Put $q=e^{2\pi i\tau }$ and ${\sigma }_j(n)={\sum }_{d\mid n}{d}^j$. Separate the $m=0$ terms in the lattice Eisenstein series. Differentiating the cosecant partial-fraction identity $2k-2$ times gives, for $\Im z>0$,The initial identity is ${\sum }_n(z+n{)}^{-2}={\pi }^2{\csc }^2(\pi z)=-4{\pi }^2{\sum }_{r\ge 1}r{e}^{2\pi irz}$; differentiation accounts for the factorial and the sign. Pair $m$ and $-m$ and collect the products $mr$. Absolute convergence of the lattice sum for $k\ge 2$, and local uniform convergence of the resulting power series, justify these operations. Thus the requested Fourier expansions areHere $E_{2k}=G_{2k}/(2\zeta (2k))$, and the Bernoulli numbers have convention $B_2=1/6$. The second formula uses Euler's even-value identity for the Riemann zeta function, $2\zeta (2k)=(-1{)}^{k+1}{B}_{2k}(2\pi {)}^{2k}/(2k)!$. In particular $E_4=1+240\sum {\sigma }_3(n)q^n$ and $E_6=1-504\sum {\sigma }_5(n)q^n$.

We need the rational, rather than merely complex, version of the polynomial ring of level-one modular forms. Recall how the valence formula for the modular group supplies its proof. For a nonzero weight-$w$ modular form, with $\rho =e^{2\pi i/3}$,Integrate $f^{\prime }/f$ around a truncated standard fundamental domain of the modular group, indenting around boundary zeros. The paired vertical sides cancel by translation; pair the two halves of the circular edge using $f(-1/\tau )={\tau }^{w}f(\tau )$. The logarithmic derivative of ${\tau }^w$ supplies $w/12$, while the top edge supplies $v_{\infty }(f)$. The half-disc and third-disc at the two elliptic points account for the fractions $1/2,1/3$. The argument principle then gives the formula. All orders in it are nonnegative for a holomorphic modular form. It follows that negative weights have no forms, $M_0=\mathbb{C}$, and $M_2=0$: a weight-two form vanishes at $i$, contributing at least $1/2>2/12$.

Define ${\Delta }_0=(E_4^3-E_6^2)/1728$. It is a weight-twelve cusp form and its expansion starts with $q$. The valence formula for the modular group forces its order at infinity to be exactly one and leaves no interior zeros. Thus division by ${\Delta }_0$ identifies $S_w$ with $M_{w-12}$. Every nonnegative even $w\ne 2$ has a monomial $E_4^a{E}_6^b$ of weight $w$: solve $2a+3b=w/2$, taking $b=0$ when $w/2$ is even and $b=1$ when it is odd and at least three. Subtract the constant coefficient of a modular form times such a monomial and divide by ${\Delta }_0$. Induction on weight proves generation by $E_4,E_6$.

If the original Fourier coefficients are rational, every step preserves rationality: both generators and ${\Delta }_0$ have rational expansions, and formal division by ${\Delta }_0=q+O(q^2)$ does too. This proves the rational structure of level-one modular forms. Write $E_{2k}=\sum r_{a,b}{E}_4^a{E}_6^b$ with $r_{a,b}\in \mathbb{Q}$ and $4a+6b=2k$. Rescaling givesThe Bernoulli number formula above makes each zeta ratio rational: its powers of $\pi$ cancel precisely because the weights agree. This proves the required rational coefficients for the unnormalized lattice Eisenstein sums.

Finally $S_8=S_{10}=0$ because the quotient weights are negative, and $S_{14}=0$ because $M_2=0$. In each of these weights a form is determined by its constant coefficient. The indicated monomials have that coefficient equal to one, so

Fix a positive integer $q$ and an integer $a$ coprime to $q$. We use Dirichlet characters modulo $q$, extended by zero on integers not coprime to $q$, and write ${\chi }_0$ for the principal Dirichlet character. For a nonprincipal Dirichlet character, the sum over one complete period is zero. Its partial sums $A_{\chi }(u)={\sum }_{n\le u}\chi (n)$ are therefore $O(q)$. Partial summation givesThe integral converges locally uniformly there, so these Dirichlet L-functions are holomorphic at one. On the other hand,has a simple pole at one, with residue $\phi (q)/q$.

