Let $K$ be a number field of degree $n$ and $\mathcal{O}_K$ denote the ring of integers of $K$. For any complex number $s=\sigma +it$ with $\sigma > 1$, we define the Dedekind zeta function by the formula $$ \zeta_K(s)=\sum_{\mathfrak{a}}\frac{1}{N(\mathfrak{a})^s}, $$ where $\mathfrak{a}$ runs over all non-zero ideals of $\mathcal{O}_K$.
For each positive integer $n$, let $a(n)$ denote the number of non-zero ideals $\mathfrak{a} \subseteq \mathcal{O}_K$ with $N(\mathfrak{a})=n$. Then, we have $$ \zeta_K(s)=\sum_{n=1}^{\infty}\frac{a(n)}{n^s}, $$ with $\sigma > 1$.
We know the fact that the number $I(x)$ of ideals $\mathfrak{a}$ with $N(\mathfrak{a})\leq x$ is
$$
I(x)=cx + O(x^{1-\frac{1}{d}}),
$$
where $c=c(K)$ is positive constant and d is a positive integer greater than 1.
Using this fact and the Abel's summation formula, I want to obtain the meromorphic continuation of the Dedekind zeta function for all $s$ with $\sigma > 1 - \frac{1}{d}$.
By the Abel's summation formula, I've obtained $$ \sum_{n\leq x}\frac{a(n)}{n^s} = \frac{I(x)}{x^s} + s\int_1^x \frac{I(t)}{t^{s+1}}dt, $$ where $I(x) = cx + O(x^{1 - \frac{1}{d}})$.
Keeping $\sigma > 1$, let $x \longrightarrow \infty$. Then, we have \begin{align} \zeta_K(s) &= s\int_1^{\infty}\frac{I(t)}{t^{s+1}}dt \\ &=cs\int_1^{\infty} \frac{dt}{t^s} + s\int_1^{\infty}\frac{O(t^{1-\frac{1}{d}})}{t^{s+1}}dt \\&= \frac{cs}{s-1}+ s\int_1^{\infty}\frac{O(t^{1-\frac{1}{d}})}{t^{s+1}}dt \end{align}
Since the latter integral is absolutely and uniformly convergent for $\sigma > 1 - 1/d$, we have $$ \zeta_K(s) = \frac{cs}{s-1}+ s\int_1^{\infty}\frac{O(t^{1-\frac{1}{d}})}{t^{s+1}}dt $$ is analytic in the half plane $\sigma > 1 - 1/d$ apart from a simple pole at $s=1$ with residue $c$.