# Tricky InverseLaplaceTransform problem

Is there a way to get an expression for $$g(t)$$ defined as Laplace Transform below? $$g(t)=\mathcal{L}^{-1}\left[\frac{x^2 \tan^{-1} x}{x+\tan^{-1} x}\right](t)$$ where $$x=\sqrt{\frac{2}{s}}$$

This gives symbolic solution to NDSolve problem solved numerically by Ulich here, worked out by user here

Putting it into Mathematica gives an expression containing both $$t$$ and $$s$$

x = Sqrt[2/s];
InverseLaplaceTransform[(x^2 ArcTan[x])/(x + ArcTan[x]), s, t]

• With version "13.2.1 for Mac OS X ARM (64-bit) (January 27, 2023)" this returns (after use of Simplify) (4 - 4 E^(-2 t) + 4 t Log^2 - 4 Log[1/8 (-2 - s)] + Log[1/8 (-2 - s)]^2 - 2 t Log Log[-2 - s] + 4 t Log[-2 - s]^2)/(2 t (-2 + Log[1/8 (-2 - s)])^2) The presence of s in the result indicates that this is wrong. May 7 at 14:19
• I'm on very shaky ground here (mathematically), but it is possible to do a series expansion around s==0. Can the transform be applied term by term in some way? May 7 at 19:13
• @mikado well, there this result, from Ch 2 of "Numerical Methods For Inverse Laplace" book, $\bar{f}$ refers to Laplace transform of $f$ May 7 at 20:39
• @mikado The first obstacle of using this approach is finding symbolic expression for $\lambda_n$ in Eq 2.40. $g(t)$ comes from diff-eq corresponding to the DPR1 problem described in community post May 7 at 20:47
• There exist formulas for Laplace transforms of functions of sqrt s, and the tan^.1 1/s =cot^-1 s. But the formula is an integral of the Laplace transform of f(s) with a Gaussian. Unfortunately Mathematica cannot perform the transform of the expression with arccot s nor the direct integral. So I don't expect a closed formula besides the usual series expressions. May 14 at 12:54

Another way.

I use Post - Widder formula to compute InverseLaplaceTransform. $$\mathcal{L}_s^{-1}[F(s)](t)=\underset{n\to \infty }{\text{lim}}\frac{\frac{\partial ^n}{\partial t^n}\left(t^{n-1} F\left(\frac{n}{t}\right)\right)}{\Gamma (n)}$$ I only use n=20, because I can't find closed-form for n-th derivative.

 g = (2 ArcTan[Sqrt Sqrt[1/s]])/(
s (Sqrt Sqrt[1/s] + ArcTan[Sqrt Sqrt[1/s]]));
U[t_] := InverseLaplaceTransform[g, s, N@t];
F[s_] := (2 ArcTan[Sqrt Sqrt[1/s]])/(
s (Sqrt Sqrt[1/s] + ArcTan[Sqrt Sqrt[1/s]]));
G[t_, n_] := 1/Gamma[n]*D[t^(n - 1)*F[n/t], {t, n}]

{Plot[{U[t], Evaluate@G[t, 1]}, {t, 0, 10}], Plot[{U[t], Evaluate@G[t, 20]}, {t, 0, 10}]} would a numerical solution help?

f = x^2 ArcTan[x]/(x + ArcTan[x]) /. x -> Sqrt[2/s] // FullