[proofplan] We expand the $s$th power of the finite Weyl sum as a finite sum over all tuples $(x_1,\dots,x_s)\in\{1,\dots,P\}^s$. After multiplying by $e(-N\alpha)$ and interchanging the finite sum with the integral, each tuple contributes an integral of the additive character $e(m\alpha)$ with integer frequency $m=x_1^k+\cdots+x_s^k-N$. Direct orthogonality on $[0,1]$ shows that this integral is $1$ when $m=0$ and $0$ otherwise, so the surviving tuples are exactly the representations counted by $R_{s,k}(N;P)$. [/proofplan]

[step:Expand the Weyl sum into a finite tuple sum]For every $\alpha\in\mathbb R$, finite distributivity gives \begin{align*} S_k(\alpha;P)^s=\sum_{(x_1,\dots,x_s)\in\{1,\dots,P\}^s} e(\alpha x_1^k)\cdots e(\alpha x_s^k). \end{align*} Since $e(u)e(v)=e(u+v)$ for all $u,v\in\mathbb R$, this becomes \begin{align*} S_k(\alpha;P)^s=\sum_{(x_1,\dots,x_s)\in\{1,\dots,P\}^s} e(\alpha(x_1^k+\cdots+x_s^k)). \end{align*} Multiplying by $e(-N\alpha)$ and using the same multiplicative identity, \begin{align*} S_k(\alpha;P)^s e(-N\alpha)=\sum_{(x_1,\dots,x_s)\in\{1,\dots,P\}^s} e((x_1^k+\cdots+x_s^k-N)\alpha). \end{align*}[/step]

[guided]The Weyl sum is a finite sum, so raising it to the $s$th power is just repeated finite multiplication. For each choice of one summation variable from each copy of $S_k(\alpha;P)$, we obtain one tuple $(x_1,\dots,x_s)\in\{1,\dots,P\}^s$. Hence \begin{align*} S_k(\alpha;P)^s=\sum_{(x_1,\dots,x_s)\in\{1,\dots,P\}^s} e(\alpha x_1^k)\cdots e(\alpha x_s^k). \end{align*} The function $e$ satisfies $e(u)e(v)=e(u+v)$ because \begin{align*} \exp(2\pi i u)\exp(2\pi i v)=\exp(2\pi i(u+v)). \end{align*} Applying this identity repeatedly gives \begin{align*} e(\alpha x_1^k)\cdots e(\alpha x_s^k)=e(\alpha(x_1^k+\cdots+x_s^k)). \end{align*} After multiplying by $e(-N\alpha)$, the same identity combines the phase with the factor $-N\alpha$: \begin{align*} e(\alpha(x_1^k+\cdots+x_s^k))e(-N\alpha)=e((x_1^k+\cdots+x_s^k-N)\alpha). \end{align*} Therefore \begin{align*} S_k(\alpha;P)^s e(-N\alpha)=\sum_{(x_1,\dots,x_s)\in\{1,\dots,P\}^s} e((x_1^k+\cdots+x_s^k-N)\alpha). \end{align*} This is the key reduction: the integral of the original product is now a finite sum of one-dimensional character integrals indexed by the possible representations.[/guided]

[step:Interchange the finite tuple sum with the integral]Because $\{1,\dots,P\}^s$ is finite, linearity of the [Lebesgue integral](/page/Lebesgue%20Integral) gives \begin{align*} \int_0^1 S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha)=\sum_{(x_1,\dots,x_s)\in\{1,\dots,P\}^s}\int_0^1 e((x_1^k+\cdots+x_s^k-N)\alpha)\,d\mathcal L^1(\alpha). \end{align*} For each tuple $x=(x_1,\dots,x_s)\in\{1,\dots,P\}^s$, define the integer \begin{align*} m_x:=x_1^k+\cdots+x_s^k-N. \end{align*} Then the last expression is \begin{align*} \sum_{x\in\{1,\dots,P\}^s}\int_0^1 e(m_x\alpha)\,d\mathcal L^1(\alpha). \end{align*}[/step]

[guided]The sum obtained in the previous step is indexed by the finite set $\{1,\dots,P\}^s$, whose cardinality is $P^s$. Therefore no convergence theorem is needed to exchange summation and integration; ordinary finite linearity of the Lebesgue integral applies. Thus \begin{align*} \int_0^1 S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha)=\sum_{(x_1,\dots,x_s)\in\{1,\dots,P\}^s}\int_0^1 e((x_1^k+\cdots+x_s^k-N)\alpha)\,d\mathcal L^1(\alpha). \end{align*} For a fixed tuple $x=(x_1,\dots,x_s)\in\{1,\dots,P\}^s$, define \begin{align*} m_x:=x_1^k+\cdots+x_s^k-N. \end{align*} Because each $x_i$, $k$, and $N$ is an integer, $m_x\in\mathbb Z$. With this notation the integral is a character integral with integer frequency, and the expression becomes \begin{align*} \sum_{x\in\{1,\dots,P\}^s}\int_0^1 e(m_x\alpha)\,d\mathcal L^1(\alpha). \end{align*} This isolates the only analytic input needed in the proof: orthogonality of integer additive characters on the unit interval.[/guided]

