= Solution
{c}
Use first-order <time-dependent perturbation theory>, setting $c_1=1$ in the equation for $c_2$ and retaining the generally complex <matrix element> $d=\langle\psi_2|\widehat D_0|\psi_1\rangle$. With $\delta=\omega_0-\omega$ and $\Sigma=\omega_0+\omega$, the <probability amplitude> is
$$
c_2(t)=-\frac{id}{2\hbar}\int_{-\infty}^t e^{-b|s|}\{e^{i\delta s}+e^{i\Sigma s}\}\,ds.
$$
For $t>0$, split the <integral> at the pulse maximum. For either $x=\delta$ or $x=\Sigma$,
$$
\begin{aligned}
F(x,t)&=\int_{-\infty}^0e^{(b+ix)s}ds+\int_0^te^{(-b+ix)s}ds\\
&=\frac1{b+ix}+\frac{1-e^{(-b+ix)t}}{b-ix}
=\frac{2b}{b^2+x^2}-\frac{e^{(-b+ix)t}}{b-ix}.
\end{aligned}
$$
Thus the full first-order answer, without discarding the counter-rotating contribution, is
$$
\boxed{c_2(t)=-\frac{id}{2\hbar}\{F(\delta,t)+F(\Sigma,t)\},\qquad t>0.}
$$
At long times it gives
$$
\boxed{P_2(\infty)=\frac{|d|^2b^2}{\hbar^2}
\left(\frac1{b^2+\delta^2}+\frac1{b^2+\Sigma^2}\right)^2.}
$$
This full expression is not exactly one squared <Lorentzian line shape>. To obtain the intended near-<resonance> <spectral line> profile, additionally assume $b,|\delta|\ll\omega_0$, so the pulse has a narrow positive-<frequency> lobe and $F(\Sigma,t)$ is negligible. Then
$$
c_2(t)\simeq-\frac{id}{2\hbar}\left[\frac{2b}{b^2+\delta^2}-\frac{e^{(-b+i\delta)t}}{b-i\delta}\right],
\qquad
\boxed{P_2(\infty)\simeq\frac{|d|^2}{\hbar^2b^2}\frac1{[1+(\delta/b)^2]^2}.}
$$
The weak-excitation condition near <resonance> is $|d|/(\hbar b)\ll1$. The squared <Lorentzian line shape> occurs because the symmetric exponential pulse has a <Lorentzian line shape> \i[amplitude] <Fourier transform>, which is then squared to obtain <probability>.
At half the maximum, $[1+(\delta/b)^2]^2=2$, so
$$
\boxed{\Delta\omega_{\rm FWHM}=2b\sqrt{\sqrt2-1}\simeq1.2872b,
\qquad\Delta\nu_{\rm FWHM}=\frac b\pi\sqrt{\sqrt2-1}.}
$$
The <full width at half maximum> refers to varying the drive <frequency> through the positive-<frequency> <resonance>. For a concrete failure of the unqualified squared-<Lorentzian line shape> claim, set $b=\omega=\omega_0$ while keeping $|d|/(\hbar b)\ll1$. The full first-order resonant <probability> is $36/25$ times the one-lobe prediction. Weak radiation alone therefore does not justify dropping the second <Fourier transform> lobe.
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