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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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Does $\displaystyle\sin u=\frac{e^{iu}-e^{-iu}}{2i}$ give you any ideas? Where are you getting $\delta$'s from? – anon May 17 '12 at 17:02
    
@anon From computing inverse Laplace transforms of each term of the stated series... – Sasha May 17 '12 at 17:12
    
@Sasha: Oh, duh. I'm confusing $\mathcal{L}$ with $\mathcal{L}^{-1}$. – anon May 17 '12 at 17:13
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Direct Laplace transform, $F(s) = \mathcal{L}_s(f(x)) = \int_0^\infty \mathrm{e}^{-s x} f(x) \mathrm{d} x$, of an integrable function $f(x)$, has the property that $F(s)$, if exists, vanishes for large positive $s$. Hence $\sin(a s)$ can not be a direct Laplace transform of any integrable function. – Sasha May 17 '12 at 17:46
    
@Sasha i'll settle for a distribution, but not the ugly Dirac delta and its derivatives !! – Mohammad Al Jamal May 17 '12 at 20:26
up vote 1 down vote accepted

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.

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

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