[step:Evaluate the unit interval character integral]For every integer $m\in\mathbb Z$, the character integral satisfies \begin{align*} \int_0^1 e(m\alpha)\,d\mathcal L^1(\alpha)=1 \end{align*} if $m=0$, and \begin{align*} \int_0^1 e(m\alpha)\,d\mathcal L^1(\alpha)=0 \end{align*} if $m\ne 0$. Indeed, if $m=0$, then $e(m\alpha)=1$ for every $\alpha\in[0,1]$, so the integral is $\mathcal L^1([0,1])=1$. If $m\ne 0$, then the continuous complex-valued function $\alpha\mapsto e(m\alpha)$ has antiderivative $\alpha\mapsto (2\pi i m)^{-1}e(m\alpha)$ on $\mathbb R$. Applying the [fundamental theorem of calculus](/theorems/632) to its real and imaginary parts gives \begin{align*} \int_0^1 e(m\alpha)\,d\mathcal L^1(\alpha)=\frac{e(m)-e(0)}{2\pi i m}. \end{align*} Since $m\in\mathbb Z$, one has $e(m)=\exp(2\pi i m)=1=e(0)$, so the integral is $0$.[/step]

[guided]The point of this step is that integration over one full period kills every non-zero integer frequency. Fix $m\in\mathbb Z$. If $m=0$, then the integrand is the constant function $1$, and therefore \begin{align*} \int_0^1 e(m\alpha)\,d\mathcal L^1(\alpha)=\mathcal L^1([0,1])=1. \end{align*} Now suppose $m\ne 0$. The map $\alpha\mapsto e(m\alpha)$ is continuous as a function $[0,1]\to\mathbb C$, and the map $\alpha\mapsto (2\pi i m)^{-1}e(m\alpha)$ is an antiderivative on $\mathbb R$. Applying the fundamental theorem of calculus to the real and imaginary parts of this complex-valued function yields \begin{align*} \int_0^1 e(m\alpha)\,d\mathcal L^1(\alpha)=\frac{e(m)-e(0)}{2\pi i m}. \end{align*} Because $m$ is an integer, $e(m)=\exp(2\pi i m)=1$, while $e(0)=1$. Hence the numerator is zero, and \begin{align*} \int_0^1 e(m\alpha)\,d\mathcal L^1(\alpha)=0. \end{align*} Thus the integral is exactly the detector for whether the integer frequency $m$ is zero.[/guided]

[step:Identify the surviving tuples with the representation count]Let $\mathbb{1}_{\{0\}}:\mathbb Z\to\{0,1\}$ denote the indicator function of the subset $\{0\}\subset\mathbb Z$. Applying the preceding character integral formula with $m=m_x$ gives \begin{align*} \int_0^1 S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha)=\sum_{x\in\{1,\dots,P\}^s}\mathbb{1}_{\{0\}}(m_x). \end{align*} By the definition of $m_x$, the condition $m_x=0$ is exactly \begin{align*} x_1^k+\cdots+x_s^k=N. \end{align*} Therefore \begin{align*} \sum_{x\in\{1,\dots,P\}^s}\mathbb{1}_{\{0\}}(m_x)=\#\{(x_1,\dots,x_s)\in\{1,\dots,P\}^s:x_1^k+\cdots+x_s^k=N\}. \end{align*} The right-hand side is $R_{s,k}(N;P)$ by definition, and thus \begin{align*} \int_0^1 S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha)=R_{s,k}(N;P). \end{align*} This is the desired identity.[/step]

[guided]Define $\mathbb{1}_{\{0\}}:\mathbb Z\to\{0,1\}$ to be the indicator function of $\{0\}\subset\mathbb Z$, so $\mathbb{1}_{\{0\}}(m)=1$ when $m=0$ and $\mathbb{1}_{\{0\}}(m)=0$ when $m\ne 0$. The character integral formula from the previous step says exactly that \begin{align*} \int_0^1 e(m_x\alpha)\,d\mathcal L^1(\alpha)=\mathbb{1}_{\{0\}}(m_x) \end{align*} for each tuple $x\in\{1,\dots,P\}^s$. Substituting this into the finite sum gives \begin{align*} \int_0^1 S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha)=\sum_{x\in\{1,\dots,P\}^s}\mathbb{1}_{\{0\}}(m_x). \end{align*} Now unpack the definition of $m_x$. For $x=(x_1,\dots,x_s)$, the equality $m_x=0$ is equivalent to \begin{align*} x_1^k+\cdots+x_s^k-N=0, \end{align*} and hence to \begin{align*} x_1^k+\cdots+x_s^k=N. \end{align*} Thus the sum of the indicator function counts precisely the tuples in $\{1,\dots,P\}^s$ satisfying the representation equation: \begin{align*} \sum_{x\in\{1,\dots,P\}^s}\mathbb{1}_{\{0\}}(m_x)=\#\{(x_1,\dots,x_s)\in\{1,\dots,P\}^s:x_1^k+\cdots+x_s^k=N\}. \end{align*} By the definition of $R_{s,k}(N;P)$, this cardinality is $R_{s,k}(N;P)$. Therefore \begin{align*} \int_0^1 S_k(\alpha;P)^s e(-N\alpha)\,d\mathcal L^1(\alpha)=R_{s,k}(N;P), \end{align*} which is the asserted identity.[/guided]