Right choice is (a) \(\frac{1}{a+jω}\)
The explanation is: Given f(t)= e^-at u(t)
We know that \(
u(t)=\begin{cases}
0 &\text{\(t<0\)} \\
1 &\text{\(t>0\)} \\
\end{cases}\)
Fourier transform,
\(F(ω) = \int_{-∞}^∞ f(t)e^{-jωt} \,dt = \int_{-∞}^∞ e^{-at} u(t)e^{-jωt} \,dt = \int_0^∞ e^{-(a+jω)t} \,dt\)
F(ω) = \(\frac{1}{a+jω}\), a>0.