(cod: P-76-83-8) The Laplace transform of a function \(f: [0,+\infty) \to \mathbb{R}\) is the function given by \[\mathcal{L}\{f(t)\}(s) = \int_0^{+\infty} e^{-st}f(t) \, dt,\] which is defined only for the values of \(s\) where the improper integral converges.

For example, consider the constant function \(f(t) = 1, t \geq 0\). In this case, for \(s>0\), we have: \begin{eqnarray*}\mathcal{L}\{f(t)\} &=& \int_0^{+ \infty} e^{-st} \, dt \\ &=& \lim\limits_{A \to + \infty} \int_0^A e^{-st} \, dt \\ &=& \lim\limits_{A \to + \infty} \left(\frac{1}{s} - \frac{e^{-sA}}{s}\right) \\ &=& \frac{1}{s}. \end{eqnarray*} Consider the following questions:

(1) Determine \(\mathcal{L}\{\sin(at)\}\), where \(a\) is a non-zero constant.

(2) Let \(f\) be a differentiable function with a continuous derivative on \([0,+\infty)\). Suppose that the transform of \(f\) is defined for \(s>0\) and that \[\lim\limits_{t \to + \infty} f(t)e^{-st} = 0,\] for all \(s>0\). In this case, it can be verified that the Laplace transform of \(f'(t)\) is also defined for \(s>0\). Select the option that presents a correct relationship between the transforms of \(f\) and \(f'\).