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I am given a question of Laplace Transform which is as follows:

$$ e^{t}sin2t $$ for $$t\leq0$$

Now I know that by using the transform pair we get:

$$ [e^{-at}sin\omega_0 t]u(t) \rightarrow \frac{\omega_0}{(s+\alpha)^2 +(\omega_0)^2} $$

Using this pair we:

$$ [e^{t}sin2t]u(t) \rightarrow \frac{2}{(s+1)^2 +(2)^2} $$

Now for $u(-t)$ what I think should come is: $$ -\frac{2}{(s-1)^2 +(2)^2} $$

Is it correct?

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Start first at $f(t)=sin(2t)u(t)$, which has a Laplace transform of $$\frac{2}{s^2+2^2}$$ then apply time-scaling property $f(\alpha t)=\frac{1}{|\alpha|}F(\frac{s}{\alpha})$ with $a=-1$. Now, $f(t)$ becomes $sin(-2t)u(-t)$, but this is just equal to $-sin(2t)u(-t)$ because $sin(x)$ is odd function.

Lastly, apply $e^t$ to $f(t)=-sin(2t)u(-t)$, and then remove the negative.

Edit: By the way, the answer you gave is correct.

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