Question

The Fourier transform \(X(\omega)\) of \(x(t)=e^{-t^{2}}\) is \[ X(\omega)=\sqrt{\pi}\,e^{-\tfrac{\omega^{2}}{4}}. \]

• A. $\sqrt{\pi}\,e^{\frac{\omega^{2}}{2}}$
• B. $\dfrac{e^{-\frac{\omega^{2}}{4}}}{2\sqrt{\pi}}$
• C. $\sqrt{\pi}\,e^{-\frac{\omega^{2}}{4}}$
• D. $\sqrt{\pi}\,e^{\frac{\omega^{2}}{2}}$

Hint:

For $X(\omega)=\int x(t)e^{-j\omega t}dt$, the transform of $e^{-at^{2}}$ is $\sqrt{\frac{\pi}{a}}\,e^{-\frac{\omega^{2}}{4a}}$ for $\Re\{a\}>0$. Set $a=1$ to get $\sqrt{\pi}\,e^{-\omega^{2}/4}$.

Answer 1

The Correct Option is C

Assume the Fourier transform \[ X(\omega)=\int_{-\infty}^{\infty} x(t)\,e^{-j\omega t}\,dt. \] Given \(x(t)=e^{-t^{2}}\), we have \[ X(\omega)=\int_{-\infty}^{\infty} e^{-t^{2}}\,e^{-j\omega t}\,dt =\int_{-\infty}^{\infty} \exp\!\big(-t^{2}-j\omega t\big)\,dt. \]

Complete the square:
\[ -t^{2}-j\omega t = -\Big(t+\frac{j\omega}{2}\Big)^{2}-\frac{\omega^{2}}{4}. \]

Hence,
\[ X(\omega)=e^{-\frac{\omega^{2}}{4}} \int_{-\infty}^{\infty}\exp\!\Big[-\Big(t+\frac{j\omega}{2}\Big)^{2}\Big]\,dt. \]

The shift by a constant in the complex plane does not change the Gaussian integral value. Using \[ \int_{-\infty}^{\infty} e^{-y^{2}}\,dy=\sqrt{\pi}, \] we obtain
\[ X(\omega)=e^{-\frac{\omega^{2}}{4}}\;\sqrt{\pi}. \]

Final Answer:
\[ \boxed{X(\omega)=\sqrt{\pi}\,e^{-\frac{\omega^{2}}{4}}} \] which corresponds to option (C).