The assumed nonvanishing of a nonprincipal Dirichlet L-function at one for real Dirichlet characters also excludes vanishing for nonreal ones. To prove this, form the product over all Dirichlet charactersUse the logarithm supplied by the absolutely convergent Euler product. The Orthogonality of Dirichlet characters yieldsThus $\mathcal{F}(\sigma )\ge 1$. If a nonreal Dirichlet character $\chi$ had $L(1,\chi )=0$, its distinct conjugate $\overline{\chi }$ would also have a zero there, since $L(1,\overline{\chi })=\overline{L(1,\chi )}$. These two zeros would outweigh the single pole of the principal factor; all other factors are bounded near one. It would follow that $\mathcal{F}(\sigma )\to 0$ as $\sigma \downarrow 1$, contradicting the lower bound. Together with the assumption for real Dirichlet characters, this proves $L(1,\chi )\ne 0$ for every nonprincipal Dirichlet character.

The first-degree part of the Euler product logarithm givesThe omitted terms are bounded uniformly, since ${\sum }_p{\sum }_{m\ge 2}(m{p}^{m\sigma }{)}^{-1}\ll {\sum }_p{p}^{-2}<\infty$. For each nonprincipal factor, its logarithm remains bounded: choose a holomorphic logarithm near the now established nonzero value $L(1,\chi )$; the Euler-product branch differs from it by a fixed integral multiple of $2\pi i$ on a sufficiently small real interval. For the principal factor,Applying Orthogonality of Dirichlet characters again givesThe right side diverges as $\sigma \downarrow 1$, so the arithmetic progression contains infinitely many primes. This proves Dirichlet's theorem on arithmetic progressions: every reduced residue class modulo $q$ contains infinitely many primes. The argument also covers $q=1$ and $q=2$, when there need not be a nonreal character.

For the sharper statement about reciprocal primes in a fixed arithmetic progression, use the given prime-counting function asymptotic with $q$ fixed:The second equivalence follows by integration by parts in the logarithmic integral function. Partial summation yieldsFor any $\epsilon >0$, choose a fixed $T$ so large that the relative error in $\pi (t;q,a)$ is at most $\epsilon$ for $t\ge T$. The initial integral is a constant, while the remaining integral lies betweenThe boundary term is $O_q(1/\log x)$. Divide by $\log \log x$, let $x\to \infty$, and then let $\epsilon \downarrow 0$. We obtainOnly a relative asymptotic is deduced from the given information; an additive $O(1)$ error, or uniformity for a growing modulus $q$, would require additional estimates.

The Selberg upper-bound sieve replaces the indicator function of a sifted set by the square of a divisor sum. If the real Selberg sieve weights satisfy ${\lambda }_1=1$, thenWhen $F(n)$ has no forbidden prime factors, the divisor sum is exactly one; otherwise its square is still nonnegative. Taking the weights to vanish for $d>z$ controls the error in summing this majorant. The main term is a positive quadratic form that can be diagonalized and minimized explicitly. This optimization, together with a bound for the remainder, is the central idea of the Selberg sieve.

For twin primes, take $F(n)=n(n+2)$ and let $\rho (d)$ be the number of its roots modulo a squarefree integer $d$. The Chinese remainder theorem makes $\rho$ a multiplicative arithmetic function, with $\rho (2)=1$ and $\rho (p)=2$ for odd primes. Thus the polynomial root density in a sieve is $g(d)=\rho (d)/d$, andSumming the majorant giveswhere $Q(\lambda )={\sum }_{d,e\le z}{\lambda }_d{\lambda }_{e}g([d,e])$, with all indices squarefree. Here $\rho ([d,e])\le \rho (d)\rho (e)$ controls the error.

