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Let $$ K(x, y)=\frac{2 x_{n}}{n \alpha(n)} \frac{1}{|x-y|^n} $$ 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 $$ \int_{\partial \mathbb{R}_{+}^{n}} K(x, y) \, d y=1 \text{?} $$

There was an answer provided by Evaluating the Poisson Kernel in the upper half space in $n$-dimensions

I just didn't understand why the following step is correct. What is the justification? \begin{equation} \int_{\mathbb{R}^{n-1}} \frac{2 x_{n}}{n \alpha(n)} \frac{1}{|x-y|^n} d y=\frac{2 x_{n}}{n \alpha(n)} \int_{\mathbb{R}^{n-1}} \frac{1}{\left(x_n^2+y_1^2+\ldots+y_{n-1}^2\right)^{\frac{n}{2}}} \, d y \end{equation}

Michael Hardy
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  • If the first n-1 components of x are 0, then it's by Pythagorean theorem. Otherwise, recenter y to make it so the first n-1 components of x are 0. – extremeaxe5 May 27 '21 at 03:26

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