To find the Fourier transform X(ω) of the given function \(\rm x(t)=e^{-t^2}\), we will use the standard definition of the continuous-time Fourier Transform (CTFT).

Fourier Transform Definition

The Fourier transform of a time-domain function x(t) is mathematically defined as:

\(\displaystyle \mathcal{F}\{x(t)\} = X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt\)

Applying Fourier Transform to Gaussian Function

Substitute the given function \(\rm x(t)=e^{-t^2}\) into the Fourier transform integral:

\(\displaystyle X(\omega) = \int_{-\infty}^{\infty} e^{-t^2} e^{-j\omega t} dt\)

We can combine the exponential terms by adding their exponents:

\(\displaystyle X(\omega) = \int_{-\infty}^{\infty} e^{-(t^2 + j\omega t)} dt\)

Completing the Square for Integration

To solve this integral, we need to transform the exponent \(-(t^2 + j\omega t)\) into a form that resembles the argument of a standard Gaussian integral. We do this by completing the square for the term inside the parenthesis: \(t^2 + j\omega t\).

Recall the algebraic identity for completing the square: \(a^2 + 2ab + b^2 = (a+b)^2\). Here, a = t, and we have jωt as the middle term. To match 2ab, we can set \(2b = j\omega\), which implies \(b = \frac{j\omega}{2}\).

So, we add and subtract \(\left(\frac{j\omega}{2}\right)^2\) to complete the square:

\(\displaystyle t^2 + j\omega t = t^2 + j\omega t + \left(\frac{j\omega}{2}\right)^2 - \left(\frac{j\omega}{2}\right)^2\)

\(\displaystyle t^2 + j\omega t = \left(t + \frac{j\omega}{2}\right)^2 - \frac{j^2\omega^2}{4}\)

Since \(j^2 = -1\) in complex numbers:

\(\displaystyle t^2 + j\omega t = \left(t + \frac{j\omega}{2}\right)^2 - \frac{(-1)\omega^2}{4}\)

\(\displaystyle t^2 + j\omega t = \left(t + \frac{j\omega}{2}\right)^2 + \frac{\omega^2}{4}\)

Now, substitute this completed square form back into the exponent of our integral:

\(\displaystyle -(t^2 + j\omega t) = -\left[ \left(t + \frac{j\omega}{2}\right)^2 + \frac{\omega^2}{4} \right]\)

\(\displaystyle -(t^2 + j\omega t) = -\left(t + \frac{j\omega}{2}\right)^2 - \frac{\omega^2}{4}\)

Substitute this back into the Fourier transform integral:

\(\displaystyle X(\omega) = \int_{-\infty}^{\infty} e^{-\left(t + \frac{j\omega}{2}\right)^2 - \frac{\omega^2}{4}} dt\)

Using the property \(e^{A+B} = e^A e^B\), we can separate the exponential terms:

\(\displaystyle X(\omega) = \int_{-\infty}^{\infty} e^{-\left(t + \frac{j\omega}{2}\right)^2} \cdot e^{-\frac{\omega^2}{4}} dt\)

Since \(e^{-\frac{\omega^2}{4}}\) does not depend on the integration variable t, it can be taken out of the integral:

\(\displaystyle X(\omega) = e^{-\frac{\omega^2}{4}} \int_{-\infty}^{\infty} e^{-\left(t + \frac{j\omega}{2}\right)^2} dt\)

Evaluating the Gaussian Integral Using the Hint

To evaluate the remaining integral, let's perform a substitution. Let \(y = t + \frac{j\omega}{2}\).

Then, the differential \(dy = dt\).

The limits of integration remain unchanged. As \(t \to -\infty\), \(y \to -\infty\). As \(t \to \infty\), \(y \to \infty\).

So the integral part becomes:

\(\displaystyle \int_{-\infty}^{\infty} e^{-y^2} dy\)

The question explicitly provides a hint for this standard Gaussian integral: \(\rm \int_{-\infty}^{\infty}e^{-y^2}dy=\sqrt{\pi}\).

Substitute this value back into our expression for X(ω):

\(\displaystyle X(\omega) = e^{-\frac{\omega^2}{4}} \cdot \sqrt{\pi}\)

Rearranging the terms, we get:

\(\displaystyle X(\omega) = \sqrt{\pi} e^{-\frac{\omega^2}{4}}\)

Comparing with Options

Let's compare our derived Fourier transform with the given options:

Our calculated result, \(\rm \sqrt{\pi}e^{-\frac{\omega^2}{4}}\), perfectly matches Option 3.