Change of Scale Property | Laplace Transform

Change of Scale Property
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then,
 

$\mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a} F \left( \dfrac{s}{a} \right)$

 

Proof of Change of Scale Property
$\displaystyle \mathcal{L} \left\{ f(at) \right\} = \int_0^\infty e^{-st} f(at) \, dt$
 

Let
$z = at$

$t = z/a$

$dt = \dfrac{dz}{a}$
 

when   $t = 0, \, z = 0$

when   $t = \infty, \, z = \infty$
 

$\displaystyle \mathcal{L} \left\{ f(at) \right\} = \int_0^\infty e^{-s(z/a)} f(z) \, \dfrac{dz}{a}$

$\displaystyle \mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a}\int_0^\infty e^{-(s/a)z} f(z) \, dz$
 

Hence,
$\mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a} F\left( \dfrac{s}{a} \right)$       okay
 

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