[proofplan]
We expand the $s$-th power of the finite exponential sum as a sum over all tuples in $\{1,\dots,N\}^s$. After multiplying by $e(-m\alpha)$, each tuple contributes the additive character $e(\alpha t)$ with integer frequency $t=n_1^k+\cdots+n_s^k-m$. The integral over $[0,1]$ is then computed directly: it equals $1$ when $t=0$ and $0$ otherwise. Thus the integral is exactly the number of tuples whose $k$-th powers sum to $m$.
[/proofplan]
[step:Expand the power as a finite sum over all tuples]
Let
\begin{align*}
A_m:=\{(n_1,\dots,n_s)\in\{1,\dots,N\}^s:n_1^k+\cdots+n_s^k=m\}.
\end{align*}
This is a finite set, so its cardinality is well-defined. For each $\alpha\in\mathbb R$, the distributive law for the finite product gives
\begin{align*}
S_k(\alpha;N)^s=\sum_{(n_1,\dots,n_s)\in\{1,\dots,N\}^s}e(\alpha n_1^k)\cdots e(\alpha n_s^k).
\end{align*}
Since $e(x+y)=e(x)e(y)$ for all $x,y\in\mathbb R$, this becomes
\begin{align*}
S_k(\alpha;N)^s=\sum_{(n_1,\dots,n_s)\in\{1,\dots,N\}^s}e(\alpha(n_1^k+\cdots+n_s^k)).
\end{align*}
Multiplying by $e(-m\alpha)$ and using the same exponential law gives
\begin{align*}
S_k(\alpha;N)^s e(-m\alpha)=\sum_{(n_1,\dots,n_s)\in\{1,\dots,N\}^s}e(\alpha(n_1^k+\cdots+n_s^k-m)).
\end{align*}
[/step]
[step:Move the finite tuple sum outside the integral]
The functions
\begin{align*}
\alpha\mapsto e(\alpha(n_1^k+\cdots+n_s^k-m))
\end{align*}
from $[0,1]$ to $\mathbb C$ are continuous for every tuple $(n_1,\dots,n_s)\in\{1,\dots,N\}^s$, hence Lebesgue integrable on $[0,1]$ with respect to $\mathcal L^1$. Since the tuple sum is finite, linearity of the [Lebesgue integral](/page/Lebesgue%20Integral) gives
\begin{align*}
\int_0^1 S_k(\alpha;N)^s e(-m\alpha)\,d\mathcal L^1(\alpha)=\sum_{(n_1,\dots,n_s)\in\{1,\dots,N\}^s}\int_0^1 e(\alpha(n_1^k+\cdots+n_s^k-m))\,d\mathcal L^1(\alpha).
\end{align*}
[/step]
[step:Compute the unit interval integral of each integer character]
Define the map
\begin{align*}
I:\mathbb Z &\to \mathbb C
\end{align*}
by
\begin{align*}
I(t):=\int_0^1 e(\alpha t)\,d\mathcal L^1(\alpha).
\end{align*}
If $t=0$, then $e(\alpha t)=e(0)=1$ for every $\alpha\in[0,1]$, so
\begin{align*}
I(0)=\int_0^1 1\,d\mathcal L^1(\alpha)=1.
\end{align*}
If $t\ne 0$, then the function $\alpha\mapsto e(\alpha t)$ is continuously differentiable on $[0,1]$, and
\begin{align*}
\int_0^1 e(\alpha t)\,d\mathcal L^1(\alpha)=\frac{e(t)-e(0)}{2\pi i t}.
\end{align*}
Because $t\in\mathbb Z$, one has $e(t)=\exp(2\pi i t)=1=e(0)$. Therefore
\begin{align*}
I(t)=0.
\end{align*}
Thus, for every $t\in\mathbb Z$, we have $I(t)=1$ if $t=0$, and $I(t)=0$ if $t\ne 0$.
[guided]
The integral we need is the orthogonality relation for additive characters, but here it is short enough to prove directly. Define the map
\begin{align*}
I:\mathbb Z &\to \mathbb C
\end{align*}
by
\begin{align*}
I(t):=\int_0^1 e(\alpha t)\,d\mathcal L^1(\alpha).
\end{align*}
There are two cases. If $t=0$, the integrand is constantly equal to $e(0)=\exp(0)=1$, and the interval $[0,1]$ has one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) $1$. Hence
\begin{align*}
I(0)=\int_0^1 1\,d\mathcal L^1(\alpha)=1.
\end{align*}
Now suppose $t\ne 0$. Since $e(\alpha t)=\exp(2\pi i t\alpha)$, the map $\alpha\mapsto e(\alpha t)$ is continuously differentiable on $[0,1]$. Its antiderivative is
\begin{align*}
\alpha\mapsto \frac{e(\alpha t)}{2\pi i t}.
\end{align*}
Therefore
\begin{align*}
I(t)=\frac{e(t)-e(0)}{2\pi i t}.
\end{align*}
The integer condition on $t$ is exactly what makes the boundary values cancel: $e(t)=\exp(2\pi i t)=1$ and $e(0)=1$. Hence
\begin{align*}
I(t)=0.
\end{align*}
Combining the two cases gives $I(t)=1$ if $t=0$, and $I(t)=0$ if $t\ne 0$.
This is the mechanism that filters the expanded sum: only tuples for which the integer frequency $n_1^k+\cdots+n_s^k-m$ is zero survive the integration.
[/guided]
[/step]
[step:Identify the surviving terms with the representation count]
Define the map
\begin{align*}
t:\{1,\dots,N\}^s \to \mathbb Z
\end{align*}
by
\begin{align*}
t(n_1,\dots,n_s)=n_1^k+\cdots+n_s^k-m
\end{align*}
for each tuple $(n_1,\dots,n_s)\in\{1,\dots,N\}^s$.
By the previous step,
\begin{align*}
\int_0^1 e(\alpha t(n))\,d\mathcal L^1(\alpha)=1
\end{align*}
exactly when $t(n)=0$, and it equals $0$ otherwise. Since $t(n)=0$ is precisely the condition
\begin{align*}
n_1^k+\cdots+n_s^k=m,
\end{align*}
we obtain
\begin{align*}
\int_0^1 S_k(\alpha;N)^s e(-m\alpha)\,d\mathcal L^1(\alpha)=\sum_{n\in\{1,\dots,N\}^s}\mathbb{1}_{A_m}(n).
\end{align*}
Because $A_m$ is finite, the last sum is exactly $\#A_m$. Therefore
\begin{align*}
\int_0^1 S_k(\alpha;N)^s e(-m\alpha)\,d\mathcal L^1(\alpha)=\#\{(n_1,\dots,n_s)\in\{1,\dots,N\}^s:n_1^k+\cdots+n_s^k=m\}.
\end{align*}
This is the claimed formula.
[/step]
