Classical Zero Free Region and the Prime Number Theorem — Examples

Using $\cos (2u)=2{\cos }^{2}u-1$, the left side is $2+4\cos u+2{\cos }^{2}u=2(1+\cos u{)}^2$. The square vanishes precisely for $\cos u=-1$.

In the logarithmic-derivative expansion the nth contribution is $2\mathrm{\Lambda }(n)n^{-\sigma }(1+\cos (t\log n){)}^2$. It is zero for non-prime-powers and nonnegative otherwise. At t=0 its weight is eight, and at t log n equal to an odd multiple of pi its weight is zero, exactly as required by the inequality.

Write $L=\log (\mathrm{\mid }\gamma \mathrm{\mid }+2)$. Here $L>0$ and $C>0$, so the chosen $\delta$ is positive. Apply the displayed assumed inequality at this value of $\delta$. Substitution makes $3\mathrm{/}\delta +CL=7CL$, so $1+\delta -\beta \ge 4\mathrm{/}(7CL)$; the denominator is positive because $\beta <1$.

Subtract $\delta =1\mathrm{/}(2CL)$ to obtain $(4\mathrm{/}7-1\mathrm{/}2)\mathrm{/}(CL)=1\mathrm{/}(14CL)$. A smaller constant proves exclusion on a closed boundary. This calculation applies only to $\mathrm{\mid }\gamma \mathrm{\mid }\ge 3$; under countable choice, [F1] states the all-height zero-free region, whose small-height conclusion is not supplied by this conditional calculation.

Under countable choice, put $u=\sqrt{\log x}$. The error-balance lemma gives respectively $O(x{u}^2{e}^{-(c_0\mathrm{/}A)u+o(1)})$, $O(x{u}^4{e}^{-Au})$, and $O(u^2)$. Thus any $c<\min (c_0\mathrm{/}A,A)$ absorbs all polynomial factors. Equality of the two exponential rates occurs at $A=\sqrt{c_0}$; a fixed positive $A$ already suffices.

Thus the choice $A=\sqrt{c_0}$ balances the two exponential rates at $\sqrt{c_0}$. More generally any fixed $A>0$ gives a positive decay rate $\min (c_0\mathrm{/}A,A)$; bounded x can use $T=3$ and an enlarged constant. This is the square-root-logarithm decay scale recorded in the theorem-level estimate [F1].

Under countable choice, the supplied bounds $\psi (t)-t=O(t{e}^{-c\sqrt{\log t}})$ and $0\le \psi (t)-\theta (t)=O(\sqrt{t}\log t)$ give $E(t)=O(t{e}^{-c^{\mathrm{\prime }}\sqrt{\log t}})$ after decreasing the positive constant. The ratio of the prime-power error to $t{e}^{-c^{\mathrm{\prime }}\sqrt{\log t}}$ is $(\log t)e^{-\log t\mathrm{/}2+c^{\mathrm{\prime }}\sqrt{\log t}}$, which is bounded.

Substitute $\theta (t)=t+E(t)$ into the partial-summation identity. The main term $x\mathrm{/}\log x+{\int }_2^{x}dt\mathrm{/}{\log }^{2}t$ equals $\mathrm{Li}(x)+2\mathrm{/}\log 2$ by integration by parts. In the E-integral, [2,square root x] contributes O(square root x), and [square root x,x] contributes $O(x{e}^{-(c^{\mathrm{\prime }}\mathrm{/}\sqrt{2})\sqrt{\log x}})$. The endpoint term satisfies the same bound. At x=2 the empty integral leaves $2\mathrm{/}\log 2+(\log 2-2)\mathrm{/}\log 2=1=\pi (2)$.

The character-transform lemma at q=1 proves boundedness, local integrability, the displayed transform and its continuation through the closed boundary, including the cancellation at s=0.

Newman gives convergence of ${\int }_0^{\mathrm{\infty }}f(t)dt={\int }_1^{\mathrm{\infty }}(\psi (x)-x)x^{-2}dx$. Since psi is nonnegative, nondecreasing and O(x), desmoothing with a=1 yields $\psi (x)\mathrm{/}x\to 1$. This is the qualitative assertion; no estimate of a uniform continuation width was needed.

The displayed character values give the two residue indicators as $({\chi }_0+{\chi }_4)\mathrm{/}2$ and $({\chi }_0-{\chi }_4)\mathrm{/}2$, including the zero values on even numbers. Multiply by Lambda and sum to obtain both psi identities.

The principal sum is ${\mathrm{\Psi }}_{{\chi }_0}(x)=\psi (x)-{\sum }_{2^k\le x,k\ge 1}\log 2=\psi (x)-\lfloor \log x\mathrm{/}\log 2\rfloor \log 2$ for x at least one. Its correction is O(log x). Since $\phi (4)=2$, the fixed-progression theorem gives each asserted prime-count asymptotic, including exclusion of the single prime two.

Write $s=1+ϵ>1$. For a decreasing function $t^{-s}$, the discrepancy between its sum and integral on the kth block is at most ${10}^{-ks}$. Summing these discrepancies is O(1) uniformly as $ϵ\downarrow 0$. The integrals form a geometric series, giving ${\sum }_{n\in S}{n}^{-s}=(1-2^{-ϵ})\mathrm{/}[ϵ(1-{10}^{-ϵ})]+O(1)$. Since $ϵ\zeta (1+ϵ)\to 1$, the relative density is $\log 2\mathrm{/}\log 10$.

For primes, the Euler logarithm gives ${\sum }_p{p}^{-s}=\log (1\mathrm{/}ϵ)+O(1)$: all terms of exponent at least two sum to at most ${\sum }_{n\ge 2}{\sum }_{j\ge 2}{n}^{-j}<\mathrm{\infty }$, and the zeta pole determines its logarithm. Finite sets have zero relative density, and finite additivity for disjoint sets follows by taking limits of their finite sum identities.

The qualitative prime number theorem and $\mathrm{Li}(x)\sim x\mathrm{/}\log x$ give $E(x):=\pi (x)-\mathrm{Li}(x)=o(x\mathrm{/}\log x)$; the Li asymptotic follows by integration by parts and splitting its integral at $\sqrt{x}$. Put $a_k={10}^k$, $b_k=2a_k$, and ${\eta }_k={\sup }_{x\ge a_k}\mathrm{\mid }E(x)\mathrm{\mid }\log x\mathrm{/}x$, so ${\eta }_k\to 0$. For $s=1+ϵ$ and $k\ge 1$, Stieltjes integration by parts on $[a_k,b_k)$ gives ${\sum }_{a_k\le p<b_k}{p}^{-s}-{\int }_{a_k}^{b_k}\frac{x^{-s}}{\log x}dx=O\text{\hspace{0.17em}⁣}\left(\frac{{\eta }_k{e}^{-ϵk\log 10}}{k}\right)\mathrm{.}$ Indeed each endpoint term is at most ${\eta }_k{x}^{-ϵ}\mathrm{/}\log x$, and the interior term is bounded by $s{\eta }_k{\int }_{a_k}^{b_k}{x}^{-1-ϵ}\mathrm{/}\log xdx$, of the same order for $1<s\le 2$. Summing over $k$ and absorbing the initial block yields ${\sum }_{p\in S}{p}^{-s}={\int }_2^{\mathrm{\infty }}w(\log x)\frac{x^{-s}}{\log x}dx+o\text{\hspace{0.17em}⁣}(\log \frac{1}{ϵ}),$ because ${\sum }_{k\ge 1}{\eta }_k{e}^{-ϵk\log 10}\mathrm{/}k=o(\log (1\mathrm{/}ϵ))$: split at a fixed large $K$ and use ${\eta }_k\to 0$ on the tail. Here $w$ is the periodic indicator of $[0,\log 2)$ modulo $\log 10$.

Up to an endpoint error at most one, the counts at ${10}^m$ and $2\cdot {10}^m$ are respectively $({10}^m-1)\mathrm{/}9$ and $({10}^{m+1}-1)\mathrm{/}9$. Dividing by the two endpoints yields $1\mathrm{/}9$ and $5\mathrm{/}9$. Thus an integer counting asymptotic already fails despite the density limit.

Set $y=\log x$. The primitive of $w(y)-d$ is bounded because its integral over a period is zero. Integration by parts against $e^{-ϵy}\mathrm{/}y$ therefore bounds its contribution by $O(1)$, uniformly as epsilon tends to zero. The mean contribution is $d{\int }_{\log 2}^{\mathrm{\infty }}{e}^{-ϵy}dy\mathrm{/}y=d\log (1\mathrm{/}ϵ)+O(1)$: split at $1\mathrm{/}ϵ$ and substitute $ϵy$ on the tail. The $o(\log (1\mathrm{/}ϵ))$ error in step 1.3 vanishes after division by the prime denominator from step 1.2, proving the claimed relative Dirichlet density.

For the prime counting limits, fix $J$. At $X={10}^m$, the last $J$ completed blocks, indexed $k=m-j$ with $1\le j\le J$, have counts $\pi (2\cdot {10}^k)-\pi ({10}^k)\sim {10}^k\mathrm{/}\log ({10}^k)$. Relative to $\pi (X)\sim X\mathrm{/}\log X$, their limits sum to ${\sum }_{j=1}^J{10}^{-j}$. Earlier blocks contribute at most $\pi ({10}^{m-J})\mathrm{/}\pi ({10}^m)\to {10}^{-J}$. Let $J$ tend to infinity to get $1\mathrm{/}9$. At $X=2\cdot {10}^m$, the current block contributes relative limit $1\mathrm{/}2$, and the preceding blocks contribute $(1\mathrm{/}2){\sum }_{j\ge 1}{10}^{-j}=1\mathrm{/}18$, totaling $5\mathrm{/}9$. Endpoints are composite for $m$ positive, so they add no ambiguity. These unequal limits refute the inference for primes too.