# How to calculate the Fourier transform of the Poisson kernel

We know that the Fourier transform of the Poisson kernel $P(x,t)$ \begin{equation} \frac{\Gamma(\frac{(n+1)}{2})}{\pi^{\frac{(n+1)}{2}}}\frac{t}{(t^2+\lvert x\rvert^2)^{\frac{(n+1)}{2}}} \end{equation} is the Abel kernel $K(x,t)$ \begin{equation} e^{-2\pi t \lvert \xi \rvert}. \end{equation} However, I have just seen one method of proving it from Stein's Introduction to Fourier Analysis on Euclidean Spaces. The key of the proof is to use \begin{equation} e^{-\beta}=\frac{1}{\sqrt \pi} \int_0^\infty \frac{e^{-u}}{\sqrt u} e^{-\frac{\beta^2}{4u}} \, \mathrm{d} u. \end{equation} And it start with the Abel kernel to Poisson kernel. But I feel that this proof is a little trick. So is there any other proof of it?

Thank you very much!

First note that for $$f$$ a Schwartz function on $$\mathbb{R}^n$$ there is a unique $$u\in C^{\infty}(\mathbb{R}^n\times \mathbb{R}_+)$$ satisfying $$\begin{cases} \frac{\partial^2 u}{\partial t^2}+\Delta_x u=0\\ \lim_{t^2+|x|^2\to \infty} u=0\\ \lim_{t\to 0}||u(\cdot, t)-f(\cdot)||_{\infty}=0. \end{cases}$$ Uniqueness of the above solution follows from the maximum principle, and existence follows from choosing $$u(x,t)=P(x,t)*f(x)$$.
One can also show existence and uniqueness for the following problem. $$\begin{cases} \frac{\partial^2 u}{\partial t^2}-4\pi|\xi|^2 u=0\\ \lim_{t^2+|\xi|^2\to \infty} u=0\\ \lim_{t\to 0}||u(\cdot, t)-f(\cdot)||_{\infty}=0. \end{cases}$$ In this case, uniquness follows from the fact that the first two conditions imply for each fixed $$\xi_0$$, $$u(t,\xi_0)=Ce^{-2\pi t |\xi_0|}$$. The last condition then implies that these constants are determined by $$f$$, so we have $$u(t,\xi)=e^{-2\pi t |\xi|}\cdot f(\xi)$$, and this formula gives existence.
Finally, if $$u$$ solves the first system with initial data $$f$$, then $$\hat{u}$$ solves the second system with initial data $$\hat{f}$$, so we have that by uniqueness of solutions, $$\hat{f}\cdot e^{-2\pi t |\xi|}=(f*P(x,t))^\wedge=\hat{f}\cdot \hat{P}(x,t).$$ As this holds for all $$f$$ Schwartz functions, $$\hat{P}(x,t)=e^{-2\pi t |\xi|}$$.