= Solution
\b[First alternative.] The <Euler product> of the <Riemann zeta function> is absolutely convergent for $\sigma>1$. Its <logarithm> and the nonnegative <trigonometric polynomial>
$$
3+4\cos u+\cos2u=2(1+\cos u)^2
$$
give
$$
3\log\zeta(\sigma)+4\log|\zeta(\sigma+it)|+\log|\zeta(\sigma+2it)|
=\sum_p\sum_{m\ge1}\frac{3+4\cos(mt\log p)+\cos(2mt\log p)}{mp^{m\sigma}}\ge0.
$$
Consequently $\zeta(\sigma)^3|\zeta(\sigma+it)|^4|\zeta(\sigma+2it)|\ge1$. Suppose $t\ne0$ and the <Riemann zeta function> had a <zero> of order $m\ge1$ at $1+it$. As $\sigma\downarrow1$, its simple <pole> at one gives $\zeta(\sigma)=O((\sigma-1)^{-1})$, the alleged <zero> gives $\zeta(\sigma+it)=O((\sigma-1)^m)$, and $\zeta(\sigma+2it)$ remains bounded. The product would be $O((\sigma-1)^{4m-3})\to0$, contradicting its lower bound. Thus \b[there are no zeta <zeros> on $\Re s=1$]; the point $s=1$ itself is a <pole>, not a <zero>.
For the <first integral of the Chebyshev function>, define
$$
\Psi_1(x)=\int_0^x\psi(u)\,du=\sum_{n\le x}(x-n)\Lambda(n).
$$
The <logarithmic derivative> identity $-\zeta'/\zeta(s)=\sum_n\Lambda(n)n^{-s}$, valid for $\Re s>1$, leads to
$$
\boxed{\Psi_1(x)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}
-\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s+1}}{s(s+1)}\,ds,\qquad c>1.}
$$
To verify the kernel rather than assume an inversion formula, integrate $y^s/(s(s+1))$ on the vertical line $\Re s=c$. Closing to the left when $y>1$ gives the <residues> at <zero> and minus one, namely $1-y^{-1}$; closing to the right when $0<y<1$ gives <zero>. At $y=1$ the integral is <zero> as well. The horizontal integrals tend to <zero> because of the quadratic denominator, and the distant vertical integrals vanish in the indicated direction. Multiplying this kernel by $x$ with $y=x/n$ gives $(x-n)^+$. The <absolute convergence> of the <Dirichlet series> on $\Re s=c$, together with the integrable denominator, permits interchange with the integral and proves the formula.
Here is a quantitative contour route from this relation to the <Prime number theorem>. Question 3 establishes a <zero-free region of the Riemann zeta function> and the accompanying bounds for its <logarithmic derivative>, independently of the <Prime number theorem>. In a sufficiently narrow half-width region they imply
$$
\left|\frac{\zeta'(s)}{\zeta(s)}\right|\ll\log^2(T+3)
$$
on the left and horizontal sides of the rectangle with heights $\pm T$ and left edge $\sigma_L=1-b/(2\log(T+3))$, after making $b>0$ small enough also for bounded heights. The explanation of this bound is included at the end of Question 3. Take $c=1+1/\log x$ and $T=\exp(\sqrt{\log x})$. On the original line, the omitted tails are
$$
O\left(\frac{x^2\log x}{T}\right),
$$
since the <Euler product> bounds $|\zeta'/\zeta(c+it)|$ by $-\zeta'/\zeta(c)=O(\log x)$. Shift the truncated integral to the left edge. The only enclosed <pole> is at $s=1$, with <residue> $x^2/2$; the kernel <poles> at <zero> and minus one lie outside the rectangle. The left edge contributes $O(x^{2-b/(2\log(T+3))}\log^2(T+3))$, since $|s(s+1)|^{-1}\ll(1+t^2)^{-1}$ there. The two horizontal edges contribute $O(x^2\log^2(T+3)/T^2)$. In particular, for some $a>0$,
$$
\Psi_1(x)=\frac{x^2}{2}+O\left(x^2e^{-a\sqrt{\log x}}\right),
\qquad\text{so}\qquad\Psi_1(x)\sim\frac{x^2}{2}.
$$
It remains to remove the smoothing; differentiating an asymptotic formula would not justify this step. The <monotonicity> of the <Second Chebyshev function> gives, for $h=\varepsilon x$ with $0<\varepsilon<1$,
$$
\frac{\Psi_1(x)-\Psi_1(x-h)}h\le\psi(x)
\le\frac{\Psi_1(x+h)-\Psi_1(x)}h.
$$
First let $x\to\infty$ with $\varepsilon$ fixed. After division by $x$, the lower and upper bounds tend to $1-\varepsilon/2$ and $1+\varepsilon/2$. Let $\varepsilon\downarrow0$ to obtain $\psi(x)\sim x$. Higher <prime powers> contribute $O(\sqrt x\log x)$ by the <Chebyshev estimate>, so $\theta(x)\sim x$. Finally <partial summation> gives
$$
\pi(x)=\frac{\theta(x)}{\log x}+\int_2^x\frac{\theta(t)}{t(\log t)^2}\,dt
\sim\frac{x}{\log x}.
$$
The integral is $O(x/\log^2x)$, by splitting at $\sqrt x$ and using $\theta(t)=O(t)$. This proves the <Prime number theorem> with \b[$\pi(x)\sim x/\log x$].
