Classical Zero Free Region and the Prime Number Theorem

For bounded $s$ away from zeros, the terms $1\mathrm{/}(s-\rho )+1\mathrm{/}\rho$ are $O_s(\mathrm{\mid }\rho {\mathrm{\mid }}^{-2})$. The unit-interval count, reflected using conjugate zeros, makes their tails normally convergent. Logarithmically differentiating the canonical product therefore gives ${\xi }^{\mathrm{\prime }}\mathrm{/}\xi =B+{\sum }_{\rho }(1\mathrm{/}(s-\rho )+1\mathrm{/}\rho )$.

Put $z=1+s\mathrm{/}2$. In a wider fixed sector containing these high-height points, Stirling gives $\mathrm{\Gamma }(z)=\sqrt{2\pi }\exp ((z-1\mathrm{/}2)\mathrm{Log}z-z)(1+r(z))$ with $r(z)=O(\mathrm{\mid }z{\mathrm{\mid }}^{-1})$. On discs of radius $ϵ\mathrm{\mid }z\mathrm{\mid }$ in that sector, Cauchy gives $r^{\mathrm{\prime }}(z)=O(\mathrm{\mid }z{\mathrm{\mid }}^{-2})$. Thus ${\mathrm{\Gamma }}^{\mathrm{\prime }}\mathrm{/}\mathrm{\Gamma }(z)=\mathrm{Log}z-1\mathrm{/}(2z)+O(\mathrm{\mid }z{\mathrm{\mid }}^{-2})$, whose real part is $\log \mathrm{\mid }t\mathrm{\mid }-\log 2+O(1\mathrm{/}\mathrm{\mid }t\mathrm{\mid })$. Remaining bounded heights are compact.

In the defining formula for $\xi$, use $\mathrm{\Gamma }(1+s\mathrm{/}2)=(s\mathrm{/}2)\mathrm{\Gamma }(s\mathrm{/}2)$ to write $\xi =(s-1){\pi }^{-s\mathrm{/}2}\mathrm{\Gamma }(1+s\mathrm{/}2)\zeta (s)$. The Gamma recurrence follows by integration by parts on its defining integral and meromorphic continuation. Differentiating this equality proves the first formula wherever its factors are nonzero, hence meromorphically.

For fixed $s$ the real summands have tails $O_s(\mathrm{\mid }\mathrm{Im}\rho {\mathrm{\mid }}^{-2})$, since $0<\mathrm{Re}\rho <1$. The same estimate applies to $\mathrm{Re}(1\mathrm{/}\rho )$. Their absolute convergence follows from the unit-band count. Absorb the constant $\mathrm{Re}B+\sum \mathrm{Re}(1\mathrm{/}\rho )$ and the bounded pole term into $O(1)$, obtaining the second formula.

The negative logarithmic derivative has coefficients $\mathrm{\Lambda }(n)\ge 0$. Its series is absolutely convergent for $\sigma >1$ (also $\mathrm{\Lambda }(n)\le \log n$), so the displayed expression equals ${\sum }_{n\ge 1}\mathrm{\Lambda }(n)n^{-\sigma }[3+4\cos (t\log n)+\cos (2t\log n)]$.

For real $u$, $3+4\cos u+\cos (2u)=2(1+\cos u{)}^2\ge 0$. Every summand is nonnegative, so the convergent sum is nonnegative.

For $1<\sigma \le 2$, the simple pole gives $-{\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta (\sigma )=1\mathrm{/}(\sigma -1)+O(1)$. If ${\rho }_0=\beta +i\gamma$ is a zero with $\mathrm{\mid }\gamma \mathrm{\mid }\ge 3$, positivity of the real zero summands gives $-\mathrm{Re}({\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta )(\sigma +i\gamma )\le 1\text{\hspace{0.17em}⁣}\cdot C_1\log (\mathrm{\mid }\gamma \mathrm{\mid }+2)-1\mathrm{/}(\sigma -\beta )$ and $-\mathrm{Re}({\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta )(\sigma +2i\gamma )\le C_1\log (\mathrm{\mid }\gamma \mathrm{\mid }+2)$.

