= Laplace transform of a derivative
{c}
{title2=$\mathcal L\{f'\}(s)=sF(s)-f(0+)$}
Suppose $f$ is an <absolutely continuous function> on every finite <interval> $[0,T]$, both $e^{-st}f(t)$ and $e^{-st}f'(t)$ are <Lebesgue integrable functions>, and $e^{-sT}f(T)\to0$. Then <integration by parts> on $[0,T]$ followed by a <limit> gives
$$
\int_0^T e^{-st}f'(t)\,dt=e^{-sT}f(T)-f(0+)+s\int_0^Te^{-st}f(t)\,dt,\qquad \mathcal L\{f'\}(s)=sF(s)-f(0+).
$$
Here $f(0+)$ is the right <one-sided limit>. <Exponential order> bounds on $f$ and $f'$ suffice when $\operatorname{Re}s$ exceeds both bounds. For $f(t)=e^{-2t}$, the identity reads $-2/(s+2)=s/(s+2)-1$ on $\operatorname{Re}s>-2$. This is the ordinary <derivative> on nonnegative time; differentiating a <causal extension by zero> at its jump instead introduces a <Dirac delta function>. https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/22aba4126352ce0f76930d858d8dffa5_MIT18_03SCF11_s29_1text.pdf[MIT 18.03SC, derivative rules and initial data]{ref}.
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