Evaluating the Poisson Kernel in the upper half space in $n$-dimensions

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?

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.