Scaling property of Laplace transform

Question

I am not sure how to do the following problem:

Let $$\hat{F}(s)=\mathfrak{L}(f(t))$$ be the Laplace transform of $f(t)$. Show that:

$$\mathfrak{L}(f(at))=\frac{1}{a}\hat{F}\left(\frac{s}{a}\right) $$

Answer 1

Make a change of variables in the integral. Scale with $a$: $$ \mathcal L (f(at))(s)=\int_0^{+\infty}f(at)e^{-st}\,dt=\{u=at\}=\frac{1}{a}\int_0^{+\infty}f(u)e^{-(s/a)u}\,du=\frac{1}{a}\mathcal L(f(t))(s/a). $$

Answer 2

Hint:

From the definition, $\mathcal{L}\{f(at)\}$ is an improper integral, try this with the sustitution $r=at$, then $$\int_0^{\infty}e^{-st}f(at)\,dt=\int_0^{\infty}e^{-\frac{s}{a}r}f(r)\frac{1}{a}dr$$

Answer 3

$$\mathfrak{L}(f(at))=\int_0^\infty f(at)e^{-st}dt$$

Let $u=at$ then $dt=du/a$, so: $$\mathfrak{L}(f(at))=\int_0^\infty f(u)e^{-(s/a)u}\frac {du} a=\frac1 a \hat{F}(\frac s a)$$