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Put $F(s)=(s-1)\zeta (s)$. The fractional-part continuation formula for the Riemann zeta function gives $|F(s)|\ll (|s|+1{)}^2$ for $\mathrm{Re}s\ge 1/4$. At $2+ij$, the reciprocal Euler product bounds $|\zeta (2+ij)|$ below by $1/\zeta (2)$. Applying the Jensen zero-count bound with outer radius $7/4$ and inner radius $8/5$ gives $O(\log (|j|+2))$ zeros in the latter disk. Those disks cover the half-strip $1/2\le \mathrm{Re}s\le 1$. Reflecting zeros using the functional equation of the Riemann zeta function completes the $O(T\log T)$ bound without requiring left-half-plane growth estimates.

A nontrivial zero of the Riemann zeta function lies in the critical strip $0<\Re s<1$. The functional equation of the Riemann zeta function reflects such zeros across the critical line $\Re s=1/2$.

First alternative. The Euler product of the Riemann zeta function is absolutely convergent for $\sigma >1$. Its logarithm and the nonnegative trigonometric polynomialgiveConsequently $\zeta (\sigma {)}^3|\zeta (\sigma +it){|}^4|\zeta (\sigma +2it)|\ge 1$. Suppose $t\ne 0$ and the Riemann zeta function had a zero of order $m\ge 1$ at $1+it$. As $\sigma \downarrow 1$, 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 $O((\sigma -1{)}^{4m-3})\to 0$, contradicting its lower bound. Thus 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, defineThe logarithmic derivative identity $-{\zeta }^{\prime }/\zeta (s)={\sum }_n\Lambda (n)n^{-s}$, valid for $\Re s>1$, leads toTo 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 implyon 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 aresince the Euler product bounds $|{\zeta }^{\prime }/\zeta (c+it)|$ by $-{\zeta }^{\prime }/\zeta (c)=O(\log x)$. 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 $a>0$,

It remains to remove the smoothing; differentiating an asymptotic formula would not justify this step. The monotonicity of the Second Chebyshev function gives, for $h=\epsilon x$ with $0<\epsilon <1$,First let $x\to \infty$ with $\epsilon$ fixed. After division by $x$, the lower and upper bounds tend to $1-\epsilon /2$ and $1+\epsilon /2$. Let $\epsilon \downarrow 0$ 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 givesThe integral is $O(x/{\log }^{2}x)$, by splitting at $\sqrt{x}$ and using $\theta (t)=O(t)$. This proves the Prime number theorem with $\pi (x)\sim x/\log x$.

Second alternative. We prove the functional equation of the Riemann zeta function from a theta function. Set $\Theta (u)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}u}$ for $u>0$. With Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(v)e^{-2\pi iv\xi }dv$, the Gaussian function $f(v)=e^{-\pi u{v}^2}$ hasIndeed, differentiation under the integral and integration by parts give ${\hat{f}}^{\prime }(\xi )=-2\pi \xi \hat{f}(\xi )/u$, while the Gaussian integral gives $\hat{f}(0)=u^{-1/2}$. Periodize $f$: the smooth period-one function ${\sum }_{n}f(v+n)$ has Fourier coefficients $\hat{f}(k)$, 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. HenceFor $\Re s>1$, termwise integration and the defining integral of the Gamma function give the Mellin representation of the completed Riemann zeta function:Split at one and substitute $u=1/v$ in the lower integral. The elementary part isand the remaining part becomes an integral over $[1,\infty )$. ThusThe 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 $s\mapsto 1-s$. ThereforeThe 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 isMore 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 $2\to 2+iT\to 1/2+iT$, 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

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 usinggives ${\gamma }_n\le (c+o(1))n/\log n$. The last elementary estimate follows by integral comparison with $1/\log t$ and one integration by parts. The Riemann–von Mangoldt formula would implya 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 ${\gamma }_{n+1}-{\gamma }_n>c/\log n$ infinitely often, for any fixed $0<c<2\pi$.

