= Solution
For $t>0$, close the real-axis <contour> in the upper half-plane: $|e^{izt}|=e^{-t\operatorname{Im}z}$ supplies decay. The only enclosed <simple pole> is $z=i$, with <residue> $e^{-t}/(2i)$. Thus $h(t)=2\pi i\,e^{-t}/(2i)=\pi e^{-t}$.
For $t<0$, close downwards. The clockwise orientation contributes $-2\pi i$ times the <residue> $e^t/(-2i)$ at $z=-i$, giving
$$
\boxed{h(t)=\pi e^{-|t|}.}
$$
At $t=0$, direct integration of $1/(1+p^2)$ gives $\pi$, consistent with the continuous extension. The large arc tends to zero by <Jordan's lemma> for nonzero $t$, or the denominator estimate when $t=0$.
Back to article page
