Laplace transform of scaling function

Question

I'm currently doing a class in differential equations and the first subject is laplace transforms - I'm struggling with this task however;

Let $f$ be a continuous function with scaling symmetry given by $f(mt)=m^kf(t)$ for any $m>0$ Show that $(\mathcal{Lf})(s) = m^{-k-1}\frac{s}{m}(\mathcal{Lf})$

I don't see a way to apply the laplace transforms I know already - is there a certain theorem that simplifies this a bit?

Answer 1

From the definition of the Laplace transform, $$ \mathcal L[f]\left(\frac{s}{m}\right)= \int_0^\infty f(t) e^{-st/m}dt = m\int_0^\infty f(m u) e^{-su}du = m^{k+1}\int_0^\infty f(u) e^{-su}du = m^{k+1}\mathcal{L}[f](s). $$ Dividing through by $m^{k+1}$ gives $$ \mathcal{L}[f](s) = m^{-k-1}\mathcal{L}[f]\left(\frac{s}{m}\right). $$