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I haven't been able to find the inverse of the Laplace transform image $$F(s)=\exp\left(\sqrt{s^3}-\sqrt{(s+1)^3}\right)$$ in the tables or using Mathematica. Computing the contour integral around the branch cut seems difficult. Has anyone seen this one before or know a trick to attack it?

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Contour integration yields the following integral representation: $$\mathcal{L}^{-1}(F)(t)=$$ $$\frac{1}{\pi}\left(\int_0^1 \sin\left(s^{3/2}\right)\exp\left(-(1-s)^{3/2}-st\right)ds+\int_1^\infty\sin\left(s^{3/2}-(s-1)^{3/2}\right)\exp(-st)ds\right)$$ which agrees with the numerical inverter maintained by Peter Valkó. These integrals appear hopeless to compute explicitly, although I would be interested in getting an asymptotic as $t\searrow 0$. The asymptotic as $t\to\infty$ is $\frac{3}{4e\sqrt{\pi}}t^{-\frac{5}{2}}$ which follows from Watson's lemma applied to the first term. I'm not sure how to obtain the small $t$ asymptotic, though.

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    any reason to believe that there exists a nice closed form? – 2017-02-04
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    [This answer](http://math.stackexchange.com/a/348021/301977) might give you some ideas – 2017-02-04
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    @tired All I know is that modulo a normalization factor, $F$ is the Laplace transform of a probability density that has a finite mean but infinite variance. – 2017-02-05

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