Let $ K_R (y) \subset \mathbb{R}^n $ be an n-dimensional Ball with radius $R>0$ and midpoint $y \in \mathbb{R}^n $
Far what $ \alpha \in \mathbb{R} $ and $ n \in \mathbb{N}\{0\} $ do following integrals exist:
$ \int_{K{_R(y)}} |x-y|^{\alpha}dx $
$ \int_{ \mathbb{R}^n \setminus K{_R(y)}} |x-y|^{\alpha}dx $
$ \int_{ \mathbb{R}^n } |x-y|^{\alpha}dx $
with what criteria can i solve this task? I don't know how to begin. Thanks for any help!!
Let's take $ \int_{K{_R(y)}} |x-y|^{\alpha}dx $ ,$y=0, n=2 $
If I replace by polarcoordinates $ x= rsin \phi $ : $ \int_0^{2 \pi} \int_0^R | r|^{\alpha} r dr d\phi $..is this the right way?
Let $n\omega_n$ denote the $n-1$ dimensional surface area of the $n-1$-sphere.
Your integrals, respectively, are $$\int_{K_R(y)} |x-y|^\alpha \, dx= \int_0^R \int_{|x-y| = r} |x-y|^\alpha \, dx \, dr = n \omega_n \int_0^R |x-y|^{n+\alpha} \, dr$$ $$\int_{\mathbb R^n \setminus K_R(y)} |x-y|^\alpha \, dx= \int_R^\infty \int_{|x-y| = r} |x-y|^\alpha \, dx \, dr = n \omega_n \int_R^\infty |x-y|^{n+\alpha} \, dr$$ $$\int_{\mathbb R^n} |x-y|^\alpha \, dx= \int_0^\infty \int_{|x-y| = r} |x-y|^\alpha \, dx \, dr = n \omega_n \int_0^\infty|x-y|^{n+\alpha} \, dr$$ where $dx$ is the usual $n$-dimensional volume element in the integrals on the left and the $n-1$ dimensional surface area element in the integrals in the middle.
The remaining improper integrals are straight forward to work out.