# How to abstractly prove a Laplace transform identity?

1. Apr 8, 2017

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"Suppose that $F(s) = L[f(t)]$ exists for $s > a ≥ 0$.
(a) Show that if c is a positive constant, then
$L[f(ct)]=\frac{1}{c}F(\frac{s}{c})$

2. Relevant equations
$L[f(t)]=\int_0^\infty f(t)e^{-st}dt$

3. The attempt at a solution
$L[f(ct)]=\int_0^\infty f(ct)e^{-st}dt$

D: $e^{-st}$ I: $f(ct)$
D: $-e^{-st}$ I: $\frac{1}{c}F(ct)$
I: $F(ct)$

$L[f(ct)]=\int_0^\infty f(ct)e^{-st}dt=(\frac{1}{c}e^{-st}F(ct))+\frac{1}{c}\int_0^\infty e^{-st}F(ct)dt$

D: $F(ct)$ I: $e^{-st}$
D: $cf(ct)$ I: $-\frac{1}{s}e^{-st}$

$L[f(ct)]=\int_0^\infty f(t)e^{-st}dt=(\frac{1}{c}e^{-st}F(t))+\frac{1}{c}(-\frac{1}{s}e^{-st}F(ct))+\frac{c}{s}\int_0^\infty e^{-st}f(ct)dt$
$\frac{s-c}{s}\int_0^\infty f(ct)e^{-st}dt=(\frac{1}{c}e^{-st}F(ct))+(-\frac{1}{cs}e^{-st}F(ct))$
$L[f(ct)]=\frac{s}{s-c}(\frac{1}{c}(e^{-st}F(ct)-\frac{1}{s}e^{-st}F(ct))=\frac{s}{s-c}(e^{-st}F(ct))(\frac{1}{c}-\frac{1}{s})$
$L[f(ct)]=\frac{s}{s-c}(e^{-st}F(ct))(\frac{s}{cs}-\frac{c}{cs})=\frac{s}{c}e^{-st}F(ct)$

Can anyone tell me what it is I'm doing wrong?

2. Apr 8, 2017

### FactChecker

Try ∫0f(ct)e-stdt = ∫0f(ct)e-s/c⋅ctdt = 1/c∫0f(h)e-s/c⋅hdh, where h=ct.

Note: If c is negative, you have to take into account the change of limits of the integral. Since it was stated that c>0, that is not a concern here.

Last edited: Apr 14, 2017
3. Apr 8, 2017

### pasmith

This is a straightforward consequence of integration by substitution. First, to avoid confusion, define $g(t) = f(ct)$. Then $$L[g(t)] = G(s) = \int_0^\infty g(t)e^{-st}\,dt = \int_0^\infty f(ct)e^{-st}\,dt.$$ Now set $ct = u$ and compare to $$F(p) = \int_0^\infty f(u)e^{-pu}\,du.$$

4. Apr 11, 2017

### Eclair_de_XII

Thanks for the help, everyone. I couldn't have finished the homework without it.