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I was wondering if there is a closed-form Laplace inverse of the sine function. I have tried the following: $ \sin(as)=\sum_{n=0}^{\infty}\frac{(-1)^{n}(as)^{2n+1}}{(2n+1)!} $ an $n$-th power of $s$ contributes with an $n$-th derivative of the Dirac delta. So one expects a series expansion in terms of the Delta function and its derivatives. But that is utterly ugly! Hence the question.

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    @Sasha i'll settle for a distribution, but not the ugly Dirac delta and its derivatives !!2012-05-17

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If $\mathcal L\{f\} = F(s)$ vanishes on an infinite sequence of points that are located at equal intervals along a line parallel to the real axis $ F(s_0+n\sigma)=0\qquad (\sigma >0, n=1,\,2,\,\ldots) $ $s_0$ being a point of convergence of $\mathcal L\{f\}$; then it follows that $f(t)$ is a nullfunction.

So it follows that a Laplace transform $F(s)\neq 0$ cannot be periodic.

Thus $\sin(s)$ cannot be the Laplace transform of a function.

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Perhaps Post's inversion formula might be helpful for you.

$\mathcal{L}^{-1}_{s\to t}\{\sin as\}=\lim\limits_{k\to\infty}\dfrac{(-1)^k}{k!}\left(\dfrac{k}{t}\right)^{k+1}a^k\sin\left(\dfrac{ak}{t}+\dfrac{k\pi}{2}\right)$