Insert these bounds in the three-four-one inequality. For an absolute $C\ge 1$, put $L=\log (\mathrm{\mid }\gamma \mathrm{\mid }+2)$; then $4\mathrm{/}(\sigma -\beta )\le 3\mathrm{/}(\sigma -1)+CL$. Taking $\sigma -1=1\mathrm{/}(2CL)$ yields $1-\beta \ge 1\mathrm{/}(14CL)$. Choosing a strictly smaller constant excludes even the closed boundary of the claimed high-height region.

The function $h(s)=(s-1)\zeta (s)$ is holomorphic near the compact segment $\{1+it:\mathrm{\mid }t\mathrm{\mid }\le 3\}$, nonzero there, and $h(1)=1$. Finitely many nonvanishing neighborhoods cover this segment and contain a uniform thin rectangle about it. Shrink $c_0$ so the proposed bounded-height region to the left of one lies in that rectangle. To the right use the already proved zero-free half-plane. This proves the claim at every height, including zero.

Put $L=\log (\mathrm{\mid }t\mathrm{\mid }+2)$, $s_1=1+L^{-1}+it$. The Euler series and the real-axis simple-pole expansion give $\mathrm{\mid }{\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta (s_1)\mathrm{\mid }\le -{\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta (1+L^{-1})=O(L)$. The same comparison holds for every $\sigma \ge 1+L^{-1}$; for $\sigma \ge 2$ it is even bounded by the convergent series at two. The real-part formula now gives ${\sum }_{\rho }\mathrm{Re}(1\mathrm{/}(s_1-\rho ))=O(L)$, all summands being positive.

For $\mathrm{\mid }\mathrm{Im}\rho -t\mathrm{\mid }\le 1$, the region theorem implies $1-\mathrm{Re}\rho \ge c_0\mathrm{/}(KL)$ with an absolute $K$, since $\log (\mathrm{\mid }\mathrm{Im}\rho \mathrm{\mid }+2)\le KL$. Choose $d<c_0\mathrm{/}(2K)$. For $1-d\mathrm{/}L\le \sigma \le 1+1\mathrm{/}L$, the positive real parts of $s-\rho$ and $s_1-\rho$ are comparable, hence $\mathrm{\mid }s-\rho \mathrm{\mid }\ge c\mathrm{\mid }{s}_1-\rho \mathrm{\mid }$. Consequently $\mathrm{\mid }1\mathrm{/}(s-\rho )-1\mathrm{/}(s_1-\rho )\mathrm{\mid }\le C\mathrm{/}(L\mathrm{\mid }{s}_1-\rho {\mathrm{\mid }}^2)\le C\mathrm{Re}(1\mathrm{/}(s_1-\rho ))$.

For ordinates off the zero ordinates, subtract the two local logarithmic-derivative formulas. Their pole terms are bounded at these heights. Sum the preceding comparison over the common local zero set and use its positive-sum bound to obtain $O(L)$. The constants do not depend on the distance of t from an ordinate. Taking limits from non-ordinates extends the bound to all t, because the entire horizontal segment is zero-free. Together with the Euler-series range this proves the assertion.

Choose $c_1$ smaller than the constant in the horizontal comparison. This gives the stated derivative bound throughout the high-height region.

At $s_1=1+1\mathrm{/}L+it$, $L=\log (\mathrm{\mid }t\mathrm{\mid }+2)$, the Euler logarithm satisfies $\mathrm{\mid }\log \zeta (s_1)\mathrm{\mid }\le \log \zeta (1+1\mathrm{/}L)\le \log (1+L)$, by comparing the positive real zeta series to its integral. Integrate ${\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta$ from $s_1$ horizontally to $s$ for $1-c_2\mathrm{/}L\le \sigma \le 1+1\mathrm{/}L$. The length is $O(1\mathrm{/}L)$ and the integrand is $O(L)$, so the change in the continued logarithm is $O(1)$. Exponentiating its negative real part gives $\mathrm{\mid }1\mathrm{/}\zeta (s)\mathrm{\mid }=O(L)$. For larger sigma the Euler logarithm already gives that bound.

