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.

Let $a_0$ be the constant Fourier coefficient and suppose first that $k>0$ is even and $M_k\ne 0$; thus $k\ge 4$. Put $\epsilon =i^k=(-1{)}^{k/2}$. The precise Mellin continuation of a noncuspidal modular form isIt is meromorphic on the whole plane, with possible simple poles only at $0,k$, of residues $-a_0,\epsilon a_0$, respectively. For a cusp form it is an entire function. A nonzero constant term makes both of these poles genuine.

We first justify the initial Mellin transform. Decompose $f=a_0{E}_k+h$ with $h\in S_k$. For a cusp form, $y^{k/2}|h(x+iy)|$ is invariant under the modular group and bounded on its fundamental domain, since it tends to zero at the cusp. Therefore it is bounded throughout the complex upper half-plane. Integrating the Fourier coefficient at height $y=1/n$ gives $|a_n(h)|\le C{e}^{2\pi }{n}^{k/2}$. The Eisenstein series coefficients satisfy ${\sigma }_{k-1}(n)\le \zeta (k-1)n^{k-1}$ by replacing divisors $d$ by $n/d$. Consequently $a_n(f)=O(n^{k-1})$, and absolute termwise integration is valid for $\Re s>k$:Indeed ${\int }_0^{\infty }{e}^{-2\pi ny}{y}^{s-1}dy=(2\pi n{)}^{-s}\Gamma (s)$, and the sum of the absolute integrals converges in that half-plane.

Write $I_f(s)={\int }_1^{\infty }(f(iy)-a_0)y^{s-1}dy$. Exponential cusp decay of the difference implies uniform convergence on compact subsets of the $s$-plane, including after any number of derivatives in $s$, so $I_f$ is entire. The modular transformation $f(i/y)=(iy{)}^{k}f(iy)$ givesOn the lower half of the Mellin integral, subtract the transformed constant explicitly:Substitute $y=1/t$ in the decaying term and integrate the two powers in the constant term. This yields, initially for $\Re s>k$,The right side supplies the claimed meromorphic continuation and its residues. Replacing $s$ by $k-s$ and multiplying by $\epsilon$, with ${\epsilon }^2=1$, leaves this expression unchanged, proving the functional equation with its correct sign. In particular the residue at $k$ is tied to the residue at zero by that equation; subtracting only $a_0$ at infinity and overlooking its transformed value at zero would miss these poles.

In weight zero every modular form is constant, so $L(f,s)=0$ and its completion is identically zero. The same holds for the zero form in odd, negative, or weight-two spaces. Thus the assertion covers all weights in the notation of the paper. There is no conjugation of $f$ in this level-one functional equation.

Write $\Gamma =S{L}_2(\mathbb{Z})$ and use the determinant-normalized slash operatorThe automorphy factor identity proves $(f{|}_k\alpha ){|}_k\beta =f{|}_k(\alpha \beta )$, and positive scalar matrices act trivially. Let ${\mathcal{M}}_n$ be all integral matrices of determinant $n>0$. It is stable under left and right multiplication by $\Gamma$. Its left orbits have unique representativesIndeed Bézout's identity reduces the first column to $(a,0)$, the determinant fixes $d=n/a$, and a row shear reduces $b$ modulo $d$. These are the determinant-n matrix representatives for Hecke operators. For composite $n$, ${\mathcal{M}}_n$ can contain more than one double coset; keeping every integral matrix, including nonprimitive ones, is essential.

Define the Hecke operator byChanging a representative on the left leaves its summand unchanged. Right multiplication by $\Gamma$ permutes left orbits of ${\mathcal{M}}_n$, so the result satisfies the weight-$k$ transformation law. Each summand is holomorphic. Its expansion at infinity is bounded, and vanishes there when $f$ is a cusp form; the group has just one cusp class. Therefore $T_n$ preserves both $M_k$ and $S_k$, and $T_1$ is the identity.

For a Fourier expansion $f={\sum }_{r\ge 0}{a}_r{q}^r$, the sum over $b$ vanishes unless $d\mid r$. Substitute $r=dt$ and then collect the coefficient of $q^m$ to obtain the Fourier coefficients of a composite-index Hecke operator:Here $\gcd (0,n)=n$, so the constant coefficient is ${\sigma }_{k-1}(n)a_0$; at $m=1$ it is $a_n$. This proves preservation of integral coefficients. At a prime,with $a_{m/p}=0$ if $p\nmid m$.

