Frequency Shift Property of The Laplace Transform

If $ F(s) $ is the Laplace transform of $ f(t) $, then
$$\begin{aligned}\mathcal{L}\left[e^{-a t} f(t)\right] &=\int_{0}^{\infty} e^{-a t} f(t) e^{-s t} d t \\&=\int_{0}^{\infty} f(t) e^{-(s+a) t} d t=F(s+a)\end{aligned}$$
$$\bbox[10px,border:1px solid grey]{\mathcal{L}\left[e^{-a t} f(t)\right]=F(s+a) }\tag{1}$$
That is, the Laplace transform of $ e^{-a t} f(t) $ can be obtained from the Laplace transform of $ f(t) $ by replacing every $ s $ with $ s+a $. This is known as frequency shift or frequency translation. As an example, we know that and
$$\cos \omega t \quad \Longleftrightarrow \quad \frac{s}{s^{2}+\omega^{2}}$$
$$\sin \omega t \quad \Longleftrightarrow \quad \frac{\omega}{s^{2}+\omega^{2}}$$
Using the shift property in Eq. (1), we obtain the Laplace transform of the damped sine and damped cosine functions as
$$\begin{aligned}\mathcal{L}\left[e^{-a t} \cos \omega t\right] &=\frac{s+a}{(s+a)^{2}+\omega^{2}} \\\mathcal{L}\left[e^{-a t} \sin \omega t\right] &=\frac{\omega}{(s+a)^{2}+\omega^{2}}\end{aligned}$$

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