[proofplan] We first verify the second moment directly by additive character orthogonality, both as a base case and to identify the normalization of the Weyl sum. For the higher moments we invoke [Hua's Lemma](/theorems/9065), which is the established finite-differencing estimate for integral cutoffs. Finally, we pass from the real parameter $P$ to the integer cutoff $X=\lfloor P\rfloor$, using $X\le P$. [/proofplan]

[step:Evaluate the second moment by additive character orthogonality]Let $X:=\lfloor P\rfloor$. Since $P\ge 1$, one has $1\le X\le P$. Define \begin{align*} f_X:[0,1]\to\mathbb C,\qquad \alpha\mapsto \sum_{x=1}^{X}e(\alpha x^k). \end{align*} Then $f(\alpha;P)=f_X(\alpha)$. For $j=1$, expanding the square gives \begin{align*} \int_0^1 |f_X(\alpha)|^2\,d\mathcal L^1(\alpha) = \sum_{1\le x,y\le X}\int_0^1 e(\alpha(x^k-y^k))\,d\mathcal L^1(\alpha). \end{align*} For each integer $m$, the additive character orthogonality identity is \begin{align*} \int_0^1 e(\alpha m)\,d\mathcal L^1(\alpha)=\mathbb{1}_{\{0\}}(m). \end{align*} Equivalently, this integral is $1$ when $m=0$ and $0$ otherwise. Hence only pairs with $x^k=y^k$ contribute. Since $x,y\in\{1,\dots,X\}$ and $k\ge 2$, the equality $x^k=y^k$ implies $x=y$. Therefore \begin{align*} \int_0^1 |f_X(\alpha)|^2\,d\mathcal L^1(\alpha)=X\le P. \end{align*} Because $P\ge 1$ and $\varepsilon>0$, we also have $P\le P^{1+\varepsilon}$, so this implies the claimed bound for $j=1$.[/step]

[guided]The purpose of the first step is to identify the basic mechanism behind every later moment estimate: integration over $\alpha$ enforces an integer equation. We have declared \begin{align*} f_X:[0,1]\to\mathbb C,\qquad \alpha\mapsto \sum_{x=1}^{X}e(\alpha x^k), \end{align*} where $X=\lfloor P\rfloor$, so all sums are finite and all termwise expansions are justified. For the second moment, expand the product: \begin{align*} |f_X(\alpha)|^2=f_X(\alpha)\overline{f_X(\alpha)}=\sum_{1\le x,y\le X}e(\alpha x^k)e(-\alpha y^k). \end{align*} Thus \begin{align*} |f_X(\alpha)|^2=\sum_{1\le x,y\le X}e(\alpha(x^k-y^k)). \end{align*} Because this is a finite sum of continuous functions on $[0,1]$, we may integrate term by term with respect to one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure): \begin{align*} \int_0^1 |f_X(\alpha)|^2\,d\mathcal L^1(\alpha) = \sum_{1\le x,y\le X}\int_0^1 e(\alpha(x^k-y^k))\,d\mathcal L^1(\alpha). \end{align*} Now we use orthogonality of additive characters on the unit interval. For an integer $m$, the integral of $\alpha\mapsto e(\alpha m)$ over $[0,1]$ is \begin{align*} \int_0^1 e(\alpha m)\,d\mathcal L^1(\alpha)=\mathbb{1}_{\{0\}}(m). \end{align*} This equals $1$ when $m=0$ and $0$ otherwise. Here $m=x^k-y^k$. Therefore the integral counts exactly the pairs $(x,y)\in\{1,\dots,X\}^2$ satisfying $x^k=y^k$. Since $x$ and $y$ are positive integers and the map $t\mapsto t^k$ is injective on $\{1,\dots,X\}$, this equation is equivalent to $x=y$. There are exactly $X$ such pairs, so \begin{align*} \int_0^1 |f_X(\alpha)|^2\,d\mathcal L^1(\alpha)=X\le P. \end{align*} This proves the case $j=1$.[/guided]

