= Solution
First continue the <Riemann zeta function> into $\Re s>0$. Summing the integral of the integer-part function gives, initially for $\Re s>1$,
$$
\zeta(s)=s\int_1^\infty\lfloor u\rfloor u^{-s-1}\,du
=\frac{s}{s-1}-s\int_1^\infty\{u\}u^{-s-1}\,du.
$$
The last integral defines a <holomorphic function> for $\Re s>0$, so this continuation has only a simple <pole> at one, of <residue> one.
For $\sigma>1$, logarithmic expansion of the <Euler product> and $3+4\cos v+\cos2v=2(1+\cos v)^2\ge0$ give
$$
3\log\zeta(\sigma)+4\log|\zeta(\sigma+it)|+\log|\zeta(\sigma+2it)|
=\sum_p\sum_{j\ge1}\frac{3+4\cos(jt\log p)+\cos(2jt\log p)}{jp^{j\sigma}}\ge0.
$$
This proves the <three-four-one product proof of zeta boundary nonvanishing> inequality
$$
\zeta(\sigma)^3|\zeta(\sigma+it)|^4|\zeta(\sigma+2it)|\ge1.
$$
If $t\ne0$ and $\zeta(1+it)=0$ to order $r\ge1$, the middle factor is $O((\sigma-1)^{4r})$, the first is $O((\sigma-1)^{-3})$, and the last remains bounded. Their product tends to zero, a contradiction. At $t=0$ there is a <pole>, not a zero. \b[The zeta function has no zero on $\Re s=1$.]
Put $I(x)=\int_0^x\psi(u)\,du=\sum_{n\le x}(x-n)\Lambda(n)$, where $\Lambda$ is the <Von Mangoldt function>. Logarithmic differentiation of the <Euler product> gives $-\zeta'(s)/\zeta(s)=\sum_n\Lambda(n)n^{-s}$ for $\Re s>1$. The <first integral of the Chebyshev function> is consequently
$$
\boxed{I(x)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}
-\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s+1}}{s(s+1)}\,ds,\qquad c>1.}
$$
To check its kernel, evaluate the <contour integral> of $y^s/(s(s+1))$ by closing to the left for $y>1$ and to the right for $0<y<1$. The <residues> at zero and minus one give $1-y^{-1}$ in the first case and zero in the second. Multiplication by $x$ and substitution $y=x/n$ therefore produce $(x-n)_+$. On the line $c>1$, the <Dirichlet series> and vertical integral have <absolute convergence>, justifying their interchange; at $x=n$ the kernel is zero on either side.
Here is the smoothing step that establishes the requested asymptotic. Shift the line of this <contour integral> to the left. The <pole> of $-\zeta'/\zeta$ at one has <residue> $+1$; at each <Nontrivial zero of the Riemann zeta function> $\rho$ of <zero multiplicity> $m_\rho$ its <residue> is $-m_\rho$. The denominators also contribute at zero and minus one, and the <trivial zeros of the Riemann zeta function> occur at $-2,-4,\ldots$. Summing these <residues> yields the <smoothed explicit formula for the Chebyshev function>
$$
I(x)=\frac{x^2}2-\sum_\rho\frac{x^{\rho+1}}{\rho(\rho+1)}
-x\log(2\pi)+\frac{\zeta'(-1)}{\zeta(-1)}
-\sum_{k\ge1}\frac{x^{1-2k}}{2k(2k-1)}.
$$
Zeros are repeated with <zero multiplicity>. For the limits of the <contour integral>, choose horizontal heights avoiding zero ordinates, where the logarithmic derivative on each fixed strip is $O((\log T)^2)$. The factor $s(s+1)$ makes their integrals tend to zero. On left edges chosen a fixed distance from the trivial zeros, the <functional equation of the Riemann zeta function> controls the logarithmic derivative by $O(\log(|s|+2))$; for fixed $x>1$ the remaining vertical integral vanishes as that edge moves left. This is why the extra integration of $\psi$ is useful.
The <Riemann–von Mangoldt formula>, stated in Question 3, gives $N(T)=O(T\log T)$, hence
$$
\sum_\rho\frac1{|\rho(\rho+1)|}<\infty.
$$
Indeed each dyadic height block contributes $O((\log T)/T)$. The zero-free line just proved, the nonvanishing <Euler product> on its right, and the <functional equation of the Riemann zeta function> put every <Nontrivial zero of the Riemann zeta function> in $0<\Re\rho<1$. Therefore each $x^{\rho-1}$ tends to zero, with absolute value at most one for $x\ge1$. Divide the explicit formula by $x^2$ and use <dominated convergence> on the zero terms with <absolute convergence>. The remaining displayed terms are $o(x^2)$, and we obtain
$$
\boxed{I(x)\sim x^2/2.}
$$
Monotonicity of $\psi$ removes the smoothing. For $0<h<1$,
$$
\frac{I(x)-I((1-h)x)}{hx}\le\psi(x)
\le\frac{I((1+h)x)-I(x)}{hx}.
$$
After division by $x$, the limits of the two bounds are $1-h/2$ and $1+h/2$. Let $h\downarrow0$ to get $\psi(x)\sim x$. Question 1 gives $\psi(x)-\vartheta(x)=O(\sqrt x\log x)=o(x)$, so $\vartheta(x)\sim x$. <Partial summation> now gives
$$
\pi(x)=\frac{\vartheta(x)}{\log x}+\int_2^x\frac{\vartheta(t)}{t(\log t)^2}\,dt
\sim\frac{x}{\log x}.
$$
The integral is $O(x/(\log x)^2)$, by splitting at $\sqrt x$ and using $\vartheta(t)=O(t)$. Inverting this at $x=p_n$, we have $n\sim p_n/\log p_n$ and hence $\log n\sim\log p_n$. Thus \b[the $n$th <prime> satisfies]
$$
\boxed{p_n\sim n\log n.}
$$
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