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.

  • 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

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}

  • $\begingroup$ 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
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    $\begingroup$ 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

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