## Formalized Name Major Arc Main Term for the Ternary Goldbach von Mangoldt Integral ## Formalized Statement Let $e:\mathbb R\to\mathbb C$ be defined by $e(t):=\exp(2\pi i t)$. Let $\mathbb T:=\mathbb R/\mathbb Z$, let $\mu_{\mathbb T}$ be normalized Haar probability measure on $\mathbb T$, and let $\|\cdot\|_{\mathbb T}$ denote distance to $0$ in $\mathbb T$. Let $\mathcal L^1$ and $\mathcal L^2$ denote one-dimensional and two-dimensional Lebesgue measure. Let $\Lambda:\mathbb N\to\mathbb R$ be the von Mangoldt function, let $\mu:\mathbb N\to\{-1,0,1\}$ be the Mobius function, and let $\varphi:\mathbb N\to\mathbb N$ be Euler's totient function. For each $N\in\mathbb N$, define $S_N:\mathbb T\to\mathbb C$ by \begin{align*} S_N(\alpha):=\sum_{1\le n\le N}\Lambda(n)e(n\alpha). \end{align*} For $Q,L\ge 1$ and $N\in\mathbb N$, define \begin{align*} \mathfrak M_N(Q,L):=\bigcup_{1\le q\le Q}\bigcup_{\substack{1\le a\le q, \gcd(a, q)=1}}\left\{\alpha\in\mathbb T:\left\|\alpha-\frac{a}{q}\right\|_{\mathbb T}\le \frac{L}{qN}\right\}. \end{align*} For $q\in\mathbb N$ and $m\in\mathbb Z$, define \begin{align*} c_q(m):=\sum_{\substack{1\le a\le q, \gcd(a, q)=1}}e\left(\frac{am}{q}\right). \end{align*} For $N\in\mathbb N$, define the ternary Goldbach singular series by the absolutely convergent Euler product \begin{align*} \mathfrak S_3(N):=\prod_p\left(1-\frac{c_p(-N)}{(p-1)^3}\right), \end{align*} equivalently by the absolutely convergent Ramanujan series \begin{align*} \mathfrak S_3(N)=\sum_{q=1}^{\infty}\frac{\mu(q)^3}{\varphi(q)^3}c_q(-N). \end{align*} Then, for every $C>0$, there exist $B=B(C)>0$ and $N_0=N_0(C)\in\mathbb N$ such that, if $N\ge N_0$ is odd and $Q=L=(\log N)^B$, then \begin{align*} \int_{\mathfrak M_N(Q,L)}S_N(\alpha)^3e(-N\alpha)\,d\mu_{\mathbb T}(\alpha)=\frac{1}{2}\mathfrak S_3(N)N^2+O_C\left(\frac{N^2}{(\log N)^C}\right). \end{align*} Moreover, there exists an absolute constant $c>0$ such that \begin{align*} \mathfrak S_3(N)\ge c \end{align*} for every odd $N\in\mathbb N$. ## Proof [proofplan] We first isolate disjoint major arcs around reduced rationals and invoke the standard Siegel-Walfisz major-arc estimate for the von Mangoldt exponential sum, stated in the precise logarithmic range needed here. Replacing $S_N$ by its principal term turns the integral into a product of an arithmetic Ramanujan sum and a continuous Fourier integral. The continuous integral is the Fourier representation of the volume of the simplex $x_1+x_2+x_3=N$ in $[0,N]^3$, while the arithmetic sum is the truncated singular series. Finally, a uniform squarefree tail estimate justifies replacing the truncation by $\mathfrak S_3(N)$, and the Euler product gives a positive lower bound for odd $N$. [/proofplan] [step:Make the logarithmic major arcs disjoint] Fix $C>0$. Choose a parameter $B>0$, to be enlarged finitely many times in terms of $C$, and set \begin{align*} Q=L=(\log N)^B. \end{align*} For $1\le q\le Q$ and $1\le a\le q$ with $\gcd(a,q)=1$, define \begin{align*} \mathfrak M_N(a,q;L):=\left\{\alpha\in\mathbb T:\left\|\alpha-\frac{a}{q}\right\|_{\mathbb T}\le \frac{L}{qN}\right\}. \end{align*} If $a/q$ and $a'/q'$ are distinct reduced fractions with $1\le q,q'\le Q$, then \begin{align*} \left\|\frac{a}{q}-\frac{a'}{q'}\right\|_{\mathbb T}\ge \frac{1}{qq'}\ge \frac{1}{Q^2}. \end{align*} The sum of the two radii is at most \begin{align*} \frac{L}{qN}+\frac{L}{q'N}\le \frac{2LQ}{N}. \end{align*} Since \begin{align*} \frac{2LQ^3}{N}=\frac{2(\log N)^{4B}}{N}\to 0 \end{align*} as $N\to\infty$, there exists $N_1(B)\in\mathbb N$ such that for every $N\ge N_1(B)$ the arcs $\mathfrak M_N(a,q;L)$ with $q\le Q$ are pairwise disjoint. [/step] [step:Insert the standard Siegel-Walfisz major arc approximation] Define the continuous model integral $V_N:\mathbb R\to\mathbb C$ by \begin{align*} V_N(\beta):=\int_0^N e(\beta t)\,d\mathcal L^1(t). \end{align*} For $\beta\ne 0$, the identity \begin{align*} V_N(\beta)=\frac{e(\beta N)-1}{2\pi i\beta} \end{align*} gives \begin{align*} |V_N(\beta)|\le \min\left\{N,\frac{1}{\pi |\beta|}\right\}, \end{align*} and $|V_N(0)|=N$. We use the following standard consequence of the Siegel-Walfisz theorem for primes in arithmetic progressions. For every $A_0>0$ and $B_0>0$, there are constants $K(A_0,B_0)>0$ and $N(A_0,B_0)\in\mathbb N$ such that, whenever $N\ge N(A_0,B_0)$, $1\le q\le (\log N)^{B_0}$, $\gcd(a,q)=1$, and \begin{align*} |\beta|\le \frac{(\log N)^{B_0}}{qN}, \end{align*} one has \begin{align*} S_N\left(\frac{a}{q}+\beta\right)=\frac{\mu(q)}{\varphi(q)}V_N(\beta)+O_{A_0,B_0}\left(\frac{N}{(\log N)^{A_0}}\right). \end{align*} This is the standard major-arc approximation obtained by decomposing $\Lambda(n)$ into reduced residue classes modulo $q$ and applying the Siegel-Walfisz theorem uniformly for moduli $q\le (\log N)^{B_0}$; no exceptional-character term occurs in this logarithmic range. Apply this result with $B_0=B$ and with a saving parameter $A_0$ to be chosen later. On every arc $\mathfrak M_N(a,q;L)$ we have $q\le Q=(\log N)^B$ and $|\beta|\le L/(qN)=(\log N)^B/(qN)$, so the hypotheses are satisfied. Thus, for all sufficiently large $N$ depending on $A_0$ and $B$, \begin{align*} S_N\left(\frac{a}{q}+\beta\right)=\frac{\mu(q)}{\varphi(q)}V_N(\beta)+E_N(a,q,\beta), \end{align*} where the error function $E_N$ is defined on the set of triples $(a,q,\beta)$ with $q\le Q$, $\gcd(a,q)=1$, and $|\beta|\le L/(qN)$, and satisfies \begin{align*} |E_N(a,q,\beta)|\le K(A_0,B)\frac{N}{(\log N)^{A_0}}. \end{align*} Also \begin{align*} |S_N(\alpha)|\le \sum_{1\le n\le N}\Lambda(n)\le N\log N \end{align*} for all $\alpha\in\mathbb T$ and $N\ge 2$, because $\Lambda(1)=0$ and $\Lambda(n)\le \log N$ for $2\le n\le N$. Every term in the cubic expansion containing at least one factor $E_N$ is bounded pointwise by \begin{align*} K_1(A_0,B)\frac{N^3(\log N)^2}{(\log N)^{A_0}} \end{align*} for a constant $K_1(A_0,B)>0$. Since the arcs are disjoint and \begin{align*} \mu_{\mathbb T}(\mathfrak M_N(Q,L))\le \sum_{q\le Q}\sum_{\substack{1\le a\le q, \gcd(a, q)=1}}\frac{2L}{qN}\le \frac{2QL}{N}, \end{align*} the total contribution of all terms containing at least one $E_N$ is \begin{align*} O_{A_0,B}\left(N^2(\log N)^{2+2B-A_0}\right). \end{align*} Choose \begin{align*} A_0:=C+2B+3. \end{align*} Then this contribution is \begin{align*} O_C\left(\frac{N^2}{(\log N)^C}\right), \end{align*} after allowing the implied constant to depend on the already