Solution
= Solution
Use the <Fourier transform> convention $\widetilde h(k)=\int_{\mathbb R}e^{-ikx}h(x)\,dx$, with inverse $(2\pi)^{-1}\int_{\mathbb R}e^{ikx}\widetilde h(k)\,dk$. Splitting at the origin, with $a>0$ to ensure convergence, gives
$$
\widetilde h(k)=\int_{-\infty}^0e^{(a-ik)x}\,dx+\int_0^\infty e^{-(a+ik)x}\,dx=\frac1{a-ik}+\frac1{a+ik}.
$$
Thus
$$
\boxed{\widetilde h(k)=\frac{2a}{a^2+k^2}}.
$$