Correct Answer - Option 2 :
\({e^{ - \frac{{{\omega ^2}}}{2}}}\)
Concept:
\(\frac{{d\left[ {f\left( x \right)} \right]}}{{dx}}\;\mathop \longleftrightarrow \limits^{Fourier\;transform} \;j\omega \;F\left( \omega \right)\)
\(xf\left( x \right)\mathop \longleftrightarrow \limits^{Fourier\;transform} \frac{{jd\left[ {F\left( \omega \right)} \right]}}{{d\omega }}\)
Where F(ω) is Fourier transform of f(x)
Calculation:
We are given \(f\left( x \right) = {e^{ - \frac{{{x^2}}}{2}}}\)
\(\frac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = - x \cdot {e^{ - \frac{{{x^2}}}{2}}}\) ---(1)
Take Fourier transform of above equation (1)
\(j\omega \;F\left( \omega \right) = \frac{{ - jd\left[ {F\left( \omega \right)} \right]}}{{d\omega }}\)
\(\Rightarrow \frac{1}{{F\left( \omega \right)}} \cdot d\left[ {F\left( \omega \right)} \right] = \; - \omega \;d\omega\) ---(2)
By integrating the above equation, we get
\(ln\left[ {F\left( \omega \right)} \right] = - \frac{{{\omega ^2}}}{2}\)
\(\Rightarrow F\left( \omega \right) = {e^{ - \frac{{{\omega ^2}}}{2}}}\)
