Understanding the Concept:
A unique property of the Gaussian function is that it acts as an eigenfunction of the continuous Fourier transform operation. Transforming a Gaussian function in the time domain yields another Gaussian function in the frequency domain.
Step 1: Setting up the standard Fourier Integral.
Let the given time-domain signal be $x(t) = e^{-t^2}$. Its continuous-time Fourier transform is defined as:
\[
X(\omega) = \int_{-\infty}^{\infty} e^{-t^2} e^{-j\omega t} \, dt = \int_{-\infty}^{\infty} e^{-(t^2 + j\omega t)} \, dt
\]
Step 2: Completing the square within the exponential power.
Let us algebraicly alter the quadratic term in the exponent to create a perfect square:
\[
t^2 + j\omega t = \left(t + \frac{j\omega}{2}\right)^2 - \left(\frac{j\omega}{2}\right)^2 = \left(t + \frac{j\omega}{2}\right)^2 - \left(-\frac{\omega^2}{4}\right) = \left(t + \frac{j\omega}{2}\right)^2 + \frac{\omega^2}{4}
\]
Substituting this expression back into our integral gives:
\[
X(\omega) = \int_{-\infty}^{\infty} e^{-\left[ \left(t + \frac{j\omega}{2}\right)^2 + \frac{\omega^2}{4} \right]} \, dt
\]
Using the laws of exponents to separate the factors:
\[
X(\omega) = e^{-\frac{\omega^2}{4}} \int_{-\infty}^{\infty} e^{-\left(t + \frac{j\omega}{2}\right)^2} \, dt
\]
Step 3: Evaluation via the standard Gaussian Integral.
Perform a linear variable substitution: let $u = t + \frac{j\omega}{2}$, which means $du = dt$. The integration limits remain unchanged from $-\infty$ to $\infty$:
\[
X(\omega) = e^{-\frac{\omega^2}{4}} \int_{-\infty}^{\infty} e^{-u^2} \, du
\]
Using the standard continuous total area Gaussian integral value, $\int_{-\infty}^{\infty} e^{-u^2} \, du = \sqrt{\pi}$:
\[
X(\omega) = \sqrt{\pi} \cdot e^{-\frac{\omega^2}{4}}
\]
The resulting function, $e^{-\frac{\omega^2}{4}}$, retains the exact standard mathematical definition of a Gaussian curve along the independent frequency coordinate $\omega$.