Function modulation and dilatation in a Laplace transform My question is about something incomplete in my lesson. I learnt about the modulation and the dilatation of a function in the Laplace transform but what if I have both in the same time? For the modulation, I have the following relation: $$ \mathcal{L}[e^{-\gamma t} f(t)] = \mathcal{L}f $$ For the dilatation, I have the following relation: $$ \mathcal{L}[f(\lambda t)] = \frac{1}{\lambda} \mathcal{L}f $$ What if I have the following Laplace transform : $\mathcal{L}[e^{\gamma t} f(\frac{t}{\lambda})]$ Is the answer $\lambda \mathcal{L}f$ or $\lambda \mathcal{L}f)$?

Define modulation and dilation operators: $M_{\gamma}f(t) = e^{-\gamma t} f(t)$ and $D_{\lambda}f(t) = f(\lambda t)$. Then $t \mapsto e^{\gamma t} f(t/\lambda)$ can be expressed as either $M_{-\gamma} (D_{1/\lambda} f)$ (dilation then modulation) or $D_{1/\lambda}(M_{-\gamma \lambda} f)$ (modulation then dilation). Using either form, we obtain the same result. $$\mathcal LM_{-\gamma} (D_{1/\lambda} f) = \mathcal LD_{1/\lambda} f = \lambda \mathcal L[f] (\lambda(z - \gamma))$$ $$\mathcal LD_{1/\lambda}(M_{-\gamma \lambda} f) = \lambda \mathcal LM_{-\gamma \lambda} f = \lambda \mathcal L[f] (\lambda (z - \gamma))$$