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I know that they probably treated $\displaystyle f(s,t) = e^{-st} f(t)$ so the integration/differentiation thing doesn't matter, but what confuses me is when they got rid of the derivative and how the "$t$" pop out? It's taking the derivative with respect to $s$ not $t$.

Proof: Consider the identity $\frac{dF(s)}{ds} = \frac{d}{ds} \int_0^{\infty} e^{-st} f(t) dt.$ Because of the assumptions on $f(t)$, we can apply a theorem from advanced calculus (sometimes called Leibniz's rule) to interchange the order of integration and differentiation: $ \begin{align} \frac{dF(s)}{ds} & = \frac{d}{ds} \int_0^{\infty} e^{-st} f(t) dt\\ & = \int_0^{\infty} \frac{d \left(e^{-st} \right)}{ds} f(t) dt\\ & = - \frac{d}{ds} \int_0^{\infty} t e^{-st} f(t) dt\\ & = - \mathcal{L} \{ tf(t) \}(s). \end{align} $ Thus, $\mathcal{L} \{ tf(t) \}(s) = (-1) \frac{dF(s)}{ds}$

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    P.S. you'll sometimes see "differentiation under the integral sign" instead of "Leibniz's rule" in some contexts.2012-05-08

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In general, $\frac{d}{dx} e^{ax} = a e^{ax}$

Thus, since $t$ is independent of $s$, $\frac{d}{ds} e^{-st} = -te^{-st}$

$f(t)$ doesn't depend on $s$, so it's a constant with respect to differentiation by $s$, and they pulled the negative outside the integral.

Does that make sense?

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    Put another way: differentiating with respect to $s$ did not change the fact that $t$ was still a dummy variable within the Laplace transform integral...2012-05-08