$$ f (t) = \begin{cases} 0 & \quad \text{for } t < 0 \\ A_1 \exp( i \omega_1 t - \Gamma_1 t) & \quad \text{for } t \geq 0 \end{cases} $$
I need to Fourier-transform this using the formula
$$ g(w) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(t) e^{-iwt} dt $$
Subbing in the values and functions:
$$g(w) = \frac{1}{\sqrt{2\pi}}\int_{0}^\infty A_1 e^{iw_1t - r_1t} e^{-iwt} dt $$
I combined the exponential terms:
$$ g(w) = \frac{1}{\sqrt{2\pi}}\int_{0}^\infty e^{(i(w_1 -w)t - r_1 t)} dt $$
Factorising $t$ in the exponential term:
$$ g(w) = \frac{1}{\sqrt{2\pi}}\int_{0}^\infty e^{t(i(w_1 - w) - r_1)} $$
How do I carry on from here?
$$g(\omega) = \frac{1}{\sqrt{2\pi}}\int_{0}^\infty e^{t(i(\omega_1 - \omega) -\Gamma_1 )}dt$$
$$g(\omega) = \frac{1}{\sqrt{2\pi}}\lim\limits_{L \to \infty}\int_{0}^{L} e^{t(i(\omega_1 - \omega) - \Gamma_1 )}dt$$ $$= \frac{1}{\sqrt{2\pi}}\lim\limits_{L \to \infty}\int_{0}^{L} e^{t(i(\omega_1 - \omega) - \Gamma_1 )}dt$$ $$=\frac{1}{\sqrt{2\pi}}\lim\limits_{L \to \infty}\frac{1}{(i(\omega_1 - \omega) - \Gamma_1 )}e^{t(i(\omega_1 - \omega) - \Gamma_1 )}\bigg|_{0}^{L}$$ $$=\frac{1}{\sqrt{2\pi}(i(\omega_1 - \omega) - \Gamma_1 )}\lim\limits_{L \to \infty}[e^{L(i(\omega_1 - \omega) - \Gamma_1 )}-1]$$
$$e^{L(i(\omega_1 - \omega) - \Gamma_1 )}=e^{-L\Gamma_1 }e^{iL(\omega_1 - \omega)}$$
We can notice that: $ 0\le |e^{iL(\omega_1 - \omega)} |\le1 $ then as $L\rightarrow \infty$ we have that $e^{-L\Gamma_1 }e^{iL(\omega_1 - \omega)}\rightarrow 0$
Finally:
$$g(\omega)=\frac{1}{\sqrt{2\pi}}\frac{1}{(i(\omega_1 - \omega) - \Gamma_1 )}(0-1)=\frac{1}{\sqrt{2\pi}}\frac{1}{(i(\omega-\omega_1 ) + \Gamma_1 )}$$
If we multiply by the complex conjugate from the denominator:
$$g(\omega)=\frac{1}{\sqrt{2\pi}}\frac{\Gamma_1+i(\omega_1-\omega ) }{\omega ^2+\Gamma_1^2-\omega_1^{2}}$$