In the integrals on $\mathbb{R}^n$, are those statement true? If then, How can I prove those facts? $$ a
The integrals on $\mathbb{R}^n$
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0Switch to polar coordinates to take advantage of the geometry of your domain of integration. (Note: This is essentially the same thing that @ncmathsadist suggests) – 2012-06-10
2 Answers
Here is a less-frequently-given answer which does not use polar coordinates. Let $$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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0Actually I didn't learn that equation. Then using that equation, how can I change the range of the integration, that was B(0,1)? – 2012-06-10
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0you then have the radius $r$ raging from 0 to 1 and the angle $\sigma$ from $0$ to $2\pi$ so your integral becomes $\int_0^{2\pi} \int_0^1 r^{-a}r^{n-1} dr d\sigma$ (this is for the case of $B(0,1)$) – 2012-06-10
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0The actual surface area is immaterial. – 2012-06-10
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0I approached nice conclution by your helps. Thanks all. – 2012-06-11