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First continue the Riemann zeta function into $\Re s>0$. Summing the integral of the integer-part function gives, initially for $\Re s>1$,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 $3+4\cos v+\cos 2v=2(1+\cos v{)}^2\ge 0$ giveThis proves the three-four-one product proof of zeta boundary nonvanishing inequalityIf $t\ne 0$ and $\zeta (1+it)=0$ to order $r\ge 1$, 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. 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 $-{\zeta }^{\prime }(s)/\zeta (s)={\sum }_n\Lambda (n)n^{-s}$ for $\Re s>1$. The first integral of the Chebyshev function is consequentlyTo 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 $-{\zeta }^{\prime }/\zeta$ 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 $-2,-4,\dots$. Summing these residues yields the smoothed explicit formula for the Chebyshev functionZeros 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 $N(T)=O(T\log T)$, henceIndeed 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 $x\ge 1$. 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

Monotonicity of $\psi$ removes the smoothing. For $0<h<1$,After division by $x$, the limits of the two bounds are $1-h/2$ and $1+h/2$. Let $h\downarrow 0$ 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 givesThe 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 the $n$th prime satisfies

For $\sigma >1$, the absolutely convergent logarithm of the Euler product for the Riemann zeta function isThe elementary identity $3+4\cos v+\cos 2v=2(1+\cos v{)}^2\ge 0$ therefore givesThus 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, fromwhose integral is holomorphic there.

Suppose $t\ne 0$ and $\zeta$ has a zero of order $m\ge 1$ at $1+it$. As $\sigma \downarrow 1$, 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. HenceAt $t=0$ 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:Since $\psi (u)={\sum }_{n\le u}\Lambda (n)$, its first integral of the Chebyshev function isFor any fixed $c>1$, its Mellin inversion formula isFor justification, the elementary contour kernel is $\frac{1}{2\pi i}{\int }_{(c)}{y}^s/(s(s+1))ds=1-y^{-1}$ for $y>1$ and zero for $0<y\le 1$. Close left or right and take the residues at zero and minus one; at $y=1$ the continuous value is zero. Absolute convergence on $\mathrm{Re}s=c$ 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 ${\zeta }^{\prime }(-1)/\zeta (-1)$; 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 isFor 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 henceThe Nontrivial zeros of the Riemann zeta function satisfy $0<\mathrm{Re}\rho <1$, using the Euler product, the functional equation of the Riemann zeta function, and the nonvanishing just proved. For each such zero, $x^{\rho -1}\to 0$ as $x\to \infty$, whereas $|x^{\rho -1}|\le 1$ for $x\ge 1$. 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

It remains to unsmooth; differentiating an asymptotic without justification would not suffice. Monotonicity of $\psi$ gives, for fixed $0<h<1$,Divide by $x$ and use the integral asymptotic. The lower and upper limits lie between $1-h/2$ and $1+h/2$. Letting $h\downarrow 0$ 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 givesThe integral is $O(x/(\log x{)}^2)$, by $\vartheta (t)=O(t)$ and splitting at $\sqrt{x}$. Hencewhich is the Prime number theorem.