We can prove the entire Hecke multiplication relations directly, not just quote them. On formal power series put $U_p(\sum a_m{q}^m)=\sum a_{pm}{q}^m$ and $V_{p}f(q)=f(q^p)$. Although these individual operators need not preserve level one, $U_p{V}_p=1$ and the coefficient formula givesMultiply on the left by $T_p=U_p+p^{k-1}{V}_p$. In the first part the $j=0$ term is $U_p^{r+1}$; every $j\ge 1$ reduces by $U_p{V}_p=1$ to $p^{j(k-1)}{V}_p^{j-1}{U}_p^{r-j}$. The second part, together with the $j=0$ term, is $T_{p^{r+1}}$. The remaining sum is $p^{k-1}{T}_{p^{r-1}}$. ThusFor coprime indices, the coefficient divisors split independently in the displayed formula, proving $T_m{T}_n=T_{mn}$ when $(m,n)=1$. Induction with the prime-power recurrence gives $T_{p^r}{T}_{p^t}={\sum }_{j=0}^{\min (r,t)}{p}^{j(k-1)}{T}_{p^{r+t-2j}}$. Combining the primes provesIn particular the modular Hecke algebra is commutative and is generated by the prime operators.

For cusp forms the Petersson inner product isCusp decay makes it finite and positive definite. The Hecke operators are self-adjoint for the Petersson inner product. To see the change-of-variables mechanism, fix a double coset $\Gamma \alpha \Gamma \subset {\mathcal{M}}_n$ and put $H=\Gamma \cap {\alpha }^{-1}\Gamma \alpha$. Representatives $\gamma$ of $H\backslash \Gamma$ give left representatives $\alpha \gamma$. Unfolding the sum in $⟨{\sum }_{\gamma }f{|}_k\alpha \gamma ,g⟩$ changes the integral over $\Gamma \backslash \mathbb{H}$ to the integral over $H\backslash \mathbb{H}$ of $(f{|}_k\alpha )\overline{g}$. The substitution $w=\alpha \tau$ preserves $dxdy/y^2$ and has $\Im w=(\det \alpha )y/|c\tau +d{|}^2$; these factors move the slash from $f$ to $g{|}_k{\alpha }^{-1}$. Refold over $\alpha H{\alpha }^{-1}=\Gamma \cap \alpha \Gamma {\alpha }^{-1}$ to get the reverse double coset. Since scalar matrices act trivially, ${\alpha }^{-1}$ has the same slash action as its adjugate ${\alpha }^{\ast }=n{\alpha }^{-1}$, an integral matrix of determinant $n$. The adjugate involution permutes all the double cosets in ${\mathcal{M}}_n$. The common real factor $n^{k/2-1}$ is unchanged, giving $⟨T_{n}f,g⟩=⟨f,T_{n}g⟩$.

Since $S_k$ is finite dimensional, commuting self-adjoint linear operators have a common orthonormal basis of Hecke eigenforms: diagonalize one operator, restrict the others to its eigenspaces, and repeat until these spaces cannot split further. If $f$ is a nonzero common Hecke eigenform with $T_{n}f={\lambda }_{n}f$, the coefficient at $q$ gives $a_n={\lambda }_n{a}_1$. Were $a_1=0$, every positive coefficient would vanish, contrary to $f\ne 0$ being cuspidal. Normalize $a_1=1$; then ${\lambda }_n=a_n$, and the Hecke multiplication relations yieldThe eigenvalues are real by self-adjointness. Their multiplicativity and prime-power recurrence give the Euler product of a Hecke eigenform, initially in an absolutely convergent half-plane:The Mellin transform in Question 5 supplies its analytic continuation and functional equation.

Finally, for positive even $k\ge 4$, $M_k=\mathbb{C}{E}_k\oplus S_k$, by subtracting the constant coefficient times the normalized Eisenstein series. The divisor identityfollows by multiplying the geometric sums at each prime; it and the constant-coefficient formula prove $T_n{E}_k={\sigma }_{k-1}(n)E_k$. Thus $M_k$ has an Eisenstein series eigenline and a basis of cuspidal Hecke eigenforms. Odd and negative weights have no forms, $M_2=0$, and on $M_0=\mathbb{C}$ the operator is the scalar ${\sigma }_{-1}(n)$. These statements fix the normalization and describe the exceptional small weights as well.

