Scaling Property of The Laplace Transform

If $ F(s) $ is the Laplace transform of $ f(t) $, then
$$\mathcal{L}[f(a t)]=\int_{0}^{\infty} f(a t) e^{-s t} d t$$
where $ a $ is a constant and $ a>0 $. If we let $ x=a t, d x=a d t $, then
$$\mathcal{L}[f(a t)]=\int_{0}^{\infty} f(x) e^{-x(s / a)} \frac{d x}{a}=\frac{1}{a} \int_{0}^{\infty} f(x) e^{-x(s / a)} d x$$
$$\mathcal{L}[f(t)]=F(s)=\int_{0-}^{\infty} f(t) e^{-s t} d t \tag{1}$$
Comparing this integral with the definition of the Laplace transform in Eq. (1) shows that $ s $ in Eq. (1) must be replaced by $ s / a $ while the dummy variable $ t $ is replaced by $ x $. Hence, we obtain the scaling property as
$$\mathcal{L}[f(a t)]=\frac{1}{a} F\left(\frac{s}{a}\right)$$
For example, we know from Example (2) that
$$\mathcal{L}[\sin \omega t]=\frac{\omega}{s^{2}+\omega^{2}} \tag{2}$$
Using the scaling property in Eq. (2),
$$\mathcal{L}[\sin 2 \omega t]=\frac{1}{2} \frac{\omega}{(s / 2)^{2}+\omega^{2}}=\frac{2 \omega}{s^{2}+4 \omega^{2}} \tag{3}$$
which may also be obtained from Eq. (3) by replacing $ \omega $ with $ 2 \omega $.