chosen $B=B(C)$. [/step] [step:Extract the Ramanujan sum from the arithmetic factor] Using the replacement from the previous step and the disjointness of the arcs, we obtain \begin{align*} \int_{\mathfrak M_N(Q,L)}S_N(\alpha)^3e(-N\alpha)\,d\mu_{\mathbb T}(\alpha)=\sum_{q\le Q}\frac{\mu(q)^3}{\varphi(q)^3}\sum_{\substack{1\le a\le q, \gcd(a, q)=1}}e\left(-\frac{Na}{q}\right)I_q(N,L)+O_C\left(\frac{N^2}{(\log N)^C}\right), \end{align*} where $I_q(N,L)\in\mathbb C$ is defined by \begin{align*} I_q(N,L):=\int_{-L/(qN)}^{L/(qN)}V_N(\beta)^3e(-N\beta)\,d\mathcal L^1(\beta). \end{align*} The change from Haar measure on $\mathbb T$ to $d\mathcal L^1(\beta)$ is the translation parametrization $\alpha=a/q+\beta$ on each disjoint interval of length less than $1$. By the definition of $c_q(-N)$, \begin{align*} \sum_{\substack{1\le a\le q, \gcd(a, q)=1}}e\left(-\frac{Na}{q}\right)=c_q(-N). \end{align*} Therefore \begin{align*} \int_{\mathfrak M_N(Q,L)}S_N(\alpha)^3e(-N\alpha)\,d\mu_{\mathbb T}(\alpha)=\sum_{q\le Q}\frac{\mu(q)^3}{\varphi(q)^3}c_q(-N)I_q(N,L)+O_C\left(\frac{N^2}{(\log N)^C}\right). \end{align*} [/step] [step:Evaluate and truncate the continuous Fourier integral] Let $f_N:\mathbb R\to\mathbb R$ be the compactly supported integrable function defined by \begin{align*} f_N(t):=\mathbb 1_{[0,N]}(t). \end{align*} With the Fourier convention $\widehat f(\beta)=\int_{\mathbb R}f(t)e(\beta t)\,d\mathcal L^1(t)$, the function $V_N$ is $\widehat f_N$. The convolution theorem and Fourier inversion for compactly supported integrable functions give \begin{align*} \int_{\mathbb R}V_N(\beta)^3e(-N\beta)\,d\mathcal L^1(\beta)=(f_N*f_N*f_N)(N). \end{align*} The convolution value is \begin{align*} \mathcal L^2\left(\{(x_1,x_2)\in[0,N]^2:0\le N-x_1-x_2\le N\}\right). \end{align*} Since $x_1,x_2\ge 0$, the condition $N-x_1-x_2\le N$ is automatic, and the set is the triangle \begin{align*} \{(x_1,x_2)\in[0,N]^2:x_1+x_2\le N\}. \end{align*} Its $\mathcal L^2$-measure is $N^2/2$. Hence \begin{align*} \int_{\mathbb R}V_N(\beta)^3e(-N\beta)\,d\mathcal L^1(\beta)=\frac{N^2}{2}. \end{align*} We now estimate the part omitted from $I_q(N,L)$. Put \begin{align*} R_q:=\frac{L}{qN}. \end{align*} For $0<|\beta|\le 1/N$, use $|V_N(\beta)|\le N$. For $|\beta|>1/N$, use $|V_N(\beta)|\le 1/(\pi |\beta|)$. Since $L\ge q$ need not hold for all $q\le Q$, the following uniform bound is enough: for every $R>0$, \begin{align*} \int_{|\beta|>R}|V_N(\beta)|^3\,d\mathcal L^1(\beta)\le K_2\frac{N^2}{1+NR} \end{align*} with an absolute constant $K_2>0$. This follows by splitting the integral at $1/N$ if $R<1/N$ and otherwise integrating $|\beta|^{-3}$. With $R=R_q$ this gives \begin{align*} I_q(N,L)=\frac{N^2}{2}+O\left(\frac{qN^2}{L}\right) \end{align*} uniformly for $1\le q\le Q$. Consequently, \begin{align*} \sum_{q\le Q}\frac{\mu(q)^3}{\varphi(q)^3}c_q(-N)I_q(N,L)=\frac{N^2}{2}\sum_{q\le Q}\frac{\mu(q)^3}{\varphi(q)^3}c_q(-N)+O\left(\frac{N^2}{L}\sum_{q\le Q}\frac{q|c_q(-N)|}{\varphi(q)^3}\right). \end{align*} Because $|c_q(-N)|\le \varphi(q)$, the error is at most \begin{align*} O\left(\frac{N^2}{L}\sum_{q\le Q}\frac{q}{\varphi(q)^2}\right). \end{align*} The standard lower bound for Euler's totient function gives an