## 2 Spectral Analysis
### Main Results  
We study the eigenvalue distribution results of the two matrix-sequences \(\{h^{\alpha}D_{\mathbf{n}}(\mu)\}_{n}\) and \(\{A_{\mathbf{n}}=\hat{B}_{\mathbf{n}}+h^{\alpha}D_{\mathbf{n}}(\mu)\}_{n}\), in the sense of Definition 1. The same kind of matrices and matrix-sequences are treated in [1,2]. In [2] eigenvalues localization results are studied, while in [1] singular value and eigenvalue distribution results are obtained and in both cases, the coefficient \(\mu(x,y)\) is assumed bounded. Here, we extend the results in the quoted literature. Theorem 3: _Let \(\mu(x,y)=1/(x+\mathrm{i}y)^{\gamma}\). Then, for every \(\gamma\geq 0\) such that \(\alpha>\gamma-1\)\((\alpha\in(1,2))\), we have_  
\[a1 \{h^{\alpha}D_{\mathbf{n}}(\mu)\}_{n}\sim_{\lambda}0;\] \[a2 \{\hat{B}_{\mathbf{n}}+h^{\alpha}D_{\mathbf{n}}(\mu)\}_{n}\sim_{ \lambda}t_{\alpha}.\]  
Proof: In the proof we strongly rely on Theorem 2. Therefore, we compute  
\[\|h^{\alpha}D_{\mathbf{n}}(\mu)\|_{2}^{2} =\sum_{i,j=1}^{n}|\mu_{ij}|^{2}h^{2\alpha}\] \[=\sum_{i,j=1}^{n}\Bigl{(}|ih|^{2}+|jh|^{2}\Bigr{)}^{-\gamma}h^{2\alpha}\] \[=h^{2\alpha-2\gamma}\sum_{i,j=1}^{n}\frac{1}{(i^{2}+j^{2})^{ \gamma}}.\]Then, we estimate the quantity \(\sum_{i,j=1}^{n}\frac{1}{(i^{2}+j^{2})^{\gamma}}\), under the hypothesis that \(\gamma\in(0,1)\). \[\sum_{i,j=1}^{n}\frac{1}{(i^{2}+j^{2})^{\gamma}}=2\sum_{i=1}^{n}\sum_{j=1}^{i-1} \frac{1}{(i^{2}+j^{2})^{\gamma}}+\frac{1}{2\gamma}\sum_{i=1}^{n}\frac{1}{i^{2 \gamma}}.\]  
Now, the first sum can be estimated as  
\[2\sum_{i=1}^{n}\sum_{j=1}^{i-1}\frac{1}{(i^{2}+j^{2})^{\gamma}} \leq 2\sum_{i=1}^{n}\frac{i-1}{i^{2\gamma}}=2\sum_{i=1}^{n}\frac{1}{i^{2 \gamma-1}}-2\sum_{i=1}^{n}\frac{1}{i^{2\gamma}}. \tag{8}\]  
Therefore,  
\[\sum_{i,j=1}^{n}\frac{1}{(i^{2}+j^{2})^{\gamma}} \leq 2\sum_{i=1}^{n}\frac{1}{i^{2\gamma-1}}+\left(\frac{1}{2^{ \gamma}-2}\right)\sum_{i=1}^{n}\frac{1}{2^{\gamma}}.\]  
Note, that \(\frac{1}{2^{\gamma}-2}<0\) for every \(\gamma\in(0,1)\). A basic computation leads to  
\[2\sum_{i=1}^{n}\frac{1}{i^{2\gamma-1}}\leq\int_{0}^{n}\frac{dt}{t^{2\gamma-1} }=\frac{n^{2-2\gamma}}{1-\gamma}\]  
and  
\[\sum_{i=1}^{n}\frac{1}{2^{\gamma}}\geq\int_{0}^{n}\frac{dt}{(t+1)^{2\gamma}}= \begin{cases}\frac{(n+1)^{1-2\gamma-1}}{1-2\gamma}&\gamma\
eq\frac{1}{2},\\ \log(n+1)&\gamma=\frac{1}{2}.\end{cases} \]  
As a consequence, we conclude that  
\[\sum_{i,j=1}^{n}\frac{1}{(i^{2}+j^{2})^{\gamma}}\leq \begin{cases} \frac{n^{2-2\gamma}}{1-\gamma}+\frac{(n+1)^{1-2\gamma}-1}{1-2\gamma}&\gamma \
eq\frac{1}{2}\\ \frac{n^{2-2\gamma}}{1-\gamma}+\left(\frac{1}{2^{\gamma}-2}\right)\log(n+1)& \gamma=\frac{1}{2},\end{cases} \]  
so that \(\sum_{i,j=1}^{n}\frac{1}{(i^{2}+j^{2})^{\gamma}}\leq c_{n}(\gamma)\sim\frac{ n^{2-2\gamma}}{1-\gamma}\) for every \(\gamma\in(0,1)\). This immediately implies that  
\[h^{2\alpha-2\gamma}\sum_{i,j=1}^{n}\frac{1}{(i^{2}+j^{2})^{\gamma}}\leq\frac{ c_{n}(\gamma)}{n^{2\alpha-2\gamma}}\sim\frac{n^{2-2\alpha}}{1-\gamma}=o(n^{2})\]  
for every \(\alpha\in(1,2)\), as required to apply Theorem 1 in [4] and conclude the proof. From the computation in Equation (8), we immediately deduce the following: if \(\gamma>1\), \(\sum_{i,j=1}^{n}\frac{1}{(i^{2}+j^{2})^{\gamma}}\leq c_{\gamma}\) for every \(n\), so that \(\|h^{\alpha}D_{\mathbf{n}}(\mu)\|_{2}^{2}=o(n^{2})\) if and only if \(2\alpha-2\gamma+2>0\), that is, \(\alpha>\gamma-1\). Finally, when \(\gamma=1\), we obtain the estimate  
\[\sum_{i,j=1}^{n}\frac{1}{i^{2}+j^{2}} \leq k+\int_{1}^{n}\int_{1}^{n}\frac{1}{x^{2}+y^{2}}dxdy\] \[\leq k+\int_{1}^{n}\left(\int_{0}^{\frac{\gamma}{2}}\frac{1}{r}d \theta\right)dr\] \[=k+\frac{\pi}{2}\log(n),\]  
with \(k\) being any constant independent of \(n\) satisfying  
\[k\geq 2\sum_{j=1}^{\infty}\frac{1}{j^{2}+1}-\frac{1}{2}.\]As above, this leads to the conclusion that \(\|h^{\alpha}D_{\mathbf{n}}(\mu)\|_{2}^{2}=o(n^{2})\) if and only if \(\alpha>\gamma-1\), that is \(\gamma<\alpha+1\). Then the proof is complete by Theorem 2 with \(d_{n}=n^{2}\).