Use $\Xi (u)={\pi }^{-u/2}\Gamma (u/2)\zeta (u)$ for the completed Riemann zeta function, without the polynomial factor used to define the entire Riemann xi function. For complex $\nu$ putThis is the Modified Bessel function of the second kind. The substitution $t=e^v$ writes it as $\frac{1}{2}{\int }_{\mathbb{R}}{e}^{-Y\cosh v+\nu v}dv$, proving that it is entire in $\nu$, that $K_{\nu }=K_{-\nu }$, and that it decays exponentially as $Y\to \infty$, uniformly for $\nu$ in a compact set.

For $\Re s>1$ all the lattice sums and the integrals below converge absolutely. The $m=0$ terms in the completed nonholomorphic Eisenstein series contribute $\Xi (2s)y^s$. For $m\ne 0$, use the Gamma integral and then Poisson summation in $n$:Here the Poisson identity is ${\sum }_n{e}^{-\pi u(n+mx{)}^2}=u^{-1/2}{\sum }_h{e}^{-\pi h^2/u}{e}^{2\pi ihmx}$. At $h=0$, evaluating the Gamma integral and summing over $m$ gives $\Xi (2s-1)y^{1-s}$. At $h\ne 0$, the substitution $u=|h|t/(|m|y)$ and the integral representation of the modified Bessel function of the second kind giveCollect the frequency $r=mh\ne 0$. There are two signed choices of $m$ for each positive divisor of $|r|$. Thus, with ${\sigma }_a(n)={\sum }_{d\mid n}{d}^a$, the full Fourier expansion isThe factor $2$ is important: the lattice sum includes both signs, while the defining factor $1/2$ was already used in its Mellin integral. This expansion uses exactly the completion in the question.

We can deduce both of the offered conclusions. First recall, with a short proof, the needed functional equation of the Riemann zeta function. Put $\Theta (t)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}t}$. Poisson summation gives $\Theta (t)=t^{-1/2}\Theta (1/t)$. Splitting the Mellin integral of $\Theta -1$ at one yieldsThe integral is entire and invariant under $u\mapsto 1-u$. Hence $\Xi (u)=\Xi (1-u)$, with only simple poles at $u=0,1$, of residues $-1,1$. This is the pole-bearing completion, not the entire xi function.

For fixed $y>0$, the nonconstant part of the Eisenstein expansion is entire in $s$ and locally normally convergent: the divisor sums grow at most polynomially on a compact set of $s$, whereas $K_{s-1/2}(2\pi |r|y)$ decays exponentially in $|r|$. The same argument gives local uniformity for $z$ in compact subsets of the complex upper half-plane. Thus only the constant terms can have poles. Their apparent poles at $s=1/2$ cancel: $\Xi (2s)$ has principal part $1/(2(s-1/2))$, while $\Xi (2s-1)$ has its negative, and both multiply $y^{1/2}$ there. The remaining poles are simple, at $s=0,1$, with residues $-1/2,1/2$.

Under $s\mapsto 1-s$, the two constant terms interchange by the zeta functional equation. The nonconstant terms are unchanged because $K_{\nu }=K_{-\nu }$ andwhich follows by replacing a divisor $d$ by $|r|/d$. Therefore

For the Kronecker limit formula, evaluate the finite Laurent coefficient at $s=1$. Write ${\gamma }_E$ for the Euler--Mascheroni constant. From $\zeta (u)=1/(u-1)+{\gamma }_E+O(u-1)$ and ${\Gamma }^{\prime }(1/2)/\Gamma (1/2)=-{\gamma }_E-2\log 2$ we obtainThe gamma derivative follows by differentiating the Gamma duplication formula and using ${\Gamma }^{\prime }(1)=-{\gamma }_E$. Multiplication by $y^{1-s}$ adds $-\frac{1}{2}\log y$ to the constant term; the other constant term contributes $\Xi (2)y=\pi y/6$.