absolute constant $K_3>0$ such that \begin{align*} \varphi(q)\ge K_3\frac{q}{\log\log(q+3)} \end{align*} for every $q\ge 1$. Therefore \begin{align*} \sum_{q\le Q}\frac{q}{\varphi(q)^2}\le K_4(\log\log(Q+3))^2\sum_{q\le Q}\frac{1}{q}\le K_5(\log Q)(\log\log(Q+3))^2 \end{align*} for absolute constants $K_4,K_5>0$. Hence the truncation contributes \begin{align*} O\left(\frac{N^2(\log Q)(\log\log(Q+3))^2}{L}\right). \end{align*} Since $Q=L=(\log N)^B$, increasing $B$ in terms of $C$ makes this \begin{align*} O_C\left(\frac{N^2}{(\log N)^C}\right). \end{align*} [guided] We need to understand why the analytic factor gives exactly $N^2/2$. Define $f_N:\mathbb R\to\mathbb R$ by \begin{align*} f_N(t):=\mathbb 1_{[0,N]}(t). \end{align*} This function is compactly supported and belongs to $L^1(\mathbb R,\mathcal L^1)$. With the Fourier transform convention used in this proof, \begin{align*} \widehat f_N(\beta)=\int_{\mathbb R}f_N(t)e(\beta t)\,d\mathcal L^1(t)=\int_0^N e(\beta t)\,d\mathcal L^1(t)=V_N(\beta). \end{align*} The convolution theorem says that the product $V_N(\beta)^3$ is the Fourier transform of $f_N*f_N*f_N$. Fourier inversion, applied to the compactly supported convolution $f_N*f_N*f_N$, gives \begin{align*} \int_{\mathbb R}V_N(\beta)^3e(-N\beta)\,d\mathcal L^1(\beta)=(f_N*f_N*f_N)(N). \end{align*} The right-hand side has a geometric interpretation. By the definition of convolution and Tonelli's theorem for the non-negative integrand, \begin{align*} (f_N*f_N*f_N)(N)=\int_{\mathbb R^2}\mathbb 1_{[0,N]}(x_1)\mathbb 1_{[0,N]}(x_2)\mathbb 1_{[0,N]}(N-x_1-x_2)\,d\mathcal L^2(x_1,x_2). \end{align*} Thus it is the $\mathcal L^2$-measure of the set of pairs $(x_1,x_2)\in[0,N]^2$ for which $N-x_1-x_2\in[0,N]$. Since $x_1,x_2\ge 0$, the upper inequality $N-x_1-x_2\le N$ is automatic. The remaining condition is $x_1+x_2\le N$, so the set is the right triangle \begin{align*} \{(x_1,x_2)\in[0,N]^2:x_1+x_2\le N\}. \end{align*} Its base and height are both $N$, hence its measure is $N^2/2$. Therefore \begin{align*} \int_{\mathbb R}V_N(\beta)^3e(-N\beta)\,d\mathcal L^1(\beta)=\frac{N^2}{2}. \end{align*} It remains to justify replacing the finite interval $[-L/(qN),L/(qN)]$ by the whole real line. The bound \begin{align*} |V_N(\beta)|\le \min\left\{N,\frac{1}{\pi |\beta|}\right\} \end{align*} shows that the tail of $|V_N|^3$ is integrable. More precisely, for every $R>0$ there is an absolute constant $K_2>0$ such that \begin{align*} \int_{|\beta|>R}|V_N(\beta)|^3\,d\mathcal L^1(\beta)\le K_2\frac{N^2}{1+NR}. \end{align*} With $R=L/(qN)$ this becomes \begin{align*} \int_{|\beta|>L/(qN)}|V_N(\beta)|^3\,d\mathcal L^1(\beta)\le K_2\frac{qN^2}{L}. \end{align*} Since $|e(-N\beta)|=1$, this proves \begin{align*} I_q(N,L)=\frac{N^2}{2}+O\left(\frac{qN^2}{L}\right). \end{align*} After multiplying by the arithmetic coefficient, we use $|c_q(-N)|\le\varphi(q)$ and the standard lower bound $\varphi(q)\ge K_3q/\log\log(q+3)$ to obtain \begin{align*} \frac{N^2}{L}\sum_{q\le Q}\frac{q|c_q(-N)|}{\varphi(q)^3}\le \frac{N^2}{L}\sum_{q\le Q}\frac{q}{\varphi(q)^2}\le K_5\frac{N^2(\log Q)(\log\log(Q+3))^2}{L}. \end{align*} Because $L=(\log N)^B$, choosing $B$ sufficiently large compared with $C$ makes this at