Integrability of $\frac{x_1}{|x|^{n}}$ over the unit ball Is $\frac{x_1}{|x|^{n}}$ integrable over the unit ball of $B_1(\mathbb{R}^n)$? That is, is 
$$\int_{B_1(\mathbb{R}^n)} \frac{|x_1|}{|x|^{n}}<\infty?$$
I know that $|x|^{-a}$ is integrable over the ball if $a<n$, but what if $a<n$ in just one dimension? My heart says it probably isn’t, but I’m not sure.
 A: Just remark that $|x_1|≤|x|$, so
$$
\int_{B_1(\mathbb{R}^n)} \frac{|x_1|}{|x|^n}\,\mathrm{d}x ≤  \int_{B_1(\mathbb{R}^n)} \frac{1}{|x|^{n-1}}\,\mathrm{d}x < \infty
$$

In the more general case when $a≠n$, first remark that the only singularity is at $x=0$ so it is sufficient to prove that the integral is finite on $[-1,1]^n$. On this set, we can factorize the integral using Fubini theorem and writing $x = (x_1,\tilde{x})$ and doing the change of variable $x_1 = |\tilde{x}|\,r$ and then $s=r^2$ one gets
$$
\begin{align*}
\int_{[-1,1]^n} \frac{|x_1|}{|x|^a}\,\mathrm{d}x &= \int_{[-1,1]^{n-1}}\int_{-1}^1 \frac{|x_1|}{\left(\sqrt{|x_1|^2+|\tilde{x}|^2}\right)^a}\,\mathrm{d}x_1\,\mathrm{d}\tilde{x}
\\
&= \int_{[-1,1]^{n-1}}\int_{-1/|\tilde{x}|}^{1/|\tilde{x}|} \frac{|\tilde{x}|\,|r|}{\left(\sqrt{|r|^2+1}\right)^a|\tilde{x}|^a}|\tilde{x}|\,\mathrm{d}r\,\mathrm{d}\tilde{x}
\\
&= \int_{[-1,1]^{n-1}} \frac{1}{|\tilde{x}|^{a-2}}\int_{0}^{1/|\tilde{x}|} \frac{2\,r}{\left(\sqrt{1+r^2}\right)^a}\,\mathrm{d}r\,\mathrm{d}\tilde{x}
\\
&= \int_{[-1,1]^{n-1}} \frac{1}{|\tilde{x}|^{a-2}}\int_{0}^{|\tilde{x}|^{-2}} \frac{\mathrm{d}s}{\left(1+s\right)^{a/2}}\,\mathrm{d}\tilde{x}
\\
&= \frac{2}{a-2}\,\int_{[-1,1]^{n-1}} \frac{1}{|\tilde{x}|^{a-2}}\left(1-(1+|\tilde{x}|^{-2})^{1-a/2}\right)\mathrm{d}\tilde{x}
\\
&= \frac{2}{a-2}\,\int_{[-1,1]^{n-1}} \frac{1}{|\tilde{x}|^{a-2}}-\frac{1}{(1+|\tilde{x}|^2)^{a/2-1}}\mathrm{d}\tilde{x}
\end{align*}
$$
where $a≠ 2$ (if $a=2$ you get a $\ln$ when you compute the integral in $s$). This is integrable if and only if $a-2 <n-1$, so
$$\boxed{\int_{B_1(\mathbb{R}^n)} \frac{|x_1|}{|x|^a}\,\mathrm{d}x < ∞ \ \Leftrightarrow\ a<n+1}$$
