Integral involving distance function

Question

Let $\Omega$ be an open set in $\mathbb{R}^n$, suppose that $K\subset\Omega$ is compact. Why the following integral: $$\int_{\mathbb{R}^n\setminus\Omega}\frac{1}{\text{dist}(y,\partial K)^{n+a}}\,dy$$ is finite, with $a>0$? I think that i have to use the fact that: $\text{dist}(\partial\Omega,\partial K)\geq\alpha>0$, and: $n+a>n$. But I have no ideas on how to proceed, any help is appreciated.

Answer 1

Consider first the case of $\Omega$ be bounded. Take a ball $B_R$ centered at the origin and containing $\Omega$. The set $\mathbb{R}^n\setminus\Omega$ can be written as the union of $$ A := B_R\setminus\Omega, \qquad B := \mathbb{R}^n\setminus B_R\,. $$ As you have already observed, the integral over the bounded set $A$ is finite. On the other hand, on the set $B$ it holds that $$ d(y, B_R) \leq d(y, \partial\Omega) \leq d(y, \partial K) - \alpha, \qquad \forall y\in B, $$ so that $$ \int_B \frac{1}{d(y, \partial K)^{n+a}} dy \leq \int_B \frac{1}{[\alpha + d(y, \partial B_R)]^{n+a}} dy $$ and the last integral is finite (it can be explicitly computed passing in polar coordinates).

If $\Omega$ is unbounded, it is enough to consider an open set $\Omega'\subset \Omega$ containing $K$, and observe that the integral on $\mathbb{R}^n\setminus\Omega$ is less then or equal to the one on $\mathbb{R}^n\setminus\Omega'$.

Answer 2

Idea: Recall how integration in polar coordinates is defined on $\mathbb R^n:$

if $E\subseteq \mathbb R^n$ is measurable and $r>0$ and if we define $E_r:= \{\omega\in S^{n-1}:r\omega \in E\}$ then

$\displaystyle\int_E \frac{1}{d(y,\partial K)^{n+a}} dy = \int_0^\infty \left( \int_{E_r} \frac{1}{d(r\omega,\partial K)^{n+a}}d\omega \right) r^{n-1} dr.$

To get an idea of how to calculate these integrals, suppose $K=\{0\}$ and that we can find a ball $B_{\delta}(0)\subseteq \Omega.$ Then,

$\displaystyle\int_{\mathbb{R}^n\setminus\Omega}\frac{1}{d(y,\partial K)^{n+a}}\,dy\le\int_{\mathbb{R}^n\setminus B_{\delta}(0)}\frac{1}{d(y,\partial K)^{n+a}}\,dy= \int_0^\infty \left( \int_{(\mathbb{R}^n\setminus B_{\delta}(0))_r} \frac{1}{d(r\omega,\partial K)^{n+a}}d\omega \right) r^{n-1} dr\le\left( \int_{(\mathbb{R}^n\setminus B_{\delta}(0))_r} d\omega \right) \left( \int_0^\infty r^{-1-a} dr\right)$.

Each factor is evidently finite, so this integral converges.

For the general case, we find an open ball $K\subset B_{2R}(0)$ such that $R=\max\{|x|:x\in K\}$ and consider

$\displaystyle\int_{\mathbb R^n\setminus \Omega} \frac{1}{d(y,\partial K)^{n+a}} dy=\int_{(\mathbb R^n\setminus \Omega)\cap B_{2R}(0)} \frac{1}{d(y,\partial K)^{n+a}} dy+\int_{(\mathbb R^n\setminus \Omega)\cap B_{2R}(0)^c} \frac{1}{d(y,\partial K)^{n+a}} dy.$

The first integral converges. For the second, note that if $y\in (\mathbb R^n\setminus \Omega)\cap B_{2R}(0)^c$, such that $|y|=r,$ then $d(y,\partial K)\ge r-R$.

Thus, the second integral converges because it is

$\displaystyle \left( \int_{(\mathbb R^n\setminus \Omega)\cap B_{2R}(0)^c)_r} d\omega \right) \left( \int_{2R}^\infty \frac{r^{n-1}}{(r-R)^{n+a}}dr\right)$.