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

$L(f(ct))=\int_{0}^{\infty} e^{-s t} F(c t) d t$

$U=c t \quad \rightarrow d u=c d t$

$\int_{0}^{\infty} e^{-s \frac{u}{c}} F(u) \frac{d u}{c}=\frac{1}{c} \int_{0}^{\infty} e^{-\frac{s}{c} u} F(u) d u$

$=\frac{1}{c} \cdot F\left(\frac{s}{c}\right)$

Prove that if $L\{F(t)\}=f(s)$ then $L\left\{\mathrm{e}^{a t} F(b t)\right\}=\frac{1}{b} f\left(\frac{s-a}{b}\right), b>0$

$L\left\{e^{a t} F(b t)\right\}=\int_{0}^{\infty} e^{-s t} \cdot e^{at} F(b t) d t$

$=\int_{0}^{\infty} e^{-(s-a) t} F(b t) d t \neq f(s-a)$

put $u=b t \rightarrow d u=b d t$

$\int_{0}^{\infty} e^{-(s-a) \frac{u}{b}} F(u) \frac{1}{b} d u=\frac{1}{b} \int_{0}^{\infty} e^{-\left(\frac{s-a}{b}\right) u} F(u) d u$

$=\frac{1}{b} f\left(\frac{s-a}{b}\right), b>0$