My friend and I had an argument upon the Laplace transform of $\sin \omega t$. He's saying that its Laplace transform does not exist and only the Laplace transform of $u(t) \sin \omega t$ exist. But when I checked in my mathematics book ("Advanced Engineering Mathematics" by Greenberg), the transform of $\sin \omega t$ was given. Who is right, then?
2025-01-13 02:21:25.1736734885
Unilateral and bilateral Laplace transorm
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Your friend sort of have a point. If you're talking about one-sided laplace transform it is in effect the function $\sin(\omega t)u(t)$ that is transformed. The laplace transform is the integral:
$$\int_0^\infty sin(\omega t) e^{-st}dt = \int_{-\infty}^\infty \sin(\omega t)u(t)e^{-st} = {\omega\over s^2+\omega^2}$$
If you're on the other hand talking about double sided transform you've got to use distribution interpretation of the transform. The definition:
$$\mathcal L sin(\omega t) = \mathcal F sin(\omega t)e^{-\sigma t}$$
Which is only valid for $\sigma=0$ and the result is
$$\mathcal L sin(\omega t) = \mathcal F sin(\omega t) = {\delta_{\omega}-\delta_{-\omega}\over 2i}$$