most a constant times $N^2/(\log N)^C$. [/guided] [/step] [step:Replace the truncated Ramanujan series by the singular series] We prove the uniform tail bound needed for the singular series. For squarefree $q$, multiplicativity of Ramanujan sums in the modulus gives \begin{align*} |c_q(-N)|=\prod_{p\mid q}|c_p(-N)|. \end{align*} For a prime $p$, $|c_p(-N)|$ is either $p-1$ or $1$, so \begin{align*} \left|\frac{\mu(q)^3}{\varphi(q)^3}c_q(-N)\right|\le \frac{\mu(q)^2}{\varphi(q)^2}. \end{align*} If $q$ is not squarefree, then $\mu(q)^3=0$, so the same upper bound holds for every $q\in\mathbb N$. Using again the lower bound for Euler's totient function, for $q\ge 3$ we have \begin{align*} \frac{\mu(q)^2}{\varphi(q)^2}\le K_6\frac{(\log\log(q+3))^2}{q^2} \end{align*} with an absolute constant $K_6>0$. Hence, for $Q\ge 3$, \begin{align*} \sum_{q>Q}\left|\frac{\mu(q)^3}{\varphi(q)^3}c_q(-N)\right|\le K_6\sum_{q>Q}\frac{(\log\log(q+3))^2}{q^2}\le K_7\frac{(\log\log(Q+3))^2}{Q} \end{align*} for an absolute constant $K_7>0$. This estimate is uniform in $N$. Therefore \begin{align*} \frac{N^2}{2}\sum_{q\le Q}\frac{\mu(q)^3}{\varphi(q)^3}c_q(-N)=\frac{N^2}{2}\mathfrak S_3(N)+O\left(\frac{N^2(\log\log(Q+3))^2}{Q}\right). \end{align*} Since $Q=(\log N)^B$, increasing $B$ in terms of $C$ gives \begin{align*} \frac{N^2}{2}\sum_{q\le Q}\frac{\mu(q)^3}{\varphi(q)^3}c_q(-N)=\frac{N^2}{2}\mathfrak S_3(N)+O_C\left(\frac{N^2}{(\log N)^C}\right). \end{align*} Combining this estimate with the previous steps proves \begin{align*} \int_{\mathfrak M_N(Q,L)}S_N(\alpha)^3e(-N\alpha)\,d\mu_{\mathbb T}(\alpha)=\frac{1}{2}\mathfrak S_3(N)N^2+O_C\left(\frac{N^2}{(\log N)^C}\right). \end{align*} [/step] [step:Use the Euler product to get a uniform positive lower bound] The preceding tail estimate proves absolute convergence of the Ramanujan series uniformly in $N$. Since $q\mapsto\mu(q)^3$, $q\mapsto\varphi(q)$, and $q\mapsto c_q(-N)$ are multiplicative in $q$, the coefficient \begin{align*} q\mapsto \frac{\mu(q)^3}{\varphi(q)^3}c_q(-N) \end{align*} is multiplicative and supported on squarefree integers. Absolute convergence therefore justifies rearranging the Ramanujan series into the Euler product \begin{align*} \mathfrak S_3(N)=\prod_p\left(1-\frac{c_p(-N)}{(p-1)^3}\right). \end{align*} For a prime $p$, the Ramanujan sum satisfies \begin{align*} c_p(-N)=p-1 \end{align*} if $p\mid N$, and \begin{align*} c_p(-N)=-1 \end{align*} if $p\nmid N$. Thus the local factor is \begin{align*} 1-\frac{1}{(p-1)^2} \end{align*} when $p\mid N$, and is \begin{align*} 1+\frac{1}{(p-1)^3} \end{align*} when $p\nmid N$. If $N$ is odd, then $2\nmid N$, so the local factor at $p=2$ equals $2$. For every odd prime $p$, both possible local factors are positive, and each is at least \begin{align*} 1-\frac{1}{(p-1)^2}. \end{align*} The infinite product \begin{align*} \prod_{p>2}\left(1-\frac{1}{(p-1)^2}\right) \end{align*} converges to a positive real number, because \begin{align*} \sum_{p>2}\frac{1}{(p-1)^2}<\infty. \end{align*} Define \begin{align*} c:=2\prod_{p>2}\left(1-\frac{1}{(p-1)^2}\right)>0. \end{align*} Then for every odd $N\in\mathbb N$, \begin{align*} \mathfrak S_3(N)\ge c. \end{align*} This completes the proof. [/step]