[proofplan]
We expand the $s$-th power of the finite Weyl sum into a finite sum indexed by ordered $s$-tuples of positive integers at most $P$. Since the sum is finite, we may integrate term-by-term. The elementary orthogonality identity for the functions $\alpha\mapsto e(m\alpha)$ with $m\in\mathbb Z$ then keeps exactly those tuples satisfying $x_1^k+\cdots+x_s^k=n$. Finally, the cutoff $P=\lfloor n^{1/k}\rfloor$ is shown to lose no positive representation of $n$.
[/proofplan]
[step:Expand the Weyl sum into a finite tuple sum]
Define the finite set \begin{align*}A:=\{1,\dots,P\}^s\subset\mathbb N^s.\end{align*}
Define the map $M:A\to\mathbb Z$ by the rule that, for each tuple $x=(x_1,\dots,x_s)\in A$, its value is
\begin{align*}
M(x):=x_1^k+\cdots+x_s^k-n.
\end{align*}
Using the distributive law for the finite product $f(\alpha;P)^s$, for every $\alpha\in\mathbb R$ one has
\begin{align*}
f(\alpha;P)^s e(-n\alpha)=\sum_{x\in A} e(M(x)\alpha).
\end{align*}
Indeed, $e(a)e(b)=e(a+b)$ for all $a,b\in\mathbb R$, because both sides equal $\exp(2\pi i(a+b))$.
[/step]
[step:Interchange the finite sum and the integral]
Since $A$ is finite and each map
\begin{align*}
\alpha\mapsto e(M(x)\alpha)
\end{align*}
is continuous on $[0,1]$, finite additivity and linearity of the [Lebesgue integral](/page/Lebesgue%20Integral) give
\begin{align*}
\int_0^1 f(\alpha;P)^s e(-n\alpha)\,d\mathcal L^1(\alpha)=\sum_{x\in A}\int_0^1 e(M(x)\alpha)\,d\mathcal L^1(\alpha).
\end{align*}
[/step]
[step:Evaluate the integer-frequency integrals]
For every integer $m\in\mathbb Z$, we prove that the integral has the following two values: it equals $1$ when $m=0$, and it equals $0$ when $m\ne 0$.
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 $m\ne 0$, define
\begin{align*}
F_m:[0,1]\to\mathbb C,\qquad F_m(\alpha):=\frac{e(m\alpha)}{2\pi i m}.
\end{align*}
The function $F_m$ is continuously differentiable on $[0,1]$, and its derivative is $\alpha\mapsto e(m\alpha)$. By the [Fundamental Theorem of Calculus](/theorems/632) applied to the complex-valued continuously differentiable function $F_m$ on the compact interval $[0,1]$,
\begin{align*}
\int_0^1 e(m\alpha)\,d\mathcal L^1(\alpha)=F_m(1)-F_m(0)=\frac{e(m)-1}{2\pi i m}=0,
\end{align*}
because $m\in\mathbb Z$ implies $e(m)=1$.
[guided]
The integral is the basic finite Fourier orthogonality calculation. Fix an integer $m\in\mathbb Z$. If $m=0$, the integrand is the constant function $1$, and hence
\begin{align*}
\int_0^1 e(0\cdot\alpha)\,d\mathcal L^1(\alpha)=\int_0^1 1\,d\mathcal L^1(\alpha)=1.
\end{align*}
Now suppose $m\ne 0$. We compute the integral by exhibiting an antiderivative. Define
\begin{align*}
F_m:[0,1]\to\mathbb C,\qquad F_m(\alpha):=\frac{e(m\alpha)}{2\pi i m}.
\end{align*}
Since $e(m\alpha)=\exp(2\pi i m\alpha)$, the ordinary derivative of $F_m$ is
\begin{align*}
F_m'(\alpha)=e(m\alpha)
\end{align*}
for every $\alpha\in[0,1]$. By the [Fundamental Theorem of Calculus](/theorems/632) applied to the complex-valued continuously differentiable function $F_m$ on the compact interval $[0,1]$,
\begin{align*}
\int_0^1 e(m\alpha)\,d\mathcal L^1(\alpha)=F_m(1)-F_m(0)=\frac{e(m)-e(0)}{2\pi i m}.
\end{align*}
Because $m$ is an integer, $e(m)=\exp(2\pi i m)=1$, and also $e(0)=1$. Hence
\begin{align*}
\int_0^1 e(m\alpha)\,d\mathcal L^1(\alpha)=0.
\end{align*}
This is exactly the mechanism that makes the circle method identity a counting formula: all nonzero integer frequencies vanish after integration, while the zero frequency contributes one unit.
[/guided]
[/step]
[step:Identify the surviving tuples with Waring representations]
Applying the integer-frequency integral computation with $m=M(x)$ gives
\begin{align*}
\int_0^1 f(\alpha;P)^s e(-n\alpha)\,d\mathcal L^1(\alpha)=\#\{x\in A:M(x)=0\}.
\end{align*}
By the definition of $M(x)$, the condition $M(x)=0$ is precisely
\begin{align*}
x_1^k+\cdots+x_s^k=n.
\end{align*}
Thus
\begin{align*}
\#\{x\in A:M(x)=0\}=\#\{(x_1,\dots,x_s)\in\{1,\dots,P\}^s:x_1^k+\cdots+x_s^k=n\}.
\end{align*}
[/step]
[step:Remove the cutoff without changing the representation count]
It remains to compare the restricted tuples in $\{1,\dots,P\}^s$ with all positive tuples in $\mathbb N^s$. If $(x_1,\dots,x_s)\in\mathbb N^s$ satisfies
\begin{align*}
x_1^k+\cdots+x_s^k=n,
\end{align*}
then for each $i\in\{1,\dots,s\}$ one has $x_i^k\le n$, and hence $x_i\le n^{1/k}$. Since $x_i$ is an integer, this implies $x_i\le \lfloor n^{1/k}\rfloor=P$. Therefore every positive representation of $n$ already lies in $\{1,\dots,P\}^s$, and so
\begin{align*}
\#\{(x_1,\dots,x_s)\in\{1,\dots,P\}^s:x_1^k+\cdots+x_s^k=n\}=R_{s,k}^{+}(n).
\end{align*}
Combining this identity with the preceding steps proves
\begin{align*}
R_{s,k}^{+}(n)=\int_0^1 f(\alpha;P)^s e(-n\alpha)\,d\mathcal L^1(\alpha).
\end{align*}
[/step]
