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I want to prove that $$\int_{\mathbb R^n} (1+|x|)^{-\alpha} dx < \infty$$ if and only if $\alpha > n$, but I have no idea how to generally prove this. It is easy to see for $n=1, 2 \ldots$ but I am unable to find an elegant proof for arbitrary dimension.

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The integrand is radially symmetric, so express the integral in polar coordinates, the question is thus reduced to integration theory in one variable. – Andrew Ursitti Jan 5 '14 at 19:05
The problem is that $n$-dimensional polar coordinates feature some ugly terms in the functional determinant... Would you mind to elaborate on how to get rid of those? – AlexR Jan 5 '14 at 19:06
I will give a full answer... – Andrew Ursitti Jan 5 '14 at 19:06
Thank you, Andrew :) – AlexR Jan 5 '14 at 19:07
up vote 1 down vote accepted

Using polar coordinates in $\mathbb R^n$, we have that $dx=r^{n-1}dr\,ds$, where $ds$ is the $(n-1)$-dimensional measure of the unit sphere $S^{n-1}=\{x\in\mathbb R^n : |x|=1\}$, with area $\omega_{n-1}$. We have that $$ \int_{\mathbb R^n} f(x)\,dx=\int_0^\infty\left(\int_{S^{n-1}}f(rs)\,ds\right)r^{n-1}\, dr. $$ In your case $$ \int_{\mathbb R^n} \frac{dx}{(1+|x|)^a}=\int_0^\infty\left(\int_{S^{n-1}}\frac{ds}{(1+r)^a}\right)r^{n-1}\, dr=\omega_{n-1}\int_0^\infty\frac{r^{n-1}dr}{(1+r)^a}. $$

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Haha should have thought about shifting the angles into the surface measure myself. Thanks :) – AlexR Jan 5 '14 at 19:21
@Andrew I'm still having problems resolving the integral (Maple says it's $\frac{\Gamma(\alpha-n)\Gamma(n)}{\Gamma(\alpha)}$...) – AlexR Jan 5 '14 at 20:57
@AlexR: By using the substitution $t=\dfrac1{1+r}$, we will soon arrive at a familiar expression. Actually, now that I think about it, you could have done that from the start, in the original integral, with $|x|$ instead of r. – Lucian Jan 7 '14 at 2:59
Okay I got the desired result now by means of binomial theorem and translation $u=r-1$. – AlexR Jan 11 '14 at 11:11

The answer by Yiorgos S. Smyrlis explains what to do if you want to evaluate the integral. But if you just want to prove it's finite or infinite, there is usually no need to evaluate the integral; estimates suffice. This is what I would do: $$\int_{|x|<1 }(1+|x|)^{-\alpha}\,dx \le \int_{|x|<1 }1\,dx<\infty \tag{1}$$ so we can ignore the unit ball. Then, for $k=0,1,2,\dots$ $$\int_{2^k\le |x|<2^{k+1} }(1+|x|)^{-\alpha}\,dx \le \int_{2^k\le |x|<2^{k+1} } 2^{-k\alpha}\,dx\le 2^{-k\alpha}c_n 2^{(k+1)n} \tag{2}$$ where $c_n 2^{(k+1)n}$ is the volume of the ball of radius $2^{k+1}$. Sum (2) over $k$, and see the series convergent when $\alpha>n$. In the opposite direction, using $1+|x|\le 2|x|$, $$\int_{2^k\le |x|<2^{k+1} }(1+|x|)^{-\alpha}\,dx \ge \int_{2^k\le |x|<2^{k+1} } 2^{-(k+2)\alpha}\,dx\ge 2^{-(k+2)\alpha} c_n 2^{(k+1)n}(1-2^{-n}) $$ which diverges when summed over $k$, when $\alpha\le n$. Since the estimate here is from below, I had to account for the volume of spherical shell rather than just replace it by a larger ball; this is responsible for the factor $(1-2^{-n})$.

This idea of dyadic decomposition of scales reduces some seemingly tricky convergence problems to bookkeeping of exponents. In the context of infinite series, its analog is Cauchy condensation.

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Thanks for your answer, 900sit-ups. I will keep this approach in mind for future use, though evaluation in this specific example was about as much work, I'm sure it will prove useful for other problems. – AlexR Jul 27 '14 at 9:47

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