2
$\begingroup$

I assume only $\alpha \gt 1$ gives $\int_{\mathbb{R}^n} (1+|x|)^{-\alpha} \mathrm{d}x \lt \infty$ (simply because this is true for $n=1$). I also assume some clever transformation could be used for the proof.

I tried to proof this by induction to $n$. $n=1$ is quiet simple, for $n \mapsto n+1$ I thought about:

$$\Phi: ((0,\infty),(-\pi,\pi),\mathbb{R}^{n-2}) \rightarrow \mathbb{R}^n,\quad (r,\varphi,\overline{x}) \mapsto (r \cos \varphi, r \ sin \varphi, \overline{x}),$$

where $x = (x_1,x_2,\overline{x})$. Then $|\det \Phi'| = r$ and the transformation formula gives:

$$ \int_{R^n} (1+x)^{-\alpha} \mathrm{d}x = \int_{R^{n-2}} \int_0^\infty \int_{-\pi}^\pi \frac{r}{\left(1+\sqrt{r^2+(x_3^2+\dots+x_n^2)}\right)^\alpha} \mathrm{d}\varphi \mathrm{d}r \mathrm{d}(x_3,\dots,x_n) $$

The original idea was to somehow use the induction principle here, i. e. to somehow rip $\overline{x}$ out of $\sqrt{r^2+\overline{x}}$. But how?

Does this approach have a chance to be successful? If yes: How to follow up? If no: What else should I do?

PS: $|x|$ is the $2$-norm and all integrals are Lebesgue integrals.

$\endgroup$
  • 4
    $\begingroup$ You need $\alpha > n$. Spherical/Polar coordinates give you a factor $r^{n-1}$, so the meat of it is $$\int_0^\infty \frac{r^{n-1}}{(1+r)^\alpha}\,dr.$$ $\endgroup$ – Daniel Fischer Feb 23 '14 at 14:59
  • $\begingroup$ iI assume the one-dimension integral would not be that much of a problem, but I don‘t get how to use polar coordinates here? $\endgroup$ – Keba Feb 23 '14 at 15:05
  • $\begingroup$ With @DanielFischer advice you can see that the integrand is $\sim r^{n - 1 -\alpha}$ when $r \gg 1$. Upon integration it becomes $\sim r^{n - 1 - \alpha + 1} = r^{n - \alpha}$. Convergence needs to have $n -\alpha < 0$ which means $\alpha > n$. $\endgroup$ – Felix Marin Jul 31 '14 at 3:02
2
$\begingroup$

Answer. $$ \int_{\mathbb R^n}\frac{dx}{(1+|x|)^a}<\infty $$ if and only if $a>n$.

Proof. The easiest way to show this is using polar coordinates: $dx=r^{n-1}dr\, d\vartheta$: $$ I(M)=\int_{|x|<M}\frac{dx}{(1+|x|)^a}=\int_0^M \left(\int_{|s|=1}\frac{1}{(1+r)^{a}} ds \right)r^{n-1}dr=\omega_{n-1}\int_0^M \frac{r^{n-1}\,dr}{(1+r)^a}. $$ where $\omega_{n-1}$ is the area of the unit sphere in $\mathbb R^n$. If $a>n$, then \begin{align} I(M)&\le \omega_{n-1}\int_0^M \frac{(1+r)^{n-1}\,dr}{(1+r)^a}=\omega_{n-1}\int_0^M (1+r)^{n-a-1}dr=\omega_{n-1}\left.\frac{(1+r)^{n-a}}{n-a}\right|_0^M \\&=-\frac{\omega_{n-1}(1+M)^{-a+n}}{n-a}+\frac{\omega_{n-1}}{a-n}<\frac{\omega_{n-1}}{a-n}. \end{align} Thus $$ \int_{\mathbb R^n}\frac{dx}{(1+|x|)^a}=\lim_{M\to\infty} I(M)\le \frac{\omega_{n-1}}{a-n}<\infty. $$ On the other hand, if $a\le n$, then \begin{align} I(M)&=\omega_{n-1}\int_0^M \frac{r^{n-1}\,dr}{(1+r)^a}>\omega_{n-1}\int_1^M \frac{r^{n-1}\,dr}{(1+r)^a}\ge \omega_{n-1}\int_1^M \frac{r^{n-1}\,dr}{(2r)^a} \\&= \frac{\omega_{n-1}}{2^a} \int_1^M r^{n-1-a}\,dr\ge \frac{\omega_{n-1}}{2^a} \int_1^M r^{-1}\,dr=\frac{\omega_{n-1}}{2^a} \log M\to\infty. \end{align}

$\endgroup$
  • $\begingroup$ Thanks! Just one question: Which transformation is used for $\mathrm{d}x =r^{n-1} \mathrm{d}r \mathrm{d}{\phi}$? $\endgroup$ – Keba Feb 23 '14 at 15:52
  • 1
    $\begingroup$ These are the $n-$dimensional polar coordinates. See scipp.ucsc.edu/~haber/ph116A/volume_11.pdf $\endgroup$ – Yiorgos S. Smyrlis Feb 23 '14 at 15:59
  • $\begingroup$ Ah with that hint the problem becomes lots easier, thanks. $\endgroup$ – Keba Feb 23 '14 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.