[proofplan] We expand the cube $S_N(\alpha)^3$ as a finite triple sum and multiply each summand by $e(-N\alpha)$. Because the sum is finite, we may integrate term by term without any convergence argument. The integral over $[0,1]$ of $e(m\alpha)$ is the elementary additive-character orthogonality relation: it equals $1$ when $m=0$ and equals $0$ otherwise. Therefore only triples satisfying $n_1+n_2+n_3=N$ survive, giving exactly the definition of $R_3(N)$. [/proofplan] [step:Expand the cube as a finite triple sum] For every $\alpha\in\mathbb R$, the definition of $S_N$ gives \begin{align*} S_N(\alpha)^3=\left(\sum_{1\le n\le N}\Lambda(n)e(n\alpha)\right)^3. \end{align*} Since this is a finite product of finite sums, the distributive law gives \begin{align*} S_N(\alpha)^3=\sum_{1\le n_1,n_2,n_3\le N}\Lambda(n_1)\Lambda(n_2)\Lambda(n_3)e((n_1+n_2+n_3)\alpha). \end{align*} Multiplying by $e(-N\alpha)$ and using $e(x)e(y)=e(x+y)$ for [real numbers](/page/Real%20Numbers) $x,y$, we obtain \begin{align*} S_N(\alpha)^3e(-N\alpha)=\sum_{1\le n_1,n_2,n_3\le N}\Lambda(n_1)\Lambda(n_2)\Lambda(n_3)e((n_1+n_2+n_3-N)\alpha). \end{align*} [guided] The purpose of introducing $S_N$ is that multiplication of exponential sums records addition of the underlying indices. For a fixed $\alpha\in\mathbb R$, we have \begin{align*} S_N(\alpha)=\sum_{1\le n\le N}\Lambda(n)e(n\alpha). \end{align*} Cubing means choosing one term from each of three copies of this finite sum. Thus the indices become an ordered triple $(n_1,n_2,n_3)$ with $1\le n_i\le N$ for each $i\in\{1,2,3\}$, and the corresponding coefficient is $\Lambda(n_1)\Lambda(n_2)\Lambda(n_3)$. The exponential factor is \begin{align*} e(n_1\alpha)e(n_2\alpha)e(n_3\alpha)=e((n_1+n_2+n_3)\alpha), \end{align*} because $e(x)e(y)=e(x+y)$ follows directly from the definition $e(x)=\exp(2\pi i x)$. Therefore \begin{align*} S_N(\alpha)^3=\sum_{1\le n_1,n_2,n_3\le N}\Lambda(n_1)\Lambda(n_2)\Lambda(n_3)e((n_1+n_2+n_3)\alpha). \end{align*} The extra factor $e(-N\alpha)$ shifts the frequency by $-N$. Hence \begin{align*} S_N(\alpha)^3e(-N\alpha)=\sum_{1\le n_1,n_2,n_3\le N}\Lambda(n_1)\Lambda(n_2)\Lambda(n_3)e((n_1+n_2+n_3-N)\alpha). \end{align*} This is the exact place where the additive equation $n_1+n_2+n_3=N$ is encoded as the zero frequency. [/guided] [/step] [step:Integrate term by term over $[0,1]$] The index set \begin{align*} \{(n_1,n_2,n_3)\in\mathbb N^3:1\le n_1,n_2,n_3\le N\} \end{align*} is finite. Hence finite additivity and linearity of the [Lebesgue integral](/page/Lebesgue%20Integral) give \begin{align*} \int_{[0,1]}S_N(\alpha)^3e(-N\alpha)\,d\mathcal L^1(\alpha)=\sum_{1\le n_1,n_2,n_3\le N}\Lambda(n_1)\Lambda(n_2)\Lambda(n_3)\int_{[0,1]}e((n_1+n_2+n_3-N)\alpha)\,d\mathcal L^1(\alpha). \end{align*} [/step] [step:Evaluate the additive character integral] Let $m\in\mathbb Z$. We claim that the integral \begin{align*} I_m:=\int_{[0,1]}e(m\alpha)\,d\mathcal L^1(\alpha) \end{align*} equals $1$ when $m=0$ and equals $0$ when $m\ne 0$. If $m=0$, then $e(m\alpha)=1$ for every $\alpha\in[0,1]$, so $I_m=\mathcal L^1([0,1])=1$. If $m\ne 0$, then the function $\alpha\mapsto e(m\alpha)$ is continuously differentiable on $[0,1]$, and the [Fundamental Theorem of Calculus](/theorems/632) gives \begin{align*} I_m=\frac{e(m)-e(0)}{2\pi i m}. \end{align*} Since $m\in\mathbb Z$, we have $e(m)=e(0)=1$, and the integral is $0$. Applying this with \begin{align*} m:=n_1+n_2+n_3-N\in\mathbb Z \end{align*} shows that the inner integral is $1$ exactly when $n_1+n_2+n_3=N$, and is $0$ otherwise. [/step] [step:Identify the surviving terms with the weighted representation count] Substituting the orthogonality computation into the term-by-term integral gives \begin{align*} \int_{[0,1]}S_N(\alpha)^3e(-N\alpha)\,d\mathcal L^1(\alpha)=\sum_{\substack{1\le n_1, n_2, n_3\le N, n_1+n_2+n_3=N}}\Lambda(n_1)\Lambda(n_2)\Lambda(n_3). \end{align*} If $n_1,n_2,n_3\in\mathbb N$ and $n_1+n_2+n_3=N$, then automatically $1\le n_i\le N$ for each $i\in\{1,2,3\}$. Therefore the last sum is precisely \begin{align*} \sum_{\substack{n_1+n_2+n_3=N, n_1, n_2, n_3\in\mathbb N}}\Lambda(n_1)\Lambda(n_2)\Lambda(n_3)=R_3(N). \end{align*} Thus \begin{align*} R_3(N)=\int_{[0,1]} S_N(\alpha)^3 e(-N\alpha)\,d\mathcal L^1(\alpha). \end{align*} This proves the theorem. [/step]