Laplace transform of a periodic function 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
 A: I assume you want to prove that for a periodic function,
$$L[f(t)]=\frac1{1-e^{-sp}}\int_0^{p} e^{-st}f(t)dt$$
Please let me know if I correctly understand the question. As written, it's not what you are asking for, but it would make no sense to prove $L[f(ct)]=(\frac{1}{c})F(\frac{s}{c})$ for a periodic function, since it has nothing to do with periodic functions as Harald remarked.

So, you have
$$F(s)=\int_0^\infty \exp(-st) \,f(t) \,\mathrm{d}t=\sum_{k=0}^{\infty}\int_{kp}^{(k+1)p} \exp(-st) \,f(t) \,\mathrm{d}t$$
With change of variable $t=kp+u$, and since $f$ is $p$-periodic,
$$F(s)=\sum_{k=0}^{\infty}\int_{0}^{p} \exp \left[-s (kp+u)\right] \,f(kp+u) \,\mathrm{d}u=\sum_{k=0}^{\infty} \exp(-kps) \int_{0}^{p} \exp (-su) \,f(u) \,\mathrm{d}u\\=\frac{1}{1-\exp(-sp)}\int_{0}^{p} \exp (-su) \,f(u) \,\mathrm{d}u$$
A: $$
\mathcal{F}(s) = \mathcal{L}[f(t)] = \int\limits_{0}^\infty f(t) e^{-st} \mathrm{d}t\\
\mathcal{L}[f(ct)] = \int\limits_{0}^\infty f(ct) e^{-st} \mathrm{d}t\\
\text{Make the substitution $ct = u$} \\
\mathcal{L}[f(ct)] = \frac{1}{c}\int\limits_{0}^\infty f(u) e^{-\frac{su}{c}} \mathrm{d}u\\
\mathcal{L}[f(ct)] = \frac{1}{c}\int\limits_{0}^\infty f(u) e^{\left(-\frac{s}{c}\right)u} \mathrm{d}u\\
\mathcal{L}[f(ct)] = \frac{1}{c}\mathcal{F}(\frac{s}{c})\\
$$
A: To make the connection between periodic functions and the desired theorem, we can do a proof similar to Priyatham's but using the formula for the Laplace transform of a periodic function rather than the general formula for the Laplace transform of anything.
So first, what is the period of $f(ct)$?  It is $p/c$ (or rather $p/|c|$, but since you later state $s > ca$, I'll take it as understood that $c$ is positive; negative $c$ works similarly).  This is because $f(c(t+p/c)) = f(ct+p) = f(ct)$.
So $L[f(ct)] = \frac{1} {1 - e^{-s(p/c)}} \int_0^{p/c} e^{-st} f(ct) \,dt$.  As in Priyatham's answer, let $u = ct$, so that $t = u/c$ and $dt = 1/c \,du$.  While we're at it, let $r = s/c$, so that $s = rc$.  Then
$$\eqalign{ L[f(ct)] & = \frac1 {1 - e^{-sp/c}} \int_{t=0}^{p/c} e^{-st} f(ct) \,dt \\
& = \frac1 {1 - e^{-(rc)p/c}} \int_{u=c(0)}^{c(p/c)} e^{-(rc)(u/c)} f\bigl(c(u/c)\bigr) \,\Bigl(\frac1 c \,du\Bigr) \\
& = \frac1 c \,\frac1 {1 - e^{-pr}} \int_0^p e^{-ru} f(u) \,du \\
& = \frac1 c F(r) = \frac1 c F\Bigl(\frac{s} c\Bigr) \text.}$$
Also note that if the original Laplace transform is restricted to $s > a$, then the one here is restricted to $r > a$, so $s/c > a$, so $s > ca$.  However, this is not really necessary, since the Laplace transform of a periodic function (at least if it's piecewise-continuous, which I assume is what you mean by ‘a continuous function by segments’) is defined everywhere (as can be seen from the formula, because the integral is proper).
