Asymptotics of $\sum_{\mathfrak{a}}\frac{n^{k-\epsilon}}{\mathfrak{N}\left(\mathfrak{a}\right)^{r\left(k-\epsilon\right)}}$ In this paper by Brian D. Sittinger, the following claim is made:
For an algebraic number field $K$ with norm $\mathfrak{N}$, let $\epsilon=\left[K:\mathbb{Q}\right]^{-1}$. Then, taking the sum over ideals $\mathfrak{a}$ of the ring of algebraic integers $\mathcal{O}$ of $K$,
$$\sum_{\mathfrak{a}}O\left(\frac{n^{k-\epsilon}}{\mathfrak{N}\left(\mathfrak{a}\right)^{r\left(k-\epsilon\right)}}\right)=\begin{cases}
O\left(n^{k-\epsilon}\right), & {\rm if}\; k>2,\;{\rm or}\; k=2\;{\rm and}\; r\geq2,\\
O\left(n^{2-\epsilon}\log n\right), & {\rm if}\; k=2\;{\rm and}\; r=1,\\
O\left(n^{1-\epsilon}\log n\right), & {\rm if}\; k=1\;{\rm and}\;\epsilon=\frac{r-2}{r-1},\\
O\left(n^{1-\epsilon}\right) & {\rm if}\; k=1\;{\rm and}\;\epsilon<\frac{r-2}{r-1},\\
O\left(n^{\left(2-r-\epsilon+r\epsilon\right)/r}\right) & {\rm if}\; k=1\;{\rm and}\;\epsilon>\frac{r-2}{r-1}.
\end{cases}$$
In the paper, the following estimations are made:
$$\sum_{\mathfrak{a}}\frac{n^{k-\epsilon}}{\mathfrak{N}\left(\mathfrak{a}\right)^{r\left(k-\epsilon\right)}}\leq n^{k-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{cj^{1-\epsilon}}{j^{r\left(k-\epsilon\right)}}\leq n^{k-\epsilon}\left(1+\int_{1}^{\sqrt[r]{n}}\frac{dx}{x^{r\left(k-1\right)}}\right).$$
From here, the author "[leaves] the tedious details to the reader" when coming to the claim's conclusion. As usual, I take this as a challenge to be completed.
The first two cases aren't hard, but I am puzzled by the cases with $\epsilon$. I would have thought that, when $k=1$, there would necessarily have to be a $\sqrt[r]{n}$ factor coming from the fact that $x^{r\left({1-1}\right)}=1$, but it is nowhere to be seen. In fact, when $\epsilon<\frac{r-2}{r-1}$, it seems that the integral contributes $O\left(1\right)$. What am I missing here?
 A: I have made some progress since posting this question, and I hope this is helpful for anyone else who may stumble upon this post looking for answers to this same question.
The first thing to note is that we cannot use the last part of the estimations given for the last few cases, as this leads, in the end, to $n^{1/r}\to\infty$, which does not get the desired result. So, we move back one step to the sum.
Here is the case $\epsilon = \frac{r-2}{r-1}$:
$$n^{1-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{cj^{1-\epsilon}}{j^{r\left(1-\epsilon\right)}}=cn^{1-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{1}{j^{\left(r-1\right)\left(1-\epsilon\right)}}=cn^{1-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{1}{j^{r-r\epsilon+\epsilon-1}}=\\ cn^{1-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{1}{j^{r-r\frac{r-2}{r-1}+\frac{r-2}{r-1}-1}}=cn^{1-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{1}{j^{r-\left(r-2\right)-1}}=cn^{1-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{1}{j}=O\left(n^{1-\epsilon}\log n\right).$$
Here is the case $\epsilon < \frac{r-2}{r-1}$:
Note that, in this case, $r-r\epsilon+\epsilon-1\geq1+\epsilon$ so that
$$n^{1-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{cj^{1-\epsilon}}{j^{r\left(1-\epsilon\right)}}=cn^{1-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{1}{j^{\left(r-1\right)\left(1-\epsilon\right)}}=cn^{1-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{1}{j^{r-r\epsilon+\epsilon-1}}\leq\\ cn^{1-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{1}{j^{1+\epsilon}}=O\left(n^{1-\epsilon}\right).$$
Here is the case $\epsilon > \frac{r-2}{r-1}$:
Note that, in this case, $\epsilon=1$ or $r=2$, as $r\geq3$ and $\epsilon\leq \frac{1}{2}$ gives $\epsilon \leq \frac{r-2}{r-1}$. In either case, $r-r\epsilon+\epsilon-1=1-\epsilon$. First, when $r=2$ and $\epsilon\leq\frac{1}{2}$, we have
$$n^{1-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{cj^{1-\epsilon}}{j^{r\left(1-\epsilon\right)}}=cn^{1-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{1}{j^{\left(r-1\right)\left(1-\epsilon\right)}}=cn^{1-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{1}{j^{r-r\epsilon+\epsilon-1}}=\\ cn^{1-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{1}{j^{1-\epsilon}}\leq cn^{1-\epsilon}\left(1+\int_{1}^{\sqrt[r]{n}}\frac{dx}{x^{1-\epsilon}}\right) = O\left(n^{\left(2-\epsilon\right)/r}\right).$$
Next, when $\epsilon=1$, we have
$$n^{1-\epsilon}\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}\frac{cj^{1-\epsilon}}{j^{r\left(1-\epsilon\right)}}=c\sum_{j=1}^{\lfloor\sqrt[r]{n}\rfloor}1=c\lfloor n^{1/r}\rfloor=O\left(n^{\left(2-\epsilon\right)/r}\right),$$
as required.
I am not sure how to optimize this further for the $O\left(n^{\left(2-r-\epsilon+r\epsilon\right)/r}\right)$ as claimed in the paper, but the $O\left(n^{\left(2-\epsilon\right)/r}\right)$ is enough to get the desired result. In fact, it is inevitable that $O\left(n^{\left(2-\epsilon\right)/r}\right)$ shows up in the $\epsilon>\frac{r-2}{r-1}$ case in the final asymptotics of $Q\left(n\right)$ due to the first sum that is estimated just before this section of the paper.
I am still curious about how one would go about refining the estimation for the stronger bound, though, if anyone wants to take a shot at it.
EDIT: Since posting this, I have had a chance to look back at this. I noticed one minor error with the last estimation, and have provided (what I hope to be) a fix.
