 Fourier transform of a two-sided exponential

For $a>0$, use the Fourier transform convention $\hat{f}(s)={\int }_{\mathbb{R}}f(x)e^{-isx}dx$. Splitting the integral at zero gives $1/(a+is)+1/(a-is)=2a/(a^2+s^2)$. By the Fourier inversion theorem, the dual identity isThis pair connects an exponential function with the kernel of a Cauchy distribution and produces the Poisson kernel for the upper half-plane by inversion of $e^{-y|s|}$.