The half-order Modified Bessel function of the second kind is $K_{1/2}(Y)=\sqrt{\pi /(2Y)}{e}^{-Y}$. For example, in its even integral representation set $v/2$ as variable and then $u=\sinh (v/2)$: $\cosh v=1+2u^2$ and the $\cosh (v/2)$ factor converts the integral to a Gaussian. At $s=1$ the nonconstant Eisenstein coefficients are consequently ${\sigma }_{-1}(|r|)e^{-2\pi |r|y}$. On the other hand, the absolutely convergent logarithmic series givesAdding this identity to its conjugate identifies the nonconstant sum as $-2\log |{\prod }_{n\ge 1}(1-q^n)|$. With the analytic Dedekind eta function product $\eta (z)=q^{1/24}{\prod }_{n\ge 1}(1-q^n)$, the resulting answer isThis derivation of the Kronecker limit formula uses the analytic eta product only; it does not assume its modular transformation. The lattice series is invariant for $\Re s>1$ by an integral change of the pair $(m,n)$ and $\Im (\gamma z)=y/|cz+d{|}^2$. Analytic continuation preserves this invariance, so its finite Laurent coefficient is invariant too. This supplies the independent weight-two transformation argument used in Question 2.

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

Use the Jacobi theta function $\Theta (t)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}t}$, $t>0$. Its transformation law follows directly from the Gaussian Fourier transform. With the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(u)e^{-2\pi i\xi u}du$,For example, differentiating the transform and using integration by parts gives ${\hat{f}}^{\prime }(\xi )=-(2\pi \xi /t)\hat{f}(\xi )$; its value at zero is the Gaussian integral $t^{-1/2}$. Periodize the Gaussian. Its Fourier coefficients are these transforms at integer frequencies, and both series converge absolutely, so evaluating its Fourier series at zero proves the Poisson summation formula here:

For $\Re s>1$, integrating the exponentially convergent series termwise gives the Mellin representation of the completed Riemann zeta functionSplit at one and substitute $u=1/t$ in the lower part, using the transformation just proved. The contribution of $t^{-1/2}-1$ is $1/(s-1)-1/s$. We obtain the pole-subtracted theta integral for the completed zeta functionThe integral is entire: $\Theta (t)-1$ decays exponentially, uniformly dominating the integrand and all its $s$ derivatives on compact sets. Both the rational term and the integral are invariant under $s\mapsto 1-s$. This proves the meromorphic identityEquivalently the Riemann xi functionis entire and satisfies $\xi (s)=\xi (1-s)$. The reflection and duplication formulas for the Gamma function turn the same identity intounderstood by meromorphic continuation at removable exceptional points. This establishes the functional equation of the Riemann zeta function, rather than assuming it.

If $N(T)$ counts zeros $\rho =\beta +i\gamma$ in the critical strip with $0<\gamma \le T$, including zero multiplicity, the Riemann–von Mangoldt formula isAt a zero ordinate one may instead use the conventional half-weight boundary count; that only alters the stated error. In particular $N(T)\sim T\log T/(2\pi )$.

To invert without assuming simple zeros, fix $0<\epsilon <1$ and set $T_n^{\pm }=(1\pm \epsilon )2\pi n/\log n$. Then $\log T_n^{\pm }/\log n\to 1$, soFor large $n$, $N(T_n^{-})<n<N(T_n^{+})$, and hence $T_n^{-}<{\gamma }_n\le T_n^{+}$. Let $\epsilon \downarrow 0$. The asymptotic inversion of the zeta zero count givesAs in the zero-counting formula, the sequence lists ordinates with zero multiplicity; “increasing” means nondecreasing here. The argument does not require the Riemann hypothesis or simplicity of zeros.

