In the integrals on $\mathbb{R}^n$, are those statement true? If then, How can I prove those facts? $ a
The integrals on $\mathbb{R}^n$
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0Switch to polar coordinates to take advantage o$f$ the geometry o$f$ your domain of integration. (Note: This is essentially the same thing that @ncmathsadist suggests) – 2012-06-10
2 Answers
Here is a less-frequently-given answer which does not use polar coordinates. Let $I=\int_{1<|x|<2}\frac{dx}{|x|^{\alpha}}$ The simple change of variables $x=2^{-k}u$ ($k$ any integer) shows that $\int_{2^k<|u|<2^{k+1}}\frac{du}{|u|^{\alpha}} = 2^{(n-\alpha)k}I$ The integral over $B(0,1)$ is the sum of this [geometric] series over integers $k\le -1$, so it converges iff $n-\alpha>0$. Similarly, the integral over $B(0,1)^c$ is the sum over integers $k\ge 0$, so it converges iff $n-\alpha<0$.
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0This is another nice proof. It gives me great intuiion. Thanks – 2012-06-11
Use the fact that $dx = r^{n-1}\,dr \,d\sigma$, where $d\sigma$ is surface area measure on $S^{n-1}$. Then these become univariate integrals and the question is easy to answer.
For the first integral you have
$\int_{B_1(0)} {dx\over |x|^\alpha} = \int_{S^{n-1}}\int_0^1 {r^{n-1}\,dr\,d\sigma \over r^\alpha} = \sigma(S^{n-1})\int_0^1 r^{n-\alpha - 1}\,dr $ This last integral is finite iff $n-\alpha - 1 > -1$, or when $n > \alpha$.
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0I approached nice conclution by your helps. Thanks all. – 2012-06-11