Extend $\chi$ by zero to integers not coprime to $N$, and write $g(\chi )={\sum }_{a\mathrm{mod}N}\chi (a)e^{2\pi ia/N}$ for its Gauss sum of a Dirichlet character. We use the analytic branch of $\sqrt{-i\tau }$ that is positive when $\tau$ is positive imaginary. With the convention $\hat{F}(t)={\int }_{\mathbb{R}}F(x)e^{-2\pi ixt}dx$, the Gaussian Fourier transform and its differentiated version areThe first formula follows by completing the square, first on the imaginary axis and then by analytic continuation; the second follows by differentiating the Fourier transform in $t$. All the functions are Schwartz functions on the real axis.

For completeness, the finite Fourier transform of a primitive Dirichlet character satisfiesFor a unit $r$ this is substitution by its inverse. If $p\mid (r,N)$, primitivity supplies a unit $u\equiv 1(\mathrm{mod}N/p)$ with $\chi (u)\ne 1$: otherwise the character would factor through modulus $N/p$. Multiplication by $u$ fixes the additive exponential but multiplies the sum by $\chi (u)$, so the sum vanishes. This proves the identity also on nonunits. Finite Parseval identity now gives $|g(\chi ){|}^2=N$, since exactly $\phi (N)$ frequencies have nonzero character values. In particular this Gauss sum of a Dirichlet character is nonzero.

Apply Poisson summation to each residue class: ${\sum }_{j}F(a+Nj)=N^{-1}{\sum }_r{e}^{2\pi ira/N}\hat{F}(r/N)$. Multiply by $\chi (a)$ and sum over $a$. The formulas above give the two character theta transformationsandIn the odd case the prefactor originally reads $g(\chi )/(N^2\tau \sqrt{-i\tau })$; using $\tau =i(-i\tau )$ explains the factor $i$. Opposite parity makes the corresponding series identically zero by pairing $n,-n$. This is why the weight-three-halves series is needed for odd Dirichlet characters.

To deduce the products from these transformations, use two specific primitive Dirichlet characters. The even character ${\chi }_{12}$ takes values $1,-1,-1,1$ on $1,5,7,11$ modulo twelve. It is primitive: its values do not factor through six or four, hence through no proper divisor of twelve. Its Gauss sum of a Dirichlet character is $2\sqrt{3}=\sqrt{12}$. PutThe even transformation gives $A(-1/\tau )=\sqrt{-i\tau }A(\tau )$; also $A(\tau +1)=e^{\pi i/12}A(\tau )$, since each nonzero term has $n^2\equiv 1(\mathrm{mod}24)$. Thus $A^{24}$ transforms as a weight-twelve modular form under the generators $S,T$ of the modular group. It is holomorphic and starts with $q$, so it is a normalized cusp form. The proof in Question 1 shows $S_{12}=\mathbb{C}{\Delta }_0$, giving $A^{24}={\Delta }_0$. In particular $A$ has no interior zeros, by the valence formula for the modular group.

The Serre derivative argument in Question 2 gives $D{\Delta }_0=E_2{\Delta }_0$. Hence $DA/A=E_2/24$. The analytic product $P(\tau )=q^{1/24}{\prod }_{n\ge 1}(1-q^n)$ is nonzero and has the same logarithmic derivative, since $DP/P=1/24-\sum {\sigma }_1(n)q^n$. Therefore $A/P$ is constant on the connected complex upper half-plane; its limit at infinity is one. This proves $A=P$, without assuming either of the product identities being deduced. Pair the terms $n,-n$ in $A$, choosing representatives $n=6r+1$. Since ${\chi }_{12}(6r+1)=(-1{)}^r$, we getThis is the Euler pentagonal identity, derived by the character theta proof of the eta product.

For the second product use the odd primitive character ${\chi }_4(1)=1$, ${\chi }_4(3)=-1$, with Gauss sum of a Dirichlet character $2i$. SetThe odd transformation gives $B(-1/\tau )=(-i\tau {)}^{3/2}B(\tau )$ and the series gives $B(\tau +1)=e^{\pi i/4}B(\tau )$. Thus $B^8$ is another normalized weight-twelve cusp form, and $B^8={\Delta }_0=A^{24}$. Because $A$ is nonzero, $B/A^3$ has eighth power one, so it is constant; its leading coefficient fixes it to one. Pairing positive and negative odd integers now givesSince $B=A^3=q^{1/8}{\prod }_{n\ge 1}(1-q^n{)}^3$, the Jacobi cubic product identity follows:All pairings and differentiations take place in locally uniformly convergent theta functions or products on $|q|<1$.

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 ${\sum }_{d\le z}{d}^2=O(z^3)$.