Showing $\sup\limits_{y\neq 0}\int\limits_{|x|\ge2|y|}\frac{|y|^{\delta}}{|x|^{n+\delta}}dx< \infty$

Question

Showing $\sup\limits_{y\neq 0}\int\limits_{|x|\ge2|y|}\frac{|y|^{\delta}}{|x|^{n+\delta}}dx< \infty$, where $\delta>0$ and $n$ is the dimension

First, if $n\ge2$ then under normal circumstances the integral should be finite, but what happens if you take the supremum, does l'Hôpital imply that there's also then no problem ? And for $n=1$ I'm not sure if the claim is still true

Answer 1

We have $\displaystyle\int_{|x|\geq 2|y|}\dfrac{|y|^{\delta}}{|x|^{n+\delta}}dx\\ =\omega_{n-1}\displaystyle\int_{2|y|}^{\infty}\dfrac{|y|^{\delta}}{r^{n+\delta}}r^{n-1}dr\\ =\omega_{n-1}|y|^{\delta}\dfrac{1}{\delta}\dfrac{1}{2^{\delta}|y|^{\delta}}\\ =\omega_{n-1}\dfrac{1}{\delta 2^{\delta}}$