Im having a problem integrating $u(-t)$ so I can get the Laplace transform. My table shows $\mathcal {L}\{-u(-t)\} = 1/s$ but I'm not sure if there's a property I should be using for negative arguments or if it just changes what values would be allowed as inputs. I've tried Wolfram and MATLAB and they both keep giving me zero, which I assume to be because the value of the step function for a negative time is zero. It is my assumption that $\mathcal{L} \{ u(-t) \} = -1/s$ but I get a stuck trying to evaluate the definite integral from $ -\infty $ to zero.
2026-04-20 09:59:04.1776679144
Laplace transform of unit step with negative arguments
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The Laplace tranform ignores what has happened the time before $t=0.$ The Fourrier Transform, on the other hand, considers time in both directions.
But you could consider a function like $u(a-t)$
$\mathcal L \{u(a-t)\} = \int_0^{a} e^{-st}\ dt = \frac {1 - e^{-as}}{s}$
vs.
$\mathcal L \{u(t-a)\} = \int_a^{\infty} e^{-st}\ dt = \frac {e^{-as}}{s}$