Use the Fourier transform convention $\hat{f}(u)={\int }_{\mathbb{R}}f(x)e^{-2\pi iux}dx$. A version of the Poisson summation formula sufficient here isBoth sums converge absolutely. To prove it, form the periodization of a Schwartz function $P(x)={\sum }_{n\in \mathbb{Z}}f(x+n)$. The rapid decay of $f$ and all its derivatives makes this series and its differentiated series uniformly convergent on $[0,1]$, so $P$ is smooth and one-periodic. Its Fourier coefficients areRepeated integration by parts makes these coefficients rapidly decreasing. Thus ${\sum }_m\hat{f}(m)e^{2\pi imx}$ converges absolutely and uniformly and has the same coefficients as $P$. By uniqueness of Fourier series it equals $P$. Evaluation at zero proves the formula.

For $u>0$, the Gaussian $f(x)=e^{-\pi u{x}^2}$ has transform $u^{-1/2}{e}^{-\pi m^2/u}$. This follows from the Gaussian integral at frequency zero and integration by parts, which gives the differential equation ${\hat{f}}^{\prime }(v)=-(2\pi v/u)\hat{f}(v)$. Hence the Jacobi theta functionsatisfies $\Theta (u)=u^{-1/2}\Theta (1/u)$.

Use the normalization of the completed Riemann zeta functionInitially, for $\Re s>1$, the Gamma function integral and absolute convergence give its Mellin transform representationIndeed, the two equal terms for $n$ and $-n$ cancel the factor $1/2$, and the $n$th integral is ${\pi }^{-s/2}\Gamma (s/2)n^{-s}$.

Split at $u=1$ and use the theta transformation in the integral over $(0,1)$. Its elementary part isIn the remaining part substitute $v=1/u$. DefineThe pole-subtracted theta integral for the completed zeta function is thereforeSince $\Theta (u)-1=O(e^{-\pi u})$ for $u\ge 1$, both integrals and every derivative in $s$ converge uniformly on compact subsets of $\mathbb{C}$. Thus $J$ is an entire function. The boxed identity supplies the meromorphic continuation of $\Xi$ to the whole plane, with no other singularities. Its poles really are simple, because the residues of the displayed rational terms are nonzero:The identity also gives $\Xi (s)=\Xi (1-s)$, the completed functional equation of the Riemann zeta function.

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.

Use the meromorphic completionThe functional equation of the Riemann zeta function isIts entire version is $\xi (s)=\frac{1}{2}s(s-1)Z(s)$ with $\xi (s)=\xi (1-s)$. The completion $Z$ itself has poles at zero and one; it should not be confused with the entire Riemann xi function.

We prove the equation by a theta integral. SetThe Gaussian Fourier transform, with kernel $e^{-2\pi ixy}$, isFor example, differentiation in $y$ and integration by parts give the differential equation $I^{\prime }(y)=-(2\pi y/t)I(y)$; the Gaussian integral gives $I(0)=t^{-1/2}$. Periodize the Gaussian by summing $e^{-\pi t(x+n{)}^2}$ over integers $n$. Its $k$th Fourier coefficient is the displayed transform at $k$. Both the periodized function and its Fourier series converge absolutely, so evaluation at $x=0$ gives the Poisson summation formula in this case:

For $\mathrm{Re}s>1$, integration term by term in the Gamma function integral givesIndeed, the two terms for $n$ and $-n$ cancel the factor $1/2$, and substitution $v=\pi n^{2}t$ gives ${\pi }^{-s/2}\Gamma (s/2)n^{-s}$. Split at one, and use the transformation above in the integral from zero to one. The elementary contribution isSubstitution $t=1/u$ in the remaining term therefore givesThe theta tail decays exponentially. The integral is consequently an entire function of $s$, with uniform convergence and termwise differentiation on compact subsets. The formula continues $Z$ meromorphically to the whole plane. Both its rational part and its integral are unchanged by $s\mapsto 1-s$, proving the equation and the entire xi formulation. Using the reflection and duplication identities of the Gamma function, one may equivalently writeunderstood as an identity of meromorphic functions, including points where individual factors need continuation.