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Let be $K(x,y)=\frac{2x_n}{n\alpha(n)}\frac{1}{\vert x-y\vert^n}$ where n is the dimension and $\alpha(n)$ is the volume o unitary sphere in $R^n$, $x=(x_1,...,x_n)\in R^n$ and $y=(y_1,...,y_{n-1},0)\in\partial R^n_+$. Show that:

$\int_{\partial R^n_+}K(x,y)dy=1$

Thanks in advance!

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    What did you try? Did you get stuck somewhere? Is this a homework problem?2012-11-02
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    This is a step in a demonstration in the book of Evans on Partial differential equations. I can't to do this.2012-11-02
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    As a first step, you can see that the integral does not depend on $x$, since it is invariant under translations in the first $n-1$ coordinates, and dilations, as can be seen by standard change of variables. So you can assume $x=(0,0,\ldots,0,1)$, and then use integration over spheres in $\mathbb{R}^{n-1} = \partial \mathbb{R}^n$. There are probably other ways to do it, but this one works if you crank it out.2012-11-02
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    I don't know what can I do after: $I=\lim_{R\rightarrow \infty}\int_0^R$ $(\int_{S^{n-1}(0,R)}\frac{1}{(r^2+x_n^2)^n} ds)dr$2012-11-02
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    The exponent in the integral should be $n/2$, and you can assume that $x_n = 1$. The sphere you are integrating over is $(n-2)$-dimensional, not $(n-1)$-dimensional, and its surface area is $r^{n-2} \sigma_{n-2}$, where $\sigma_{n-2}$ is the surface area of the $(n-2)$-dimensional unit sphere.2012-11-02
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    Right. I found: $I=\frac{2x_n(n-2)\alpha(n-2)}{n\alpha(n)}\lim_{\rightarrow\infty}\int_0^R\frac{r^{n-2}}{(r^2+x_n^2)^\frac{n}{2}}dr$. It looks like strange.2012-11-02

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