Functions in $\mathbb{R}^{d}$ which are integrable or non-integrable (Stein & Shakarchi) I'm reading Stein and Shakarchi's book on real analysis, and twice already they've mentioned  the following without proving it:
$$f_{a}(x) = \begin{cases}|x|^{-a} & |x| \leq 1 \\ 0 & |x|  > 1\end{cases} $$
is integrable exactly when $a < d$ (where the underlying space is $\mathbb{R}^{d}$, and
$$F_{a}(x) = \frac{1}{1 + |x|^{a}} $$ is integrable exactly when $a > d$. 
Maybe I'm wrong and they have proved it somewhere, but this is not my primary textbook. Is it easy to see that these functions are integrable under these conditions, and not integrable otherwise? This dependence on the dimension of the space is very surprising to me, and I can't begin to see why this should be the case; a plausibility argument would be enough for me.
 A: The trick is to use "polar coordinates'' to reduce to one-variable integrals.  In two dimensions, recall that the area integral $dA$ transforms to $r dr d\theta$, so an $r$ multiplier appears.  In general, you get an $r^{d-1}$ multiplier in $d$-dimensional space, which is where the difference occurs.  
Letting $\sigma(S^{d-1})$ denote the "surface measure" of the entire sphere, the first integral reduces to 
$$\sigma(S^{d-1})\int_0^1 \frac{r^{d-1}}{r^a} \, dr = \sigma(S^{d-1})\int_0^1 r^{(d-a) - 1}\, dr$$
which is integrable precisely if $d-a > 0$ by the usual one-variable arguments.
Similarly, the second integral reduces to
$$\sigma(S^{d-1})\int_0^{\infty} \frac{r^{d-1}}{1 + r^a} \, dr,$$
which (for $a > 0$) is continuous on $[0, 1]$ and on $[1, \infty)$ is comparable with $\int_1^{\infty} r^{(d-a) - 1 }\, dr$, and you need $d-a < 0$ for the power to be integrable at $\infty$.
A: You can use estimates based on breaking up $\mathbb R^d$ into differences of balls ("annuli").  E.g., for the second case, use $\mathbb R^d =\bigcup\limits_{k=1}^\infty B(0,k)\setminus B(0,k-1)$.  Note that the $k^{th}$ annulus has volume $c_d(k^d-(k-1)^d)$, roughly $Ck^{d-1}$.    Combined with upper and lower bounds for the integrand on these annuli, you get bounds for the integral above and below by series with $k^{th}$ term asymptotic to $Ck^{(d-1)-a}$.
Similarly for the first case, except using $B\left(0,\frac1k\right)\setminus B\left(0,\frac1{k+1}\right)$, you can obtain upper and lower bounding series with $n^{th}$ term asymptotic to $Ck^{a-(d+1)}$.