On the compact low-height portion choose $c_2<c_1$ sufficiently small that $h(s)=(s-1)\zeta (s)$ is holomorphic and nonvanishing on a neighborhood, including $h(1)=1$. Then $h^{\mathrm{\prime }}\mathrm{/}h$ and $1\mathrm{/}h$ are bounded there. The identities ${\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta +1\mathrm{/}(s-1)=h^{\mathrm{\prime }}\mathrm{/}h$ and $1\mathrm{/}\zeta =(s-1)\mathrm{/}h$ prove both low-height estimates and their removable interpretations.

Write $L=\log (\mathrm{\mid }t\mathrm{\mid }+2)$. For a sufficiently small absolute $a>0$, $r<a\mathrm{/}L$ makes the disc zero-free: within it $\log (\mathrm{\mid }\mathrm{Im}\rho \mathrm{\mid }+2)\le KL$, whereas $1-\mathrm{Re}\rho \le r$. This contradicts the region bound if a zero occurs.

For $\mathrm{\mid }t\mathrm{\mid }\ge 3$ and $a\mathrm{/}L\le r\le 1\mathrm{/}6$, evaluate at $s_1=1+r+it$. The Euler series gives $\mathrm{\mid }{\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta (s_1)\mathrm{\mid }=O(1\mathrm{/}r)$, using its simple-pole expansion on the real axis. The positive real zero sum is thus $O(1\mathrm{/}r+L)$. Every counted zero contributes at least $r\mathrm{/}(4r^2+r^2)=1\mathrm{/}(5r)$, since its real separation is between r and 2r and its imaginary separation at most r. Hence $n(r;t)=O(1+rL)=O(rL)$.

If $r\ge 1\mathrm{/}6$ and $\mathrm{\mid }t\mathrm{\mid }\ge 3$, finitely many adjacent unit ordinate bands give $n(r;t)=O(L)=O(rL)$. Negative bands have the same count by conjugation of zeta. For $\mathrm{\mid }t\mathrm{\mid }\le 3$, all counted zeros lie in one compact rectangle and are finite in number, while a nonempty disc must have $r\ge a\mathrm{/}L\ge a\mathrm{/}\log 5$. Enlarging the constant handles these remaining cases.

Nontrivial zeros have no accumulation in a compact subset of the plane. The finitely many with $\mathrm{\mid }\mathrm{Im}\rho \mathrm{\mid }\le 1$ have nonzero denominator: the product for xi has no zero at zero, and its factors have exactly the nontrivial zeros. Their reciprocal sum is a fixed finite constant.

For each integer $n\ge 1$, zeros with $n\le \mathrm{\mid }\mathrm{Im}\rho \mathrm{\mid }\le n+1$ each contribute at most $1\mathrm{/}n$. The count in either band is $O(\log (n+2))$; the negative band follows from $\zeta (\bar{s})=\overline{\zeta (s)}$, first on its defining half-plane and then by continuation. Therefore the remaining sum is at most $C{\sum }_{n\le T}\log (n+2)\mathrm{/}n=O({\log }^{2}T)$. Endpoint overlap only increases this upper bound.

For each zero in the finite sum $\mathrm{\mid }\mathrm{Im}\rho \mathrm{\mid }<T$, the region implies $\mathrm{\mid }{x}^{\rho }\mathrm{\mid }\le x\exp (-c_0\log x\mathrm{/}\log (T+2))$. Summing absolute values and using the reciprocal estimate, including any real zeros, bounds the entire zero sum by the first displayed error.

The supplied truncation error is at most $O(x{\log }^2(xT)\mathrm{/}T+\log x)$ because its minimum is at most one. The fixed constant ${\zeta }^{\mathrm{\prime }}(0)\mathrm{/}\zeta (0)$ and $\log (1-x^{-2})$ are bounded for $x\ge 2$. Finally $\mathrm{\mid }\psi (x)-{\psi }_0(x)\mathrm{\mid }\le (\log x)\mathrm{/}2$, so replacing the half-weighted value gives the asserted error, including prime-power endpoints. The supplied formula holds for all x,T at these bounds; if a contour construction avoids ordinates, a non-ordinate in [T,T+1] has comparable bounds.

