Integration of a radial function over a bounded domain Let $\Omega$ be a bounded domain in $\mathbb{R}^N$. Let $f(x)=|x|^\alpha$. Then $f\in L^1(\Omega)$ if $\alpha>-N$.
The above fact seems to hold for the following reason.
Since $\Omega$ is bounded, there exists $R>0$ such that $\Omega\subset B(0,R)$. Then, we have
\begin{align*}
I&=\int_{\Omega}f(x)dx\\
&\leq\int_{B(0,R)}|x|^\alpha dx\\
&=\frac{r^{\alpha+N}}{\alpha+N}\Bigg|_{0}^{R}\\
&=\frac{R^{\alpha+N}}{\alpha+N},
\end{align*}
where $\alpha+N>0$.
Then we get $f\in L^1(\Omega)$.
Kindly inform me, if the above argument seems fine with you.
Thanks.
 A: The idea behind the proof is correct, but I believe there is a computation error (off by a scalar multiple). Note that the $N$-dimensional ball is a union of concentric $(N-1)$-dimensional spherical shells, hence we have that
$$\frac{dV_N(r)}{dr}=S_{N-1}(r)=r^{N-1} S_{N-1}(1) \tag{$\ast$}$$
where $V_n(r)$ and $S_{n}(r)$ are the volume and surface area of the $n$-ball and $n$-sphere of radius $r$ respectively. We use hyperspherical coordinates $(r,\varphi_1,\varphi_2,\dots,\varphi_{N-1})$ to evaluate the integral. Note that one can simply plug in $dV=r^{N-1} S_{N-1}(1)~dr$ (see the earlier computation in $(\ast)$, and note the abuse in notation) due to radial symmetry of the integrand. Hence, for $R>0$ such that $\Omega\subset B(0,R)$, we have that
$$I=\int_{\Omega} f(x)~dV\leq \int_{B(0,R)} |x|^{\alpha}~dV=\int_{0}^R r^{\alpha}\cdot r^{N-1} S_{N-1}(1)~dr=S_{N-1}(1)\cdot \frac{R^{\alpha+N}}{\alpha+N},$$
which is finite for $\alpha>-N$, hence $f\in L^1(\Omega)$.

In case you are interested, an explicit formula for $S_n(1)$ for $n\in \mathbb{N}$ is given by
$$S_n(1)=\frac{2\pi^{(n+1)/2}}{\Gamma((n+1)/2)},$$
where $\Gamma$ is the gamma function.
Note: If radial symmetry does not hold, the volume element is given by (see the first link)
$$dV= r^{N-1}\sin^{N-2}(\varphi_1)\sin^{N-3}(\varphi_2)\cdots \sin(\varphi_{N-2})\, dr\,d\varphi_1 \, d\varphi_2\cdots d\varphi_{N-1}.$$
Note 2: Your $f$ is undefined for $\alpha<0$ at $x=0$. To resolve this problem you can set $f$ to be
$$f(x)=\begin{cases} |x|^{\alpha},&\mathrm{if}~x\neq 0, \\ 0 ,&\mathrm{if}~x=0, \end{cases}$$
and use the same proof.
