Fourier transform of the three-dimensional plane Poisson kernel

In polar coordinates, ${\int }_{{\mathbb{R}}^2}{P}_t(z)dz=t{\int }_0^{\mathrm{\infty }}r(r^2+t^2{)}^{-3\mathrm{/}2}dr=1$; the same estimate shows $P_t\in L^1$. Direct differentiation of $t(\mathrm{\mid }z{\mathrm{\mid }}^2+t^2{)}^{-3\mathrm{/}2}$ gives $({\mathrm{\Delta }}_z+{\mathrm{\partial }}_t^2)P_t=0$ for $t>0$.

Put $F(t,\xi )=\widehat{P_t}(\xi )$. For $t$ in any compact subinterval of $(0,\mathrm{\infty })$, $P_t$ and its first two $t$-derivatives are dominated by integrable radial functions, so differentiation under the Fourier integral is valid. Fourier transforming the harmonic identity in step 1.1 and using [F1] gives ${\mathrm{\partial }}_t^{2}F-(2\pi \mathrm{\mid }\xi \mathrm{\mid }{)}^{2}F=0$. Also $\mathrm{\mid }F(t,\xi )\mathrm{\mid }\le \mathrm{\parallel }{P}_t{\mathrm{\parallel }}_1=1$. For fixed $\xi$, the change $z=tw$ gives $F(t,\xi )={\int }_{{\mathbb{R}}^2}{P}_1(w)e^{-2\pi itw\cdot \xi }dw$; dominated convergence yields $F(t,\xi )\to 1$ as $t\downarrow 0$. The solutions of this constant-coefficient ODE are linear combinations of $e^{2\pi \mathrm{\mid }\xi \mathrm{\mid }t}$ and $e^{-2\pi \mathrm{\mid }\xi \mathrm{\mid }t}$ when $\xi \mathrm{\ne }0$; boundedness removes the growing term, and the boundary limit fixes the coefficient to one. When $\xi =0$, the bounded affine solutions are constant and the same limit gives one. Hence $F(t,\xi )=e^{-2\pi \mathrm{\mid }\xi \mathrm{\mid }t}$.