that $w_\tau$ is decreasing in $|x|$ and for $|x|>1$, $$\label{6.4}
w_\tau(x)=\left\{
\arraycolsep=1pt
\begin{array}{lll}
|x|^\tau   &{\rm for}\quad \tau\in(-\infty,-N+2\alpha)\setminus\{-N\},\\[2mm]
  \phantom{    }
|x|^\tau\log^{\gamma_0}(\varrho_0+|x|) \quad&{\rm for}\quad \tau=-N,
\end{array}
 \right.$$ where $\varrho_0=e^{\frac1{2\alpha}}$ and $\gamma_0=\frac{N}{2\alpha}$.

\[lm 5.1\] Assume that $\tau\in(-\infty,-N+2\alpha)$.Then\
$(i)$ for $\tau\in(-\infty,-N)$, there exist $R\ge 4$ and $c_{15}>1$ such that for $|x|>R$, $$\label{5.3}
  \frac1{c_{15}}|x|^{-N-2\alpha}\le -(-\Delta)^\alpha w_{\tau}(x)\le c_{15}|x|^{-N-2\alpha};$$ $(ii)$ for $\tau=-N$, there exist $R\ge 4$ and $c_{15}>1$ such that for $|x|>R$, $$\label{5.4}
 \frac1{c_{15}}|x|^{-N-2\alpha}\log^{\gamma_0+1}|x|\le- (-\Delta)^\alpha w_{\tau}(x)\le c_{15}|x|^{-N-2\alpha}\log^{\gamma_0+1}|x|;$$ $(iii)$ for $\tau\in(-N,-N+2\alpha)$, there exist $R\ge4$ and $c_{15}>1$ such that for $|x|>R$, $$\label{5.05}
 \frac1{c_{15}}|x|^{\tau-2\alpha}\le- (-\Delta)^\alpha w_{\tau}(x)\le c_{15}|x|^{\tau-2\alpha}.$$

[**Proof.**]{} In the following, we shall use the equivalent definition of $(-\Delta)^\alpha w_\tau$, that is, $$(-\Delta)^\alpha w_\tau(x)=-\frac12\int_{{\mathbb{R}}^N}\frac{w_\tau(x+z)+w_\tau(x-z)-2w_\tau(x)}{|z|^{N+2\alpha}}dz.$$ $(i)$ The case of $\tau\in(-\infty,-N)$. On the one hand, for $|x|>4$, we have that $$\label{5.5}
\arraycolsep=1pt
\begin{array}{lll}
 -(-\Delta)^\alpha w_{\tau }(x)=\frac12\int_{{\mathbb{R}}^N\setminus (B_1(x)\cup B_1(-x))}\frac{w_\tau(x+z)+w_\tau(x-z)-2w_\tau(x)}{|z|^{N+2\alpha}}dz
 \\[3mm]\phantom{-------\ }
 +\frac12\int_{B_1(x)\cup B_1(-x)}\frac{w_\tau(x+z)+w_\tau(x-z)-2w_\tau(x)}{|z|^{N+2\alpha}}dz
 \\[3mm]\phantom{------\ \ }
 \le\frac{|x|^{\tau -2\alpha}}2\int_{D_0}\frac{I_x(y)}{|y|^{N+2\alpha}}dy+c_{16}|x|^{-N-2\alpha}
\end{array}$$ where $c_{16}>0$ depends on ${\|w\|}_{L^1(B_1(0))}$, $e_x=\frac{x}{|x|}$, $D_0={\mathbb{R}}^N\setminus (B_{\frac1{|x|}}(e_x)\cup B_{\frac1{|x|}}(-e_x))$ and $$I_x(y) = |e_x+y|^{\tau }+|e_x-y|^{\tau }-2.$$ On the other hand, for $|x|\ge 4$, $$\label{5.6}
\arraycolsep=1pt
\begin{array}{lll}
 -(-\Delta)^\alpha w_{\tau }(x)=\frac12\int_{{\mathbb{R}}^N\setminus (B_1(x)\cup B_1(-x))}\frac{w_\tau(x+z)+w_\tau(x-z)-2w_\tau(x)}{|z|^{N+2\alpha}}dz
 \\[3mm]\phantom{-------\ }
 +\frac12\int_{B_1(x)\cup B_1(-x)}\frac{w_\tau(x+z)+w_\tau(x-z)-2w_\tau(x)}{|z|^{N+2\alpha}}dz
 \\[3mm]\phantom{------\ \ } \ge\frac{|x|^{\tau -2\alpha}}2\int_{D_0}\frac{I_x(y)}{|y|^{N+2\alpha}}dy-2\int_{B_1(x)}\frac{w_\tau(x)}{|z|^{N+2\alpha}}dz
 \\[3mm]\phantom{------\ \ }\ge \frac{|x|^{\tau -2\alpha}}2\int_{D_0}\frac{I_x(y)}{|y|^{N+2\alpha}}dy-c_{17}|x|^{\tau-N-2\alpha},
\end{array}$$ where $c_{17}>0$.

*Claim 1. There exists $c_{18}>1$ such that $$\label{15-09-1}
 \arraycolsep=1pt
 \begin{array}{lll}
  \frac1{c_{18}}|x|^{-N-\tau }\le \int_{D_1\cup D_2}\frac{I_x(y)}{|y|^{N+2\alpha}}dy\le c_{18} |x|^{-N-\tau },
 