|\zeta ({\sigma }_0+2it)|\ll L$ implyIf $1-c{L}^{-9}\le \sigma \le {\sigma }_0$, integration of the derivative along the horizontal segment changes this value by at most $C(a+c)L^{-7}$. Choose $a$ sufficiently small that $Ca<c_0{a}^{3/4}/4$, then $c\le a$ sufficiently small. The lower bound remains a positive multiple of $L^{-7}$. For ${\sigma }_0\le \sigma \le 2$, the same product inequality, $\zeta (\sigma )\le \zeta ({\sigma }_0)$, and the near-one upper bound give that lower bound directly. For $\sigma \ge 2$, the reciprocal Euler product gives $|1/\zeta (s)|\le \zeta (2)$.

Finally the no-zero result on $\Re s=1$, compactness at bounded heights and the regular reciprocal at the pole allow a further fixed reduction of $c$ to include bounded $t$. We have proved the weak logarithmic zero-free region for the Riemann zeta functionThe reciprocal at $s=1$ is its holomorphic extension, equal to zero.

Differentiate the locally uniformly convergent logarithm of the Euler product. Its logarithmic derivative isThe differentiated sum is absolutely convergent, since $\Lambda (n)\le \log n$. Taking the real parts at the three heights givesEach summand is nonnegative because its bracket is $2(1+\cos (t\log n){)}^2$. This proves the derivative form of the three-four-one zero-free-region argument.

For $\sigma >1$, the three-four-one zero-free-region argument starts fromApplying this termwise togives

Suppose $\rho =\beta +i\gamma$ is a zero with $|\gamma |\ge 4$. In the supplied Local partial-fraction expansion of the Riemann zeta logarithmic derivative, every nearby zero contributes a nonnegative real part at $\sigma +i\gamma$ when $\sigma >1$. Keeping the term from $\rho$ givesAt height zero the pole at one givesand the same local expansion at height $2\gamma$ givesSubstitution yieldsSet $\sigma =1+\eta /\log |\gamma |$, first choosing a sufficiently small absolute $\eta >0$. If $\beta >1-c/\log |\gamma |$, the left side is at least $4\log |\gamma |/(\eta +c)$. Choosing $c>0$ sufficiently small contradicts the last inequality. Therefore

For $\sigma >1$, put $F(s)=-{\zeta }^{\prime }(s)/\zeta (s)$. Its absolutely convergent Dirichlet series andgive the three-four-one zero-free-region argumentThe pole of $\zeta$ at one gives

Suppose $\rho =\beta +i\gamma$ is a zero with $\gamma \ge 4$ and $\beta$ close to one. Apply the supplied Local partial-fraction expansion of the Riemann zeta logarithmic derivative at $\sigma +i\gamma$. Every term has positive real part, so retaining the term belonging to $\rho$ givesAt $\sigma +2i\gamma$ the same expansion gives merely $\Re F(\sigma +2i\gamma )\le O(\log \gamma )$. The zero is included in the supplied disk whenever $1-\beta$ and $\sigma -1$ are sufficiently small. Hence

Set $L=\log \gamma$ and $\sigma =1+a/L$, where $a>0$ is a sufficiently small fixed constant. If $(1-\beta )L$ were smaller than a sufficiently small constant $c>0$, division by $L$ would givea contradiction. Conjugation handles negative $\gamma$. Reducing $c$ to absorb the bounded range proves the classical zero-free region of the Riemann zeta function