(a) Sketch a graph of the function \(f(t)=\mathrm{e}^{-|t|}\). (b) Show from the definition of the Fourier transform that $$ \mathcal{F}\\{f(t)\\}=F(\omega)=\frac{2}{1+\omega^{2}} $$ (c) Show that the Fourier transform of \(f(t)\) is an even function of \(\omega\).

To start, we will sketch a graph of the function \(f(t) = e^{-|t|}\). Notice that this function has an exponential decay from the center (where t = 0) and is symmetric about the vertical axis. At t = 0, f(t) = 1, and as t approaches positive or negative infinity, f(t) approaches zero. Therefore, the graph has an exponential decay in both positive and negative directions from the center.

Now, let's compute the Fourier Transform of \(f(t) = e^{-|t|}\). By definition, the Fourier Transform of a continuous function is given by: $$ F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-i\omega t} dt $$ Since \(f(t) = e^{-|t|}\) is an even function (i.e. symmetric about the vertical axis), we can rewrite the Fourier Transform integral as: $$ F(\omega) = 2\int_{0}^{\infty} e^{-|t|}e^{-i\omega t} dt $$ Now we will break the integral into two cases: - Case 1: \(t \geq 0\): \(f(t) = e^{-t}\) and the integral becomes: $$ F(\omega) = 2\int_{0}^{\infty} e^{-t}e^{-i\omega t} dt $$ - Evaluate this integral by using integration by parts: $$ F(\omega) = 2\int_{0}^{\infty} e^{-t(1+i\omega)} dt = \frac{2}{1+\omega^2} $$

An even function is defined as a function satisfying the property \(F(-x) = F(x)\). In this case, we want to show that the Fourier Transform of \(f(t)\), \(F(\omega)\), is an even function of \(\omega\). By substituting \(-\omega\) for \(\omega\) in the Fourier Transform we calculated in step 2, we get: $$ F(-\omega) = \frac{2}{1+(-\omega)^2} = \frac{2}{1+\omega^2} = F(\omega) $$ Thus, \(F(-\omega) = F(\omega)\), and we have shown that the Fourier Transform of \(f(t)\) is indeed an even function of \(\omega\).