Sanity check needed - if I have the expression

$f(t) = x$

Then taking the Laplace transform of both sides yields

$\mathcal{L}\Big(f(t)\Big) = \mathcal{L}(x)$

Is it true that if I had another expression

$f(t)g(t) = x$

The Laplace transform of both sides would give?

$\mathcal{L}\Big(f(t)g(t)\Big) = \mathcal{L}(x)$


It seems correct. But note that we don't necessarily have $$\mathcal{L}(f(t)g(t))=\mathcal{L}(f(t))\times\mathcal{L}(g(t)) $$

  • $\begingroup$ Exactly, that's what I figured. Thanks - long day. $\endgroup$ – Alex May 2 '13 at 17:40
  • $\begingroup$ Spot on, here, my friend! +1 $\endgroup$ – Namaste May 3 '13 at 0:11

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