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


$p$ 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 $\mathbb R^n$? – Erick Wong Oct 12 '12 at 16:04
up vote 1 down vote accepted

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 suppose we should can get a better result. – user39843 Oct 12 '12 at 15:47
What do you mean by a better result? – Davide Giraudo Oct 12 '12 at 15:53
that means we can get $\int_{|x|>1}|x|^{-2np}dx=\frac{w_{n-1}}{(2p-1)n}$. But I don't kown how to calculate out this result. – user39843 Oct 12 '12 at 15:57
I've added a little more details. – Davide Giraudo Oct 12 '12 at 16:00
I can't get$\int_{0}^{\infty}r^{n-1}r^{-2np}dr=\frac{1}{(2p-1)n}$ as what you imply. – user39843 Oct 12 '12 at 16:13

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