Limit with variable in integral boundary I have the following integral
$$\int_{\vert x\vert\leq\varepsilon}\frac{1}{\varepsilon}\frac{1}{\vert x\vert^2}\,\mathrm{d}x,$$
where the integral should be understood as a Lebesgue integral and where $x\in\mathbb{R}^{n}$. My question is now, can we say what happens if $\varepsilon\to 0$? I don't know how to approach this question in a mathematical way. The problem here is that on the one hand our integral region becomes smaller and smaller, which means that in the end we will end up with a Lebesgue-zero set, but in the same time our integrand becomes larger and larger.  Any ideas?
 A: Of course if $n\le2$ all the integrals are $+\infty$. Assume $n>2$.
Say $I(\epsilon)$ is the integral in question. We can use a change of variable, in particular a dilation, to rewrite $I(\epsilon)$ in terms of $I(1)$. (Similar tricks are often useful; here it's perhaps motivated by an attempt to fix the difficulty you point out, that the domain and the function are both changing.)
A simple way to get straight how an integral in $\Bbb R^n$ transforms under dilations is this:


If $\lambda<0$ then $\int f(\lambda x)\frac{dx}{|x|^n}=\int f(x)\frac{dx}{|x|^n}.$


We apply this with $f(x)=|x|^{n-2}\chi_{B_1}(x)$: $$\begin{align}I(\epsilon)&=\epsilon^{n-3}\int_{|x/\epsilon|<1}|x/\epsilon|^{n-2}\frac{dx}{|x|^n}
\\&=\epsilon^{n-3}\int f(x/\epsilon)\frac{dx}{|x|^n}
\\&=\epsilon^{n-3}\int f(x)\frac{dx}{|x|^n}
\\&=\epsilon^{n-3}I(1).\end{align}$$
So the limit as $\epsilon\to0$ is this or that, depending on the value of $n$.
A: You can definitely ask about what happens when $\epsilon \to 0$ because after all for each (fixed) $\epsilon>0$ your integral is a number -- one real number. So, there is an association: $\epsilon \to I(\epsilon)$, where $I(\epsilon)$ stands for the value of the integral. Of course, one needs to show that the integral is finite.
So, your question is
$$
\lim_{\epsilon \to 0^+} I(\epsilon)= ?
$$
As you pointed out the answer is not trivial, becasue although your set shrinks down to a null set, the interand grows to infinity. So, how do they balance each other out?
The answer can be found by explicitly computing the integral, probably using spherical coordinates. Notice that $1/\epsilon$ is a constant and can be pulled out of the integral sign.
A: Hint: Different answers for different $n.$ For example, if $n=1,2,$ you should be able to see that for any $\epsilon>0,$ the integrals equal $\infty.$ That's not true in higher dimensions however.
A: In $n$ dimensional spherical coordinates we have
$$\int_{|x|\leq \epsilon} \frac{1}{\epsilon |x|^2}dx = \int_0^\epsilon dr\frac{r^{n-3}}{\epsilon}\int_{S^{n-1}}d\Omega$$
where $d\Omega$ is the surface measure of $S^{n-1}$ embedded in $\Bbb{R}^n$. From here it is clear that as $\epsilon \to 0^+$ we have that
$$= \begin{cases} +\infty & n=1,2 \\ 4\pi & n=3 \\ 0 & n\geq 4\end{cases}$$
