# Laplace transform, proof that $L \{ \frac{1}{k}f(\frac{t}{k}) \}= F(ks)$

Let $L \{ f(t)\}=F(s)$, show that for all $k \in \mathbb{R}$, $k \neq 0$

$$L \{ \frac{1}{k}f(\frac{t}{k}) \}= F(ks)$$

if, $u=\frac{t}{k}$

$L \{ \frac{1}{k}f(\frac{t}{k}) \}= \int_0^{\infty} \exp^{-(ks)u}f(u)du \stackrel{?}{=} F(ks)$

Thanks!

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

$$L \{ \frac{1}{k}f(\frac{t}{k}) \}= \int_0^{\infty} e^{-st}\frac{1}{k}f(\frac{t}{k})dt$$

$x=\frac{t}{k}$

$dx=\frac{dt}{k}$

$dt=k dx$

$$L \{ \frac{1}{k}f(\frac{t}{k}) \}= \int_0^{\infty} e^{-skx}\frac{1}{k}f(x)k dx=\int_0^{\infty} e^{-skx}f(x) dx=F(ks)$$

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Thanks! and the another one? –  P. M. O. Oct 17 '12 at 19:32
@P.M.O. Others are very different stories. What is your aim in this question? I believe that your title proof was completed –  Mathlover Oct 17 '12 at 19:38