Does the following integral converge? Let $B_{R+\varepsilon}(0)/B_{R-\varepsilon}(0)$ be the annular region of width $\varepsilon$ centered at the origin. I'm interested if the following integral
$$\lim_{\varepsilon\to 0}\int_{B_{R+\varepsilon}(0)/B_{R-\varepsilon}(0)}\frac{1}{\varepsilon^2}\left(1+|x|^2\right)dx$$
converges to something finite. I think not, but how would I show that rigorously if that's the case?
 A: Note that your annulus actually has a width of $2\epsilon$, so I've changed the "inner ball" to be $B_R(0)$.
Working with a specific case will give us an idea of how it behaves.
Doing this question in $\mathbb{R}$ just feels degenerate, so I'll start by carrying out the integral in $\mathbb{R}^2$. (That way it actually looks like an annulus.)
Due to the rotational symmetry, the easiest way to integrate this is by using polar coordinates:
Given a point $(x, y)$ in the plane, $|(x,y)| = r$.
Your integral ends up being: $$\int_0^{2\pi}\int_{R}^{R+\epsilon}\frac{1+r^2}{\epsilon^2}r\mathrm{d}r\mathrm{d}\theta$$ Note the extra $r$ which appears in the integral (referred to as the surface element).
This evaluates to:
\begin{align*}
\int_0^{2\pi}\int_{R}^{R+\epsilon}\frac{1+r^2}{\epsilon^2}r\mathrm{d}r\mathrm{d}\theta &= \int_0^{2\pi}\int_{R}^{R+\epsilon}\frac{r+r^3}{\epsilon^2}\mathrm{d}r\mathrm{d}\theta\\
&= \int_0^{2\pi}\Big[\frac{r^2}{2\epsilon^2}+\frac{r^4}{4\epsilon^2}\Big]_{R}^{R+\epsilon}\mathrm{d}\theta\\
&= \int_0^{2\pi}\Big[\frac{(R+\epsilon)^2}{2\epsilon^2}+\frac{(R+\epsilon)^4}{4\epsilon^2}-\frac{R^2}{2\epsilon^2}-\frac{R^4}{4\epsilon^2}\Big]\mathrm{d}\theta\\
&= 2\pi\Big[stuff\Big]
\end{align*}
Sorry I couldn't be bothered figuring out precisely what $stuff$ evaluates to, but it is clear that there is a term $\frac{R}{\epsilon}$ when the first fraction is expanded. Therefore the limit does not exist.
Now we can generalise to as many dimensions as we like. Note that the expression was completely independent of angle, and only depended on radius, due to the perfect symmetry of the annulus. This means integrating over d$\theta$ only had the effect of multiplying by a constant. This can be expanded to $\mathbb{R}^n$.
Consider spherical coordinates in $\mathbb{R}^n$: $r, \phi_1, \phi_2, ..., \phi_{n-1}$. The the integral is: $$k\int_{R}^{R+\epsilon}\frac{1+r^2}{\epsilon^2}r^{n-1}\mathrm{d}r$$ where $k$ is calculated from integrating over $\phi_i$. When this is evaluated, by the binomial theorem, we can be certain that linear terms of the form $\frac{cR}{\epsilon}$ exist. Hence as $\epsilon \rightarrow 0$, the integral never converges. Unfortunately I don't know any maths outside $\mathbb{R}^n$ so my proof can't get any more general.
A: Hint: That integral is greater than the minimum value of $(1+|x|^2)/\epsilon^2$ over the annulus times the area of the annulus.