[step:Invoke Hua's Lemma for the integer cutoff] We use [citetheorem:9065]. Applied with the same integer $k\ge 2$, the integer cutoff $X\ge 1$, and the chosen integer $j$ satisfying $1\le j\le k$, Hua's Lemma gives, for every $\varepsilon>0$, a constant $C_{j,k,\varepsilon}>0$ such that \begin{align*} \int_0^1 \left|\sum_{x=1}^{X}e(\alpha x^k)\right|^{2^j}\,d\mathcal L^1(\alpha) \le C_{j,k,\varepsilon}X^{2^j-j+\varepsilon}. \end{align*} This is an external mean-value estimate for the integer cutoff $X$; the remaining step only transfers it to the real parameter $P$ through $X=\lfloor P\rfloor$. [/step]

[step:Transfer the integer cutoff estimate to $P$]Fix an integer $j$ with $1\le j\le k$ and fix $\varepsilon>0$. Applying Hua's Lemma from the previous step to the integer $X=\lfloor P\rfloor$ gives \begin{align*} \int_0^1 |f(\alpha;P)|^{2^j}\,d\mathcal L^1(\alpha)=\int_0^1 |f_X(\alpha)|^{2^j}\,d\mathcal L^1(\alpha)\le C_{j,k,\varepsilon}X^{2^j-j+\varepsilon}. \end{align*} Since $1\le X\le P$, we have \begin{align*} X^{2^j-j+\varepsilon}\le P^{2^j-j+\varepsilon}. \end{align*} Therefore \begin{align*} \int_0^1 |f(\alpha;P)|^{2^j}\,d\mathcal L^1(\alpha) \le C_{j,k,\varepsilon}P^{2^j-j+\varepsilon}. \end{align*} This is exactly \begin{align*} \int_0^1 |f(\alpha;P)|^{2^j}\,d\mathcal L^1(\alpha)\lesssim_{j,k,\varepsilon}P^{2^j-j+\varepsilon}. \end{align*} The proof is complete.[/step]

[guided]The theorem is stated for a real parameter $P\ge 1$, while Hua's Lemma is stated for an integer cutoff. We bridge that mismatch by setting $X=\lfloor P\rfloor$. Since $P\ge 1$, the integer $X$ satisfies $1\le X\le P$, so it is an admissible cutoff for Hua's Lemma. By the definition of $f_X$ and the statement notation $f(\alpha;P)$, the two functions agree for every $\alpha\in[0,1]$: \begin{align*} f(\alpha;P)=f_X(\alpha). \end{align*} Hua's Lemma applies with the fixed integers $k$ and $j$, and with the fixed real number $\varepsilon>0$. It provides a constant $C_{j,k,\varepsilon}>0$, depending only on $j$, $k$, and $\varepsilon$, such that \begin{align*} \int_0^1 |f_X(\alpha)|^{2^j}\,d\mathcal L^1(\alpha) \le C_{j,k,\varepsilon}X^{2^j-j+\varepsilon}. \end{align*} Substituting $f(\alpha;P)=f_X(\alpha)$ gives \begin{align*} \int_0^1 |f(\alpha;P)|^{2^j}\,d\mathcal L^1(\alpha) \le C_{j,k,\varepsilon}X^{2^j-j+\varepsilon}. \end{align*} The exponent $2^j-j+\varepsilon$ is positive because $j\ge 1$ and $\varepsilon>0$. Hence the inequality $X\le P$ implies \begin{align*} X^{2^j-j+\varepsilon}\le P^{2^j-j+\varepsilon}. \end{align*} Combining the two estimates yields \begin{align*} \int_0^1 |f(\alpha;P)|^{2^j}\,d\mathcal L^1(\alpha) \le C_{j,k,\varepsilon}P^{2^j-j+\varepsilon}. \end{align*} This is the asserted bound with implicit constant depending only on $j$, $k$, and $\varepsilon$.[/guided]