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The space is $n$ dimensional euclidean space.The integral is:

$\int_{|x|>1}|x|^{-2np}dx$

$p$ is just a constant. What's more, the result is $\frac{w_{n-1}}{(2p-1)n}$, where $w_{n-1}$ is the $n$-spherical surface area.

hint: let $x=\frac{1}{t}$.but mine is not the same as the answer above.

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    Why do you write $x = 1/t$, when $x$ is a point in $\mathbb R^n$?2012-10-12

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We use the following result: when $f$ is a radial integrable function, that is, we can write $f(x)=g(\lVert x\rVert)$ where $\lVert \cdot\rVert$ is the Euclidian norm, we have $\int_{\Bbb R^n}f(x)dx=nV_n\int_0^{+\infty}r^{n-1}g(r)dr,$ where $V_n$ is the volume of the unit ball of $\Bbb R^n$ for the Euclidian norm.

To see that, we can show it when $f$ is a finite sum of maps of the form $a_j\chi_{A_j}$, where $A_j=\{x\in\Bbb R^n, |x|\in B\}$ and $B$ is a Borel subset of the real line. Then we use an approximation argument.

Back to the problem: let $f(x):=|x|^{-2np}\chi_{x,|x|\geq 1}$. It's a radial function.

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    I have been forget this. Thank you very much!2012-10-12