How to abstractly prove a Laplace transform identity?

[Eclair_de_XII]

Homework Statement

"Suppose that $F(s) = L[f(t)]$ exists for $s > a ≥ 0$.
(a) Show that if c is a positive constant, then
$L[f(ct)]=\frac{1}{c}F(\frac{s}{c})$

Homework Equations

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

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?

 

[FactChecker]

Try ∫₀^∞f(ct)e^-stdt = ∫₀^∞f(ct)e^-s/c⋅ctdt = 1/c∫₀^∞f(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.

 

[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 <br /> F(p) = \int_0^\infty f(u)e^{-pu}\,du.

 

[Eclair_de_XII]

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