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.

enter image description here

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

  • $\begingroup$ Something strange. In 13.2.1 on Windows 10 g = 1/(s/2 + (5 Sqrt[s])/(Sqrt[6] ArcTan[Sqrt[2/3]/Sqrt[s]]));InverseLaplaceTransform[g, s, t] produces a mixture of t and s: 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. $\endgroup$
    – user64494
    May 11 at 9:10
  • $\begingroup$ @user64494 this also happens with simpler expressions that can be inverted exactly, some workaounds here mathematica.stackexchange.com/questions/284437/… $\endgroup$ May 11 at 9:11
  • $\begingroup$ YaroslavBulatov (@ does not work.): Sorry, this is your duty. $\endgroup$
    – user64494
    May 11 at 10:08
  • 2
    $\begingroup$ @user64494 I suspect there's no way to invert this Laplace transform symbolically, even after fixing that issue (related discussion here), so some asymptotic approach is needed here . @ doesn't work because author is automatically notified of comments on their posts $\endgroup$ May 11 at 10:20
  • 1
    $\begingroup$ Asking ChatGPT for a list of methods for this case you will find similar answers plus the Tauberian method provided by the answer in your math overflow link (which seems to also be here in a more general scenaro: en.wikipedia.org/wiki/…). ChatGPT is not good at performing math but it can be useful for finding relevant aspects of math as shown here if I recall correctly: arxiv.org/abs/2301.13867 $\endgroup$ May 11 at 23:55

1 Answer 1


Looks like the trick to get that expansion was to switch order of Laplace Transform and Asymptotic, and do asymptotic around 0

g = 1/(s/2 + (5 Sqrt[s])/(Sqrt[6] ArcTan[Sqrt[2/3]/Sqrt[s]]));
InverseLaplaceTransform[Asymptotic[g, s -> 0], s, t] (* Sqrt[(3 \[Pi])/2]/(5 Sqrt[t]) *)

The reason this works is because asymptotics of $t\to \infty$ in time domain corresponds to asymptotics of $s\to 0$ in Laplace domain, so this an gives 2 ways of getting series expansion for the same regime.

f = (Sqrt[\[Pi]/2] Erf[Sqrt[2] Sqrt[t]])/(2 Sqrt[t]);

isTimeDomain[expr_] := MemberQ[Variables@expr, t];
laplace[expr_] := If[isTimeDomain[expr],
   Distribute[LaplaceTransform[expr, t, s]],
   Distribute[InverseLaplaceTransform[expr, s, t]]];

asymp[expr_] := If[isTimeDomain[expr],
  Asymptotic[expr, {t, \[Infinity], 2}],
  Asymptotic[expr, {s, 0, 2}]

vals = Map[
   TraditionalForm, {{f, laplace@f}, {asymp@f, 
     asymp@laplace@f}, {laplace@asymp@laplace@f, 
     laplace@asymp@f}}, {2}];
 TableHeadings -> {{"original", "asymptotic1", 
    "asymptotic2"}, {"time domain", "laplace domain"}}]

enter image description here

  • $\begingroup$ All that is built on the sand. Such type answers are not a good practice though N[Sqrt[(3 \[Pi])/2]/5] results in 0.434161 which is close to 0.434. $\endgroup$
    – user64494
    May 12 at 20:09
  • $\begingroup$ Why the asymptotic as s tends to zero works for the asymptotic as t tends to infinity? $\endgroup$
    – user64494
    May 12 at 20:18
  • $\begingroup$ Tauberian type theorems give the relation between t to zero and s to infinity in your notations. $\endgroup$
    – user64494
    May 12 at 20:28
  • $\begingroup$ It should be added these theorems are about Laplace transform, not about inverse Laplace transform. $\endgroup$
    – user64494
    May 12 at 20:41
  • 1
    $\begingroup$ @user64494 Theorem 2.7 and 2.8 of Cohen's Inverse Laplace book show how Puiseux series in Laplace domain correspond to infinite series in time domain. There's also Watson's Lemma $\endgroup$ May 13 at 7:06

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