\b[Second alternative.] We prove the <functional equation of the Riemann zeta function> from a <theta function>. Set $\Theta(u)=\sum_{n\in\mathbb Z}e^{-\pi n^2u}$ for $u>0$. With <Fourier transform> convention $\widehat f(\xi)=\int_{\mathbb R}f(v)e^{-2\pi iv\xi}\,dv$, the <Gaussian function> $f(v)=e^{-\pi uv^2}$ has
$$
\widehat f(\xi)=u^{-1/2}e^{-\pi\xi^2/u}.
$$
Indeed, differentiation under the integral and <integration by parts> give $\widehat f'(\xi)=-2\pi\xi\widehat f(\xi)/u$, while the <Gaussian integral> gives $\widehat f(0)=u^{-1/2}$. Periodize $f$: the smooth period-one function $\sum_n f(v+n)$ has <Fourier coefficients> $\widehat f(k)$, by integrating on $[0,1]$ and interchanging its absolutely convergent sum with the integral. Its absolutely convergent <Fourier series>, evaluated at <zero>, proves the <Poisson summation formula> in this case. Hence
$$
\Theta(u)=u^{-1/2}\Theta(1/u).
$$
For $\Re s>1$, termwise integration and the defining integral of the <Gamma function> give the <Mellin representation of the completed Riemann zeta function>:
$$
Z(s):=\pi^{-s/2}\Gamma(s/2)\zeta(s)
=\frac12\int_0^\infty(\Theta(u)-1)u^{s/2-1}\,du.
$$
Split at one and substitute $u=1/v$ in the lower integral. The elementary part is
$$
\frac12\int_0^1(u^{-1/2}-1)u^{s/2-1}\,du
=\frac1{s-1}-\frac1s=\frac1{s(s-1)},
$$
and the remaining part becomes an integral over $[1,\infty)$. Thus
$$
Z(s)=\frac1{s(s-1)}+\frac12\int_1^\infty(\Theta(u)-1)
\left(u^{s/2-1}+u^{(1-s)/2-1}\right)\,du.
$$
The integral is an <entire function> of $s$, because $\Theta(u)-1$ decays exponentially and <uniform convergence> holds on every compact subset of the <complex plane>. This proves <meromorphic continuation> of the <completed Riemann zeta function>, with only the displayed simple <poles>, and the right side is invariant under $s\mapsto1-s$. Therefore
$$
\boxed{\pi^{-s/2}\Gamma(s/2)\zeta(s)
=\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s).}
$$
The <Gamma reflection formula> and <Gamma duplication formula> give the equivalent form $\zeta(s)=2^s\pi^{s-1}\sin(\pi s/2)\Gamma(1-s)\zeta(1-s)$, understood by <meromorphic continuation>. Multiplication by $s(s-1)/2$ gives the <entire> <Riemann xi function> $\xi(s)=\xi(1-s)$.
Let $N(T)$ count positive ordinates of <Nontrivial zeros of the Riemann zeta function> up to height $T$, including <multiplicity>. The <Riemann–von Mangoldt formula> is
$$
N(T)=\frac{T}{2\pi}\log\frac{T}{2\pi}-\frac{T}{2\pi}+O(\log T).
$$
More precisely, away from ordinates of <zeros>, the constant term is $7/8$, the additional term is $S(T)=\pi^{-1}\arg\zeta(1/2+iT)$, and the remaining error is $O(T^{-1})$. Here the argument is continued along $2\to2+iT\to1/2+iT$, and $S(T)=O(\log T)$. The coarse formula displayed above holds at all large heights with the usual right-continuous counting convention. Subtract it at $T+1$ and $T$: the main term changes by $O(\log T)$ and each error is $O(\log T)$. Hence
$$
\boxed{N(T+1)-N(T)\ll\log T.}
$$
For <infinitely many reciprocal-logarithmic gaps between zeta zeros>, fix any $0<c<2\pi$, for example $c=\pi$. Suppose all sufficiently late gaps satisfied $\gamma_{n+1}-\gamma_n\le c/\log n$. Summing and using
$$
\sum_{j=2}^{n}\frac1{\log j}\sim\frac n{\log n}
$$
gives $\gamma_n\le(c+o(1))n/\log n$. The last elementary estimate follows by integral comparison with $1/\log t$ and one <integration by parts>. The <Riemann–von Mangoldt formula> would imply
$$
n\le N(\gamma_n)\le\left(\frac c{2\pi}+o(1)\right)n,
$$
a contradiction. This reasoning is valid whether repeated ordinates are listed according to <multiplicity> or only distinct ordinates are listed, since in either case $n\le N(\gamma_n)$. Consequently \b[$\gamma_{n+1}-\gamma_n>c/\log n$ infinitely often], for any fixed $0<c<2\pi$.
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