Time Shifting property of the Fourier Transform

If $ F(\omega)=\mathcal{F}[f(t)] $, then
$$\mathcal{F}\left[f\left(t-t_{0}\right)\right]=e^{-j \omega t_{0}} F(\omega)$$
that is, a delay in the time domain corresponds to a phase shift in the frequency domain. To derive the time shifting property, we note that
$$\mathcal{F}\left[f\left(t-t_{0}\right)\right]=\int_{-\infty}^{\infty} f\left(t-t_{0}\right) e^{-j \omega t} d t$$
If we let $ x=t-t_{0} $ so that $ d x=d t $ and $ t=x+t_{0} $, then
$$\begin{aligned}\mathcal{F}\left[f\left(t-t_{0}\right)\right] &=\int_{-\infty}^{\infty} f(x) e^{-j \omega\left(x+t_{0}\right)} d x \\&=e^{-j \omega t_{0}} \int_{-\infty}^{\infty} f(x) e^{-j \omega x} d x=e^{-j \omega t_{0}} F(\omega)\end{aligned}$$
Similarly, $ \mathcal{F}\left[f\left(t+t_{0}\right)\right]=e^{j \omega t_{0}} F(\omega) $. For example, $$\mathcal{F}\left[e^{-a t} u(t)\right]=\frac{1}{a+j \omega}$$ The transform of $ f(t)=e^{-(t-2)} u(t-2) $ is
$$F(\omega)=\mathcal{F}\left[e^{-(t-2)} u(t-2)\right]=\frac{e^{-j 2 \omega}}{1+j \omega}$$

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