I have a question from Stein+ Shakarchi's Real Analysis book regarding the integrability of this particular function. (pg. 63-64)
Consider the function $$f(x)= \begin{cases} \frac{1}{ \mid x \mid ^{d+1} }\hspace{2mm} \text{if} \hspace{2mm} x \neq 0, \\ 0 \hspace{2mm} \text{otherwise} \end{cases}$$
We prove that $f$ is integrable outside any ball, $ \mid x \mid \ge \epsilon$
Ok, so at the end of the proof it is concluded that such a function is integrable using this fact. The dilation-invariance property show that $$ \int_{ \mid x \mid \ge \epsilon } \frac{dx}{\mid x \mid ^{d+1} }= \frac{1}{\epsilon} \int_{\mid x \mid \ge 1} \frac{dx}{ \mid x \mid ^{d+1}}$$
I am particularly confused about why we would ever want $\frac{1}{\epsilon}$ if we are trying to to show a function is integrable and therefore has finite integral..$\frac{1}{\epsilon}$ to me means that this integral is getting very large since $\epsilon$ can get very small.
I guess the logic of this proof is not clear to me. I would think we would want to show that the integral is bounded by something finite..