Knowing that $$L[f(t)]=\frac1{1-e^{-sp}}\int_0^{p} e^{-st}f(t)dt$$ $p$ indicates the period of the function
If $f$ is a continuous function by segments in $[0,\infty)$ and $F(s)=L[f(t)]$ exists for $s>a$, then $L[f(ct)]=(\frac{1}{c})F(\frac{s}{c})$, $s>ca.$
Could you help me with the proof of this theorem please
$$ \mathcal{F}(s) = \mathcal{L}[f(t)] = \int\limits_{0}^\infty f(t) e^{-st} \mathrm{d}t\\ \mathcal{L}[f(ct)] = \int\limits_{0}^\infty f(ct) e^{-st} \mathrm{d}t\\ \text{Make the substitution $ct = u$} \\ \mathcal{L}[f(ct)] = \frac{1}{c}\int\limits_{0}^\infty f(u) e^{-\frac{su}{c}} \mathrm{d}u\\ \mathcal{L}[f(ct)] = \frac{1}{c}\int\limits_{0}^\infty f(u) e^{\left(-\frac{s}{c}\right)u} \mathrm{d}u\\ \mathcal{L}[f(ct)] = \frac{1}{c}\mathcal{F}(\frac{s}{c})\\ $$