= Solution
With the stated <Fourier transform> convention, <Fourier inversion theorem>[Fourier inversion] gives
$$
f(x)=\frac1{2\pi}\int_{-\infty}^{\infty}
\frac{-2ik}{p^2+k^2}e^{ikx},dk.
$$
For $x>0$, close the contour in the upper half-plane. <Jordan lemma> removes the semicircle contribution, and the only enclosed pole is $k=ip$. Its residue is
$$
\operatorname{Res}_{k=ip}
\frac{-2ik e^{ikx}}{(k-ip)(k+ip)}=-i e^{-px}.
$$
The <residue theorem> therefore yields
$$
\boxed{f(x)=e^{-px},\qquad x>0}.
$$
Back to article page
