When is $ \int_{B(0;1)}{ 1 \over ∣x∣^{n} }$ is finite in $R^d$?

Question

Is $$ \int_{B(0;1)}{ 1 \over ∣x∣^{n−2} }$$ is finite for $n > 2$ ? Can we conclude this from local integrability of $\displaystyle\frac{1}{∣x∣^{n−2}}$?

I needed this in when is : $\int_{R^n\backslash\{0\}}|{f(x) \over |x|^{n-2}}| dx < \infty$ given $f$ is summable and continuous?

Answer 1

Let us denote the dimension of the space by $d$. Then the Jacobian determinant of the polar/spherical coordinate change contains a factor $r^{d-1}$ (times some sines or cosines of the angles). If we integrate a rotationally invariant function, i.e., we can write $f(x) = g(\lvert x\rvert)$, for a function $g\colon (0,\infty)\to \mathbb{R}$, then introducing polar/spherical coordinates gives

$$\int_{a < \lvert x\rvert < b} f(x)\,dx = \omega_{d-1}\int_a^b g(r)r^{d-1}\,dr,\tag{1}$$

where $\omega_{d-1}$ is the $d-1$-dimensional volume of the unit sphere $\{x\in\mathbb{R}^d : \lvert x\rvert = 1\}$. Note that $\omega_{d-1}\cdot r^{d-1}$ is the area of the $d-1$-dimensional sphere of radius $r$, so

$$\omega_{d-1}g(r)r^{d-1} = \int_{\lvert x\rvert = r} g(\lvert x\rvert)\,dS(x) = \int_{\lvert x\rvert = r} f(x)\,dS(x),$$

where $dS$ denotes the $d-1$-dimensional surface measure on the sphere, so $(1)$ transforms the $d$-dimensional integral into an iterated integration, much like one does for integrals over products of intervals.

If we look in particular at functions of the form $f_\alpha\colon x \mapsto \frac{1}{\lvert x\rvert^\alpha}$, we see that - for $0 < a < b < \infty$ - we have

$$\int_{a < \lvert x\rvert < b} f_\alpha(x)\,dx = \omega_{d-1} \int_a^b \frac{r^{d-1}}{r^\alpha}\,dr = \int_a^b r^{d-1-\alpha}\,dr = \begin{cases} \omega_{d-1} \log \frac{b}{a} &, \alpha = d\\ \dfrac{\omega_{d-1}}{d-\alpha}\left(b^{d-\alpha} - a^{d-\alpha}\right) &, \alpha\neq d.\end{cases}$$

From that we read off that the integral remains bounded as $a\to 0$ (for $b$ fixed) if and only if $\alpha < d$, since then $a^{d-\alpha} \to 0$, while $a^{d-\alpha}\to\infty$ for $a\to 0$ if $\alpha > d$, and also $\log \frac{b}{a}\to \infty$ for $a\to 0$ for the case of equality.

On the other hand, for $b\to\infty$ with a fixed $a$, the integral remains bounded if and only if $\alpha > d$.

So we have, that for arbitrary $R \in (0,\infty)$

$$\int_{\{ x\in \mathbb{R}^d :\lvert x\rvert < R\}} \frac{dx}{\lvert x\rvert^\alpha} < \infty \iff \alpha < d$$

and

$$\int_{\{ x\in \mathbb{R}^d :\lvert x\rvert > R\}} \frac{dx}{\lvert x\rvert^\alpha} < \infty \iff \alpha > d.$$

Answer 2

Expressing the integral in polar coordinates (and setting $k\equiv 2-n$) $$ \int_{B^d(0,1)}\vert x\vert^k \mathrm{dx} = C_d\int_{0}^1r^{k+d-1}\mathrm{dr}$$ where $C_d$ is a finite constant depending on the dimension [It is the surface area of the $d$-dimensional unit-sphere]. The Jacobian of the polars introduces the extra $r^{d-1}$. For the integral to exist, the overall exponent cannot be negative, so $k+d-1=d-n+1\geq 0$ is required.