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When dealing with one ODE, it happened to find the inverse Laplace transform of function

$$ F(s)=\frac{s}{\sin(2s)}. $$

I suppose it exists the inverse Laplace transform, but I could not find any standard formula from Laplace tables, either. I try to expand in Laurent series and found $$ \frac{1}{\sin(s)}={\frac {-2i{{\rm e}^{-is }}}{{{\rm e}^{-2is}}-1}} $$ from here.

But it seems incredibly difficult to proceed. Would someone has a simple way, please advise?

FD_bfa
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MathArt
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    Since $F(s)$ doesn't tend to zero as $s$ goes to infinity, its inverse Laplace transform can't be an ordinary function, but rather a distribution. And even then, it seems a bit unlikely to get infinitely many poles on the right half of the plane. Are you sure that you've done everything else up to that correctly? – S.C. Apr 21 '24 at 22:09
  • You can try mellin formula found https://en.wikipedia.org/wiki/Inverse_Laplace_transform –  Apr 21 '24 at 23:01
  • @S.C., Yes, you are right. I was vaguely feeling that not easy to find an ordinary function since it can even be growing faster than $\mathcal{L}^{-1}[s]$. I need to double check... – MathArt Apr 22 '24 at 07:06
  • It is tempting to try Post's inversion formula, but not successful. If someone has this expertise please instruct. Lorem Ipsum! – MathArt Apr 22 '24 at 07:09

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