We carry out the Selberg sieve diagonalization. PutFor squarefree integers $d,e$, multiplicativity givesThe identity behind the first equality is ${\prod }_{p\mid (d,e)}[1+h(p)]=1/g((d,e))$. Möbius inversion gives ${\lambda }_1={\sum }_r\mu (r)y_r=1$. The Cauchy-Schwarz inequality therefore impliesDefine $G(z)={\sum }_{r\le z,r\text{ squarefree}}1/h(r)$. Equality is attained at $y_r=\mu (r)/(h(r)G(z))$, giving $\min Q=1/G(z)$. Inverting again, the optimizing Selberg sieve weights areIn particular ${\lambda }_1=1$. Since $G_d(z/d)\le G(z)$,The last inequality holds separately at $p=2$, $p=3$, and $p\ge 5$. Together with $\rho (d)\le d$, this crude bound already makes the remainder $O(z^6)$, because ${\sum }_{d\le z}{d}^2=O(z^3)$.

We also need a lower bound for $G(z)$; we prove it rather than presuppose a sieve asymptotic. For odd squarefree integers $r$,where $\omega (r)$ counts distinct prime factors. Put $y=\sqrt{z}$ and $L(y)={\sum }_{a\le y,a\text{ odd}}1/a=\frac{1}{2}\log y+O(1)$. The total weight $1/(ab)$ of odd pairs $a,b\le y$ is $L(y{)}^2$. Exclude pairs for which $a$ has a square prime factor, $b$ has a square prime factor, or $a,b$ have a common prime factor. For each of these three categories the total weight is at most $L(y{)}^2{\sum }_{p\text{ odd}}{p}^{-2}$: for example, the weighted count of odd multiples of $p^2$ is $p^{-2}L(y/p^2)\le p^{-2}L(y)$, and the common-factor category uses $p^{-1}L(y/p)$ for each variable. Moreover,Hence the remaining odd coprime squarefree pairs have weight at least $L(y{)}^2/6$. Their products $r=ab\le z$ are squarefree, and $2^{\omega (r)}$ counts all ordered coprime factorizations of $r$. This proves the elementary lower bound for the twin-prime sieve denominator:

If $p>z$ and both $p,p+2$ are primes, $F(p)$ survives the sieve. The at most $z$ smaller values are the only exceptions. Thus, with $T(x)=\#\{p\le x:p,p+2\text{ prime}\}$ and $z=x^{1/10}$, the twin-prime upper bound from a quadratic sieve becomesFinally partial summation givesThe integral over $[3,\infty )$ is bounded by a constant times ${\int }_3^{\infty }dt/(t(\log t{)}^2)<\infty$. The nonnegative partial sums are therefore bounded and increasing. Thus the reciprocal series over the smaller members of twin-prime pairs converges, a form of Brun's theorem. The upper bound does not assert that infinitely many twin primes exist.

Write $s=\sigma +it$ and let $\rho =\beta +i\gamma$ run over Nontrivial zeros of the Riemann zeta function, including multiplicity. The bounds involving $\log |t|$ below are for $|t|\ge 3$. At bounded heights their uniform versions use $\log (|t|+3)$; $\log |t|$ itself cannot be the right bound near zero or $|t|=1$.

Start with the Hadamard factorization of the Riemann xi function, an entire function of order one. Its logarithmic derivative has the convergent expansionDifferentiating $\xi (s)=\frac{1}{2}s(s-1){\pi }^{-s/2}\Gamma (s/2)\zeta (s)$ givesThe real sum of $1/\rho$ converges absolutely: $0\le \beta \le 1$ makes its terms $O({\gamma }^{-2})$ at large height, and the order-one Hadamard factorization ensures the required summability. Thus these terms and $\Re B$ form a fixed constant. The Stirling formula for the digamma function gives, uniformly for $1<\sigma \le 2$ and $|t|\ge 3$,This is the useful real form of the global partial-fraction expansion of the zeta logarithmic derivative. Every term subtracted is nonnegative because the Euler product excludes zeros with $\beta >1$. At the simple pole $s=1$, one also has $F(\sigma )=1/(\sigma -1)+O(1)$ as $\sigma \downarrow 1$.

