Eigenvector of a "potential" operator

Question

Let $H=L^2(\Omega)$ with $\Omega \subset \mathbb R^n$ bounded. Define $K:H\to H$ as $$Kf(\vec y):=\int_{\Omega} \frac{f(\vec x)}{<\vec x,\vec y>^\alpha}d\vec x\qquad 0<\alpha<0.5$$ Then $K$ is symmetric, compact (Hilbert-Schmidt operator) and positive definite. Then the spectral theorem ensures that $$\|K\|=\lambda_{max}$$ and so exists a dominant eigenvector. Is there any method to find it? Maybe just for some special domain like $\Omega=(0,1)\times (0,1)$ and some value of $\alpha$? This would imply to solve $$\lambda_{max}f(\vec y)=\int_{\Omega} \frac{f(\vec x)}{<\vec x,\vec y>^2}d\vec x$$