= Solution
The inverse <Fourier transform> is
$$
f(t)=\frac1{2\pi}\int_{\mathbb R}\frac{-i\,e^{i\omega t}}{\omega-i}\,d\omega.
$$
For $t>0$, close in the upper half-plane, where $e^{i\omega t}$ decays. The single enclosed <pole> has <residue> $-ie^{-t}$, so counterclockwise integration gives $f=e^{-t}$. For $t<0$, close clockwise in the lower half-plane; there is no enclosed pole and the result is zero. Thus
$$
\boxed{f(t)=e^{-t}H(t),\qquad f(t)=0\ (t<0),\quad f(t)=e^{-t}\ (t>0).}
$$
Here $H$ is the <Heaviside step function>. At the jump, symmetric <Fourier inversion> gives $f(0)=1/2$, the average of the one-sided limits. The transform itself cannot fix a function’s value at a single point, so this is the conventional representative rather than extra frequency information.
A direct check is $\int_0^\infty e^{-(1+i\omega)t}dt=(1+i\omega)^{-1}=-i/(\omega-i)$. Distributionally, $f'=\delta(t)-e^{-t}H(t)$.
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