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Let $K(x,y) = \frac{2x_n}{n \alpha(n)} \frac{1}{|x-y|^n}$ be the Poisson Kernel, where $x \in \mathbb{R}_+^n$ (the upper half-space of in $\mathbb{R}^n$), $y \in \mathbb{R}^n$, and $\alpha(n)$ is the volume of the $n$-dimensional unit ball.

How do you show that $\int_{\partial \mathbb{R}_+^n} K(x,y) dy =1?$

I tried doing this in simple cases (e.g. two dimensions), and it can out pretty cleanly (I think you can also probably use complex analysis if we're in two-dimensions?). However, I couldn't figure how to solve it in general $n$ dimensions, because the exponent to the $n$-th power was giving me trouble. How would one go about showing the general case?

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    Hm, I guess I did mean space instead of plane. However, the formula is exactly what was written in Evans's Partial Differential Equations book, and he defined it as the Poisson formula on the upper half-space..2011-08-30

1 Answers 1

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I hope this helps you.

Possibly you've already deduced, using the fact that $\partial \mathbb{R}_+^n \cong \mathbb{R}^{n-1}$

$\int_{\partial \mathbb{R}_+^n} K(x,y) dy = \int_{\mathbb{R}^{n-1}} K(x,y) dy = \int_{\mathbb{R}^{n-1}} \frac{2x_n}{n \alpha(n)} \frac{1}{|x-y|^n} dy \\\\ = \frac{2x_n}{n \alpha(n)} \int_{\mathbb{R}^{n-1}}\frac{1}{(x_{n}^2+y_{1}^2+...+y_{n-1}^2)^{\frac{n}{2}}} dy = \frac{2x_n}{n \alpha(n)} \int_{\mathbb{R}^{n-1}}\frac{1}{(x_{n}^2+|y|^2)^{\frac{n}{2}}} dy$

Now you just factor out $x_{n}^2$ and apply the change of variables $y \mapsto \frac{y}{x_{n}}$ to the above integral. You get

$\frac{2x_n}{n \alpha(n)} \int_{\mathbb{R}^{n-1}}\frac{1}{(x_{n}^2+|y|^2)^{\frac{n}{2}}} dy = \frac{2}{n \alpha(n)} \frac{1}{x_{n}^{n-1}} \int_{\mathbb{R}^{n-1}}\frac{1}{(1+|\frac{y}{x_{n}}|^2)^{\frac{n}{2}}} dy \\\\ = \frac{2}{n \alpha(n)} \int_{\mathbb{R}^{n-1}}\frac{1}{(1+|y|^2)^{\frac{n}{2}}} dy $

Using polar coordinates you get (cf. Evans P. 628, Theorem 4)

$\frac{2}{n \alpha(n)} \int_{\mathbb{R}^{n-1}}\frac{1}{(1+|y|^2)^{\frac{n}{2}}} dy = \frac{2}{n \alpha(n)} \int_{0}^{\infty}\left(\int_{\partial B(0,r)}\frac{1}{(1+r^2)^{\frac{n}{2}}} dS \right) dr \\\\ = \frac{2(n-1)\alpha(n-1)}{n \alpha(n)} \int_{0}^{\infty}\frac{r^{n-2}}{(1+r^2)^{\frac{n}{2}}} dr $

Now $n\alpha(n)$ is just the area $S(n)$ of the $n$-sphere, which is precisely $S(n)=\frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}$ (see for example http://en.wikipedia.org/wiki/Deriving_the_volume_of_an_n-ball#General_form_and_surface_area). So your equation finally becomes

$\frac{2(n-1)\alpha(n-1)}{n \alpha(n)} \int_{0}^{\infty}\frac{r^{n-2}}{(1+r^2)^{\frac{n}{2}}} dr = \frac{2S(n-1)}{S(n)} \int_{0}^{\infty}\frac{r^{n-2}}{(1+r^2)^{\frac{n}{2}}} dr \\\\ = \frac{2\Gamma(\frac{n}{2})}{\Gamma(\frac{n-1}{2})\sqrt{\pi}}\int_{0}^{\infty}\frac{r^{n-2}}{(1+r^2)^{\frac{n}{2}}} dr$

But $\int_{0}^{\infty}\frac{r^{n-2}}{(1+r^2)^{\frac{n}{2}}} dr = \frac{\Gamma(\frac{n-1}{2})\sqrt{\pi}}{2\Gamma(\frac{n}{2})}$ (I this is most easily shown by induction, see below).

Now you've finally derived your desired result, i.e.

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

If there is anything wrong or you want to know some of the steps in more detail please do let me know via giving me a feedback.

$\textbf{Edit:}$ Claim

$\int_{0}^{\infty}\frac{r^{n-2}}{(1+r^{2})^{\frac{n}{2}}}dr=\frac{\Gamma(\frac{n-1}{2})\sqrt{\pi}}{2\Gamma(\frac{n}{2})}$

Proof:By induction on n:

For n=2 we simply have

$\int_{0}^{\infty}\frac{1}{1+r^2}dr=\int_{0}^{\frac{\pi}{2}}\frac{1}{\mathrm{cos}(r)^{2}(1+\mathrm{tan}(r)^{2})}dr$ $=\int_{0}^{\frac{\pi}{2}}\frac{\mathrm{cos}(r)^{2}}{\mathrm{cos}(r)^{2}}dr=\int_{0}^{\frac{\pi}{2}}1dr=\frac{\pi}{2}$

For the general case we use integration by parts to obtain

$\int_{0}^{\infty}\frac{r^{n-2}}{(1+r^{2})^{\frac{n}{2}}}dr=\underbrace{\left[-\frac{1}{n-2}\frac{1}{(1+r^2)^{\frac{n-2}{2}}}r^{n-3}\right]_{0}^{\infty}}_{=0}+\frac{n-3}{n-2}{\int_{0}^{\infty}\frac{r^{n-4}}{(1+r^{2})^{\frac{n-2}{2}}}dr}$

By induction hypotheses we have $\int_{0}^{\infty}\frac{r^{n-4}}{(1+r^{2})^{\frac{n-2}{2}}}dr=\frac{\Gamma(\frac{n-3}{2})\sqrt{\pi}}{2\Gamma(\frac{n-2}{2})}$

Using that for the gamma function we have $x\Gamma(x)=\Gamma(x+1)$ we obtain

$\int_{0}^{\infty}\frac{r^{n-2}}{(1+r^{2})^{\frac{n}{2}}}dr=\frac{n-3}{n-2}\int_{0}^{\infty}\frac{r^{n-4}}{(1+r^{2})^{\frac{n-2}{2}}}dr=\frac{\frac{n-3}{2}}{\frac{n-2}{2}}\frac{\Gamma(\frac{n-3}{2})\sqrt{\pi}}{2\Gamma(\frac{n-2}{2})}\\\\ =\frac{\Gamma(\frac{n-1}{2})\sqrt{\pi}}{2\Gamma(\frac{n}{2})}$

which ends the proof.