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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?

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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

$$\mathcal L sin(\omega t) = \mathcal F sin(\omega t) = {\delta_{\omega}-\delta_{-\omega}\over 2i}$$

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  • $\begingroup$ So, who is right? Is the proof given in book wrong? $\endgroup$ – Ansh Kumar Oct 15 '15 at 5:06
  • $\begingroup$ The proof in the book is probably correct given the definition that's in the book. I suspect it gives something a result like $\omega/(s^2+\omega^2)$ and in that case it's the unilateral transform it uses. For the bilateral transform you'll either have to restrict the function by multiplying with $u(t)$ or use distribution variant of it. Note that the unilateral transform will when inversely transformed give back a function only defined on one side. $\endgroup$ – skyking Oct 15 '15 at 6:03
  • $\begingroup$ What's the distribution interpretation of the transform ? $\endgroup$ – Ansh Kumar Nov 8 '15 at 15:35
  • $\begingroup$ @AnshKumar It's based on the Fourier transform for distribution (a definition that will allow Fourier transforming $e^{i\omega t}$, resulting in $\delta_\omega$). The Laplace transform is defined by Fourier transforming after multiplying with $e^{-\sigma t}$. $\endgroup$ – skyking Nov 9 '15 at 6:46
  • $\begingroup$ Could you have a look at my another question $\endgroup$ – Ansh Kumar Dec 8 '15 at 12:15

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