The three-four-one zero-free-region argument now detects a zero too close to one. The Von Mangoldt function is nonnegative and the absolutely convergent Dirichlet series givesbecause the corresponding coefficient at $n$ is $\Lambda (n)n^{-\sigma }[3+4\cos (t\log n)+\cos (2t\log n)]$. If $\beta +it$ is a zero, retain its term in the expansion at height $t$ and discard the other nonpositive terms. The expansion at height $2t$ gives an upper bound even without retaining a particular zero. We obtain, for an absolute constant $C_0>0$,Choose $0<a<1$ so small that $5/(9a)>C_0$, put $b=a/8$, and take $\sigma =1+a/\log |t|$. If $1-\beta \le b/\log |t|$, the last inequality would requirea contradiction. Thereforefor some fixed $b>0$. This proves the required zero-free region of the Riemann zeta function. Nonvanishing on $\Re s=1$ at bounded nonzero heights, proved in Question 2, and the isolation of zeros allow a decrease of $b$ giving the uniform formulation with $\log (|t|+3)$ as well. Near $s=1$, apply this observation to the nonzero holomorphic function $(s-1)\zeta (s)$.

Next evaluate the real expansion at $\sigma =2$. The absolutely convergent Dirichlet series bounds $|{\zeta }^{\prime }/\zeta (2+it)|\le {\sum }_n\Lambda (n)n^{-2}$ uniformly in $t$, soSince $1\le 2-\beta \le 2$, each term on the left is at least $1/[4+(t-\gamma {)}^2]$. The smoothed zeta zero-count bound follows:Each zero satisfying $|t-\gamma |<1$ contributes more than $1/5$, so this also gives the local zero count for the Riemann zeta function:The same kernel sum is bounded at compact heights, since its tail converges uniformly there; the bounds with $\log (|t|+3)$ are consequently valid for all real $t$.

For completeness, the contour estimate used in Question 2 follows from these very formulas. Subtract the complex logarithmic derivative expansion at $2+it$ from that at $\sigma +it$, with $1-b/(2\log (T+3))\le \sigma \le 2$ and $|t|\le T$. Terms with $|t-\gamma |>1$ have differences $O((t-\gamma {)}^{-2})$, whose sum is $O(\log (|t|+3))$ by the kernel bound. There are $O(\log (|t|+3))$ remaining terms. After reducing $b$, the zero-free region of the Riemann zeta function separates every such zero horizontally from the half-width boundary by $\gg 1/\log (T+3)$: their heights satisfy $|\gamma |\le T+1$. Thus on the left edge, and on the horizontal edges at $t=\pm T$, each reciprocal is $O(\log (T+3))$. The Gamma function and pole terms satisfy the same required bound; at small heights on the left edge the pole distance is $\gg 1/\log (T+3)$. This proves $|{\zeta }^{\prime }/\zeta |\ll {\log }^2(T+3)$ on those edges and justifies the contour shift without choosing heights that dodge unknown zeros.

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 $0<c<2\pi$.

Write $\theta (x)={\sum }_{p\le x}\log p$ for the Chebyshev theta function and $\psi (x)={\sum }_{n\le x}\Lambda (n)$ for the Second Chebyshev function, where $\Lambda$ is the Von Mangoldt function. We first establish the Chebyshev estimate. Every prime $n<p\le 2n$ divides the central binomial coefficient $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$, soApply this at successive powers of two and use monotonicity to obtain $\theta (x)=O(x)$. Sincethe terms $r\ge 2$ contribute $O(\sqrt{x}\log x)$ and therefore $\psi (x)=O(x)$. This Chebyshev estimate from central binomial coefficients is all the information about the distribution of primes needed here.

The Von Mangoldt divisor identity $\log n={\sum }_{d\mid n}\Lambda (d)$ gives the factorial proof of Mertens first theorem. With $N=\lfloor x\rfloor$, interchange the finite sums to obtainComparison of ${\sum }_{n\le N}\log n$ with ${\int }_1^N\log tdt$ gives $\log N!=N\log N-N+O(\log N)$. Dividing the preceding identity by $x$, using $N=x+O(1)$ and the Chebyshev estimate, givesThe contribution of higher prime powers is bounded independently of $x$, becauseThus the Mertens first theorem is

For the square-root logarithmic sum over primes, apply partial summation to $A(t)$ with weight $(\log t{)}^{-1/2}$. The lower endpoint is understood as $A(2^{-})=0$, soThe main terms are $\sqrt{\log x}$ and $\sqrt{\log x}-\sqrt{\log 2}$. The bounded error in $A(t)$ produces a bounded integral, since ${\int }_2^{\infty }dt/(t(\log t{)}^{3/2})<\infty$. ThereforeNeither argument uses the Prime number theorem.