= Convolution theorem for Laplace transforms
{title2=$\mathcal L(f*g)=\mathcal Lf\,\mathcal Lg$}
For <functions> $f,g$ on $[0,\infty)$ with $e^{-\sigma t}f(t)$ and $e^{-\sigma t}g(t)$ <Lebesgue integrable functions>, their unilateral <convolution>, defined <almost everywhere> if necessary, satisfies
$$
\mathcal L\{f*g\}(s)=F(s)G(s),\qquad (f*g)(t)=\int_0^tf(\tau)g(t-\tau)\,d\tau,\qquad \operatorname{Re}s=\sigma.
$$
Taking a <causal extension by zero> of both <functions> gives the existing whole-line <convolution> definition. <Absolute convergence> and <Fubini's theorem> justify the product rule: put $t=\tau+u$ in the double <integral>, so $e^{-st}=e^{-s\tau}e^{-su}$. For <piecewise continuous functions> of <exponential order> $a$ and $b$, the hypotheses hold for $\operatorname{Re}s>\max(a,b)$. The <Laplace transform> article works out the concrete example $h(t)=e^{-2t}$ and the <Heaviside step function> input, whose output is $(1-e^{-2t})/2$. https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/0d4d90053f2bb24f7551c78e2d1b21a2_MIT18_03SCF11_s30_5text.pdf[MIT 18.03SC, convolution and system response]{ref}.
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