Estimate $\int_{\|x\|\ge\delta}\frac1{\|x\|^{d+1}}\mathrm d x$ without spherical coordinates.

Question

Is it possible to estimate the following Lebesgue integral ($\|\cdot\|$ is the 2-norm) $$\int_{\|x\|\ge\delta}\frac1{\|x\|^{d+1}}\mathrm d x, \, x\in\Bbb R^d$$ in terms of $\delta$ when $\delta\to 0$? That is to say, to bound it by $O(\frac1{\delta^\alpha})$.

I know spherical coordinate transform may be a way to do it or even better yet - to give the exact value. But coordinate transform itself is quite troublesome to justify. What's more, I'm not interested in the exact value anyway, all I want is a best big O bound. So is there any good alternative, without spherical coordinate system?

What I tried, the integral is equivalent to a slightly modified one interms of the big O asymptotic of $\delta$: $$\int_{\|x\|_\infty\ge\delta}\frac1{\|x\|^{d+1}}\mathrm d x,$$ since $$\|x\|_2\ge\delta\implies \|x\|_\infty\ge \frac{\delta}{d^\frac12}\implies \|x\|_2\ge \frac{\delta}{d^\frac12}. $$ So the original integral over the space minus a ball is replaced by one over the space minus a cube - the sum of several rectangles. In dimension as low as $d=1,2$ the integral can be easily carried out over each of the rectangle to be summed up, since there are not many rectangles, but as the dimension goes higher there are as many as $3^d-1$ ones, impossible to be listed one by one. So I couldn't get any further following this seemingly hopeful thread of thought.

Answer 1

If you really want to avoid spherical coordinates, you can use the decomposition: $$ \int_{||x||\geq \delta}\frac{dx}{||x||^{d+1}}=\sum_{k=0}^{\infty}\int_{2^k\delta\leq ||x||<2^{k+1}\delta}\frac{dx}{||x||^{d+1}} $$ $$ \leq \sum_{k=0}^{\infty}(2^k\delta)^{-d-1}m(\{x:2^k\delta\leq ||x||<2^{k+1}\delta\})$$ where $m$ is Lebesgue measure. For a very crude estimate, we can say that $$ m(\{x:2^k\delta\leq ||x||<2^{k+1}\delta\})\leq m(\{x:||x||\leq 2^{k+1}\delta\})=\omega_d(2^{k+1}\delta)^d $$ where $\omega_d$ is the volume of the unit ball (it can be proved that the measure of a ball of radius $r$ is $\omega_dr^d$ just from the basic properties of Lebesgue measure, without using spherical coordinates).

Hence $$\int_{||x||\geq \delta}\frac{dx}{||x||^{d+1}}\leq \frac{2^d\omega_d}{\delta}\sum_{k=0}^{\infty}2^{-k}=\frac{2^{d+1}\omega_d}{\delta} $$

Answer 2

I don't see why you think the coordinate transformation is troublesome... Take radial and spherical coordinates, but forget about the spherical bit... i.e. $$ ||x|| = r \quad \& \quad dx = \omega(\theta) r^{d-1} dr $$ where $\omega(\theta)$ and doesn't depend on $r$ (and doesn't depend on $\delta$). Thus $$ \int_{ ||x|| \geq \delta} \frac{ dx}{||x||^{d+1}} = \Omega(d) \int_{\delta}^{\infty} \frac{ dr}{r^2}= \frac{ \Omega(d) }{ \delta} $$ where $\Omega(d)$ is the spherical component of the integral (You can show it with fubini's theorem). I think you can take it from here.