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!
 A: Here is a proof that uses PDE theory to bypass the extensive computation.
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, $\hat{P}(x,t)=e^{-2\pi t |\xi|}$.
A: In Grafakos's book 'Classical Fourier Analysis', he wrote a way of prove it Exercise 2.2.10 to Exercise 2.2.11. It is a relatively easy way. Hope things will help you.
