Time Shift Property of The Laplace Transform

$$\bbox[10px,border:1px solid grey]{\mathcal{L}[\cos w t]=\frac{1}{2}\left(\frac{1}{s-j \omega}+\frac{1}{s+j \omega}\right)=\frac{s}{s^{2}+\omega^{2}} }\tag{A}$$
If $ F(s) $ is the Laplace transform of $ f(t) $, then
$$\begin{array}{r}\mathcal{L}[f(t-a) u(t-a)]=\int_{0}^{\infty} f(t-a) u(t-a) e^{-s t} d t \\a \geq 0\end{array}$$
But $ u(t-a)=0 $ for $ t < a $ and $ u(t-a)=1 $ for $ t > a $. Hence, $$\mathcal{L}[f(t-a) u(t-a)]=\int_{a}^{\infty} f(t-a) e^{-s t} d t$$ If we let $ x=t-a $, then $ d x=d t $ and $ t=x+a $. As $ t \rightarrow a, x \rightarrow 0 $ and as $ t \rightarrow \infty, x \rightarrow \infty $. Thus,
$$\begin{aligned}\mathcal{L}[f(t-a) u(t-a)] &=\int_{0}^{\infty} f(x) e^{-s(x+a)} d x \\&=e^{-a s} \int_{0}^{\infty} f(x) e^{-s x} d x=e^{-a s} F(s)\end{aligned}$$
$$\bbox[10px,border:1px solid grey]{\mathcal{L}[f(t-a) u(t-a)]=e^{-a s} F(s)} \tag{1}$$
In other words, if a function is delayed in time by $ a $, the result in the $ s $ domain is multiplying the Laplace transform of the function (without the delay) by $ e^{-a s} $. This is called the time-delay or time-shift property of the Laplace transform. As an example, we know from Eq. (A) that
$$\mathcal{L}[\cos \omega t]=\frac{s}{s^{2}+\omega^{2}}$$
Using the time-shift property in Eq. (1),
$$\mathcal{L}[\cos \omega(t-a) u(t-a)]=e^{-a s} \frac{s}{s^{2}+\omega^{2}}$$