Right answer is (b) \(\frac{π}{a} e^{-a|ω|}\)
The best explanation: Fourier transform of e^-a|t| = \(\frac{2a}{a^2+ω^2}\)
Now, e^-a|t| = e^-at, t>0 and eat, t<0
Using Duality property, \(\frac{2a}{a^2+w^2}\) ↔ 2πe^-a|ω|
Or, \(\frac{1}{a^2+w^2} \leftrightarrow \frac{π}{a} e^{-a|ω|}\).