Put $u=\sqrt{\log x}$ and choose a fixed $A>0$. For sufficiently large x take $T=e^{Au}\ge 3$. Then $\log (T+2)=Au+O(e^{-Au})$, so the finite-zero term is $O(x{u}^2{e}^{-(c_0\mathrm{/}A)u+o(1)})$, the truncation term is $O(x{u}^4{e}^{-Au})$, and the remaining error is $O(u^2)$.

Choose $0<c<\min (A,c_0\mathrm{/}A)$. For any fixed k and positive epsilon, $u^k{e}^{-ϵu}$ is bounded; thus each error above is $O(x{e}^{-cu})$. Enlarging the constant over the initial compact x-range proves the assertion for all $x\ge 2$.

The contour interpretation is consistent with the same bound: take ${\sigma }_1=1-c_1\mathrm{/}\log (T+2)$ with a sufficiently small region constant and ${\sigma }_0=1+1\mathrm{/}\log x$. Throughout $\mathrm{\mid }t\mathrm{\mid }\le T$ the left edge stays in the proved region. At high heights the derivative is $O(\log T)$; integrating $1\mathrm{/}\mathrm{\mid }s\mathrm{\mid }$ gives a vertical contribution $O(x^{{\sigma }_1}{\log }^{2}T)$, and horizontal edges give $O(x{\log }^2(xT)\mathrm{/}T)$. At bounded height the pole-subtracted estimate bounds the derivative by $O(1+1\mathrm{/}\mathrm{\mid }s-1\mathrm{\mid })$; on the left edge its integral is $O(\log \log (T+2))$. Only the pole at one is crossed. These edge bounds explain the scale used in the finite-zero proof.

The comparison gives $0\le \psi (x)-\theta (x)\le K\sqrt{x}\log x$. Thus $\mathrm{\mid }\theta (x)-x\mathrm{\mid }\le \mathrm{\mid }\psi (x)-x\mathrm{\mid }+K\sqrt{x}\log x$.

The first term has the asserted bound. Writing $u=\sqrt{\log x}$, the ratio of the second to $x{e}^{-cu}$ is $u^2{e}^{-u^2\mathrm{/}2+cu}$, bounded on $u\ge \sqrt{\log 2}$. This proves the result after enlarging the constant.

Let $J_k(x)={\int }_2^x(\log t{)}^{-k}dt$. Integration by parts gives $J_k=x\mathrm{/}{\log }^{k}x-2\mathrm{/}{\log }^{k}2+k{J}_{k+1}$. Starting with $\mathrm{Li}=J_1$, apply this identity m times: the remainder is $m!J_{m+1}$ and the lower-end constant is $-{\sum }_{j=0}^{m-1}2j!\mathrm{/}{\log }^{j+1}2$.

For $x\ge 4$, split $J_{m+1}$ at $\sqrt{x}$. Its first part is at most $\sqrt{x}\mathrm{/}(\log 2{)}^{m+1}$ and its second at most $2^{m+1}x\mathrm{/}{\log }^{m+1}x$. The first bound and the fixed lower-end constant are also $O_m(x\mathrm{/}{\log }^{m+1}x)$. This proves the expansion for each fixed m, including m=1.

Set $E(t)=\theta (t)-t$. The exact partial-summation identity gives $\pi (x)=x\mathrm{/}\log x+{\int }_2^{x}dt\mathrm{/}{\log }^{2}t+E(x)\mathrm{/}\log x+{\int }_2^{x}E(t)\mathrm{/}(t{\log }^{2}t)dt$. Integration by parts in the definition of Li makes its main term $\mathrm{Li}(x)+2\mathrm{/}\log 2$.

For $x\ge 4$ split the error integral at $\sqrt{x}$. The initial part is $O(\sqrt{x})$ because $E(t)=O(t)$. The second is $O(x{e}^{-(c_0\mathrm{/}\sqrt{2})\sqrt{\log x}})$ using the theta error and $\log t\ge \log 2$. The endpoint error has the same form. Decrease the positive exponent constant and absorb $2\mathrm{/}\log 2$ and the compact range $2\le x\le 4$.