GATE 2023 EC Q.28: solution and answer

Official answer: CSource: Official answer key, GATE 2023 EC (IIT Kanpur)

1. Start from the definition of the Fourier transform: $$X(\omega )={\int }_{-\infty }^{\infty }{e}^{-t^2}{e}^{-j\omega t}dt={\int }_{-\infty }^{\infty }{e}^{-(t^2+j\omega t)}dt$$

2. Complete the square in the exponent: $$t^2+j\omega t={\left(t+\frac{j\omega }{2}\right)}^2+\frac{{\omega }^2}{4}$$ because ${\left(\frac{j\omega }{2}\right)}^2=-\frac{{\omega }^2}{4}$, so the square is corrected by adding back $\frac{{\omega }^2}{4}$.

3. Take the constant factor out of the integral: $$X(\omega )=e^{-\frac{{\omega }^2}{4}}{\int }_{-\infty }^{\infty }{e}^{-{\left(t+\frac{j\omega }{2}\right)}^2}dt$$

4. Substitute $y=t+\frac{j\omega }{2}$, so $dy=dt$. This moves the path of integration to a line parallel to the real axis. Because $e^{-z^2}$ is analytic everywhere and $\left|e^{-(\pm R+jv{)}^2}\right|=e^{-R^2+v^2}\to 0$ as $R\to \infty$ for any fixed $v$, the short vertical sides of the shifted contour contribute nothing, and the integral equals the given Gaussian integral: $${\int }_{-\infty }^{\infty }{e}^{-{\left(t+\frac{j\omega }{2}\right)}^2}dt={\int }_{-\infty }^{\infty }{e}^{-y^2}dy=\sqrt{\pi }$$

5. Therefore $$X(\omega )=\sqrt{\pi }{e}^{-\frac{{\omega }^2}{4}}$$

6. Quick checks on the options. First, $X(0)={\int }_{-\infty }^{\infty }x(t)dt=\sqrt{\pi }$. Option (B) gives $X(0)=\frac{1}{2\sqrt{\pi }}$, so it is out. Second, $x(t)$ is absolutely integrable, so $X(\omega )\to 0$ as $|\omega |\to \infty$. Option (A) grows without bound, so it is out. Options (C) and (D) both pass these checks, and the derivation settles it: the exponent is $-\frac{{\omega }^2}{4}$, not $-\frac{{\omega }^2}{2}$. Answer: (C).