Find the inverse Laplace transform of $\frac{1}{1+3^\sqrt{s}}$ How can I find the inverse Laplace transform of $1/(1+3^\sqrt{s})$?
or does it exist?
$\mathcal{L}^{-1}\left\{\dfrac{1}{1+3^\sqrt{s}}\right\}$

$=\mathcal{L}^{-1}\left\{\dfrac{3^{-\sqrt{s}}}{3^{-\sqrt{s}}+1}\right\}$

$=\mathcal{L}^{-1}\left\{\sum\limits_{n=0}^{\infty}(-1)^n3^{-(n+1)\sqrt{s}}\right\}$

$=\mathcal{L}^{-1}\left\{\sum\limits_{n=0}^{\infty}(-1)^ne^{-((n+1)\ln3)\sqrt{s}}\right\}$

$=\sum\limits_{n=0}^{\infty}\dfrac{(-1)^n(n+1)e^{-\frac{(n+1)^2\ln^23}{4t}}\ln3}{2\sqrt\pi t^\frac{3}{2}}$ (according to http://eqworld.ipmnet.ru/en/auxiliary/inttrans/LapInv5.pdf)

$=-\sum\limits_{n=1}^{\infty}\dfrac{(-1)^nne^{-\frac{n^2\ln^23}{4t}}\ln3}{2\sqrt\pi t^\frac{3}{2}}$
The inverse transform does not exist.  Consider an integration of the complex integral

$$\oint_C dz \frac{e^{t z}}{1+e^{a \sqrt{z}}} $$

where $a \gt 0$ and $C$ being a Bromwich contour deformed to go around the branch cut along the negative real axis and around the branch point at $z=0$.  By Cauchy's theorem, we find that the inverse transform may be written as

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{1+e^{a \sqrt{s}}} = \frac1{2 \pi} \int_0^{\infty} dx \, e^{-x t} \tan{\left ( \frac{a}{2} \sqrt{x} \right )}$$

The integral on the RHS does not converge in any sense due to the poles at $x=(2 k+1)^2 \pi^2/a^2$ for nonnegative integer $k$.  Whether the inverse transform exists as a Cauchy principal value, I leave as an exercise for the reader.