For which $\alpha$ is this integral operator compact?

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I have $Q\subset\mathbb{R}^n$

$Af(x)=\int_QK(x,y)f(y)dy$ , with $K(x,y)=\frac{K_0(x,y)}{|x-y|^\alpha}$ and $K_0\in C(Q)$

I want to estimate using an operator

$A_Mf(x)=\int_QK_M(x,y)f(y)dy$ where, $K_M=\begin{cases} K(x,y), & |x-y|\ge M \\ \frac{K_0(x,y)}{M^\alpha}, & |x-y|<M \end{cases}$.

so

$|(A-A_M)f|^2\le \int_Q|K(x,y_1)-K_M(x,y_1)|dy_1\int_Q|K(x,y_2)-K_M(x,y_2)||f(y_2)|^2dy_2$

and supposing $K(x,y)\in L^1(Q)\Rightarrow \int_Q|K(x,y_1)-K_M(x,y_1)|dy_1\le\int_Q|K(x,y_1)|dy_1\le C$

Now I have,

$\int_{|x-y|<M}|K(x,y_2)-K_M(x,y_2)|dy_2=\int_{|x-y|<M}\left | \frac{1}{|x-y|^\alpha}-\frac{1}{M^\alpha}\right|\left|K_0(x,y)\right |dy_2\le ???$ this is my issue

my aim is to have something like this

$\|(A-A_M)f\|^2\le C\int_Q|K(x,y_2)-K_M(x,y_2)||f(y_2)|^2dy_2dx\le \frac{C}{M^{g(\alpha,n)}}$