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I'm writing a dissertation about the probabilistic interpretation of Laplace's transform and I need to prove a formula about the first derivative.
Let $f \, : \, \left[0,+\infty\right)\, \to \mathbb{R}\,$ be a continuous and buonded function. We know that the Laplace's transform of $f$ is well defined in $\left[0,+\infty\right)$ since it exists a positive real number $M$ such that $ \left|f(x)\, e^{-\lambda\, x }\right| \leq M e^{-\lambda x}\, \to \, 0$ for $x \to +\infty$. Suppose now that we want to compute the first derivative of the transform $\mathcal{L}\left[f(x)\right](\lambda)$, so applying the definition we get: \begin{equation} \begin{split} \frac{\mbox{d}}{\mbox{d}\lambda}\mathcal{L}\left[f(x)\right](\lambda)&=\frac{\mbox{d}}{\mbox{d}\lambda}\int_{0}^{+\infty}f(x)\,e^{-\lambda x}\,dx. \end{split} \end{equation} Once we get to this first step, we need to justify how the next step can be done, that is \begin{equation} \frac{\mbox{d}}{\mbox{d}\lambda}\int_{0}^{+\infty}f(x)\,e^{-\lambda x}\,dx=\int_{0}^{+\infty}\frac{\partial}{\partial\lambda}\left[f(x)\,e^{-\lambda x}\right]\,dx. \end{equation} I tried to use the Dominated Convergence theorem that is often use in these cases, since all we want to prove is that a limit symbol can be swapped with the integral one, but I can't really get to a proper proof. Do you have any ideas?

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You just need to find the correct theorem for differentiation under the integral. Anyways, from your context I assume for the abscissa of absolute convergence $\lambda >0$. We can hence take any compact interval $[a,b] \subset (0,\infty)$ and use theorem 3 on this page.

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  • $\begingroup$ It is ok if I assume that I can use the theorem 3 on $f(x)e^{-\lambda x}$ even if it is not defined in a open set? I can just ignore the fact that it is also defined in $(0,0)$ since the Lebesgue integral "doesn't care" about points right? ( "doesnt care" = null measure set) $\endgroup$ Commented Oct 1, 2023 at 15:36
  • $\begingroup$ For $\lambda=0$ your integral doesn't necessarily converge with your imposed conditions. $\endgroup$
    – F. Conrad
    Commented Oct 1, 2023 at 20:45

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