= Solution
For $\sigma>1$, the absolutely convergent logarithm of the <Euler product> for the <Riemann zeta function> is
$$
\log\zeta(s)=\sum_p\sum_{k\ge1}\frac{p^{-ks}}k.
$$
The elementary identity $3+4\cos v+\cos2v=2(1+\cos v)^2\ge0$ therefore gives
$$
\log\left(\zeta(\sigma)^3|\zeta(\sigma+it)|^4|\zeta(\sigma+2it)|\right)
=\sum_p\sum_{k\ge1}\frac{3+4\cos(kt\log p)+\cos(2kt\log p)}{kp^{k\sigma}}\ge0.
$$
Thus the expression inside the logarithm is at least one. The <Meromorphic continuation of the Riemann zeta function to the right half-plane> has only a simple <pole> at one, of <residue> one. It follows, for example, from
$$
\zeta(s)=\frac{s}{s-1}-s\int_1^\infty\frac{\{u\}}{u^{s+1}}\,du\qquad(\operatorname{Re}s>0),
$$
whose integral is holomorphic there.
Suppose $t\ne0$ and $\zeta$ has a zero of order $m\ge1$ at $1+it$. As $\sigma\downarrow1$, the real factor is $O((\sigma-1)^{-3})$, the middle factor is $O((\sigma-1)^{4m})$, and the factor at $1+2it$ is bounded. Their product would tend to zero, contradicting its lower bound one. Hence
$$
\boxed{\zeta(1+it)\ne0\quad(t\ne0).}
$$
At $t=0$ there is a <pole>, not a zero; the statement about the line does not assert a finite value for $\zeta(1)$. This proves the <three-four-one product proof of zeta boundary nonvanishing>.
To relate the <Second Chebyshev function> to the <logarithmic derivative>, differentiate the <Euler product> in its half-plane of <absolute convergence>:
$$
-\frac{\zeta'(s)}{\zeta(s)}=\sum_{n\ge1}\frac{\Lambda(n)}{n^s}.
$$
Since $\psi(u)=\sum_{n\le u}\Lambda(n)$, its <first integral of the Chebyshev function> is
$$
\Psi_1(x)=\int_0^x\psi(u)\,du=\sum_{n\le x}(x-n)\Lambda(n).
$$
For any fixed $c>1$, its <Mellin inversion formula> is
$$
\boxed{\Psi_1(x)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}-\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s+1}}{s(s+1)}\,ds.}
$$
For justification, the elementary contour kernel is $\frac1{2\pi i}\int_{(c)}y^s/(s(s+1))\,ds=1-y^{-1}$ for $y>1$ and zero for $0<y\le1$. Close left or right and take the <residues> at zero and minus one; at $y=1$ the continuous value is zero. <Absolute convergence> on $\operatorname{Re}s=c$ allows termwise integration of the <Dirichlet series>, and multiplying this kernel by $x$ gives $(x-n)_+$.
Here is how the relation yields the asymptotic. Move the contour to the left, using the <functional equation of the Riemann zeta function> to control the left-hand side. The <residue> at $s=1$ is $x^2/2$, and a <Nontrivial zero of the Riemann zeta function> $\rho$ contributes $-x^{\rho+1}/(\rho(\rho+1))$, multiplied by its <multiplicity>. The kernel <poles> at zero and minus one contribute $-x\log(2\pi)$ and $\zeta'(-1)/\zeta(-1)$; a <trivial zero of the Riemann zeta function> $-2k$ contributes $-x^{1-2k}/(2k(2k-1))$. Thus the <smoothed explicit formula for the Chebyshev function> is
$$
\Psi_1(x)=\frac{x^2}{2}-\sum_\rho\frac{x^{\rho+1}}{\rho(\rho+1)}-x\log(2\pi)+\frac{\zeta'(-1)}{\zeta(-1)}-\sum_{k\ge1}\frac{x^{1-2k}}{2k(2k-1)}.
$$
For the contour argument, use heights avoiding the zero ordinates and then let those heights increase. The local zero bound from the <Riemann–von Mangoldt formula>, together with the <Local partial-fraction expansion of the Riemann zeta logarithmic derivative>, permits heights with logarithmic-derivative bound $O(\log^2 T)$ in each fixed vertical strip. The horizontal integrals then vanish because the denominator is of size $T^2$. After that, send the left edge through negative odd integers to minus infinity; the <functional equation of the Riemann zeta function> bounds the <logarithmic derivative> there by a logarithm, and $x^{s+1}$ makes the left integral vanish for fixed $x>1$. This explains why this smoothing allows a convergent contour calculation without a quantitative zero-free region.
In fact the <Riemann–von Mangoldt formula>, stated in the next solution, gives $N(T)=O(T\log T)$ and hence
$$
\sum_\rho\frac1{|\rho(\rho+1)|}<\infty.
$$
The <Nontrivial zeros of the Riemann zeta function> satisfy $0<\operatorname{Re}\rho<1$, using the <Euler product>, the <functional equation of the Riemann zeta function>, and the nonvanishing just proved. For each such zero, $x^{\rho-1}\to0$ as $x\to\infty$, whereas $|x^{\rho-1}|\le1$ for $x\ge1$. The <dominated convergence theorem> therefore makes the absolutely convergent zero sum, divided by $x^2$, tend to zero. All the other displayed terms are $o(x^2)$. We obtain
$$
\boxed{\int_0^x\psi(u)\,du\sim\frac{x^2}{2}.}
$$
It remains to unsmooth; differentiating an asymptotic without justification would not suffice. <Monotonicity> of $\psi$ gives, for fixed $0<h<1$,
$$
\frac{\Psi_1(x)-\Psi_1((1-h)x)}{hx}\le\psi(x)\le\frac{\Psi_1((1+h)x)-\Psi_1(x)}{hx}.
$$
Divide by $x$ and use the integral asymptotic. The lower and upper limits lie between $1-h/2$ and $1+h/2$. Letting $h\downarrow0$ proves $\psi(x)\sim x$. Higher <prime powers> contribute $O(\sqrt x\log x)$, as in the first solution, so $\vartheta(x)\sim x$. Finally <partial summation> gives
$$
\pi(x)=\frac{\vartheta(x)}{\log x}+\int_2^x\frac{\vartheta(t)}{t(\log t)^2}\,dt.
$$
The integral is $O(x/(\log x)^2)$, by $\vartheta(t)=O(t)$ and splitting at $\sqrt x$. Hence
$$
\boxed{\pi(x)\sim\frac{x}{\log x},}
$$
which is the <Prime number theorem>.
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