It appears that the following function, defined in terms of Inverse Laplace Transform, has a compact approximation
$$g(t)=\mathcal{L}^{-1}\left[\frac{1}{\frac{s}{2}+\frac{5 \sqrt{s}}{\sqrt{6} \tan ^{-1}\left(\frac{\sqrt{\frac{2}{3}}}{\sqrt{s}}\right)}}\right]\approx 0.434 t^{-\frac{1}{2}}.$$
How could I have obtained this with Mathematica?
In particular, I'm wondering if there's built-in functionality to let me write $f(s)=\mathcal{L}[g(t)]$, the arctan expression above, as the following series, for $0<\lambda_0<\lambda_1<\ldots$ and $\lambda_0=\frac{1}{2}$
$$f(s)=\sum_{n=0}^\infty \frac{a_n}{s^{\lambda_n}}$$
because that gives the following power series representation of $g(t)$
$$g(t)=\sum_{n=0}^\infty \frac{a_n t^{\lambda_n-1}}{\Gamma(\lambda_n)}$$
Tried Limit[InverseLaplaceTransform[...]]
and Asymptotic[InverseLaplaceTransform[...]]
with no luck.
ClearAll["Global`*"];
g = 1/(s/2 + (5 Sqrt[s])/(Sqrt[6] ArcTan[Sqrt[2/3]/Sqrt[s]]));
plot1 = DiscretePlot[
InverseLaplaceTransform[g, s, t], {t, 0., 100, 10},
PlotLegends -> {"g(t)"}];
plot2 = Plot[0.434 t^(-1/2), {t, 0, 100},
PlotLegends -> {" 0.434 t^(-1/2)"}];
Show[plot1, plot2]
Related mathoverflow discussion and mathematica.SE discussion
g = 1/(s/2 + (5 Sqrt[s])/(Sqrt[6] ArcTan[Sqrt[2/3]/Sqrt[s]]));InverseLaplaceTransform[g, s, t]
produces a mixture oft
ands
:1/t3 (-(( 5 E^(-2 t/3) Log[1/8 (-2 - 3 s)])/(-10 + Log[1/8 (-2 - 3 s)])^2) - ( 10 (-1 + E^(-2 t/3)))/(10 + Log[8] - Log[-2 - 3 s])^2 + ( 5 E^(-2 t/3) Log[1/8 (-2 - 3 s)])/(10 + Log[8] - Log[-2 - 3 s])^2 - ( E^(-2 t/3) Log[1/8 (-2 - 3 s)]^2)/(10 + Log[8] - Log[-2 - 3 s])^2 + ((-1 + E^(-2 t/3)) Log[ 1/8 (-2 - 3 s)]^2)/(10 + Log[8] - Log[-2 - 3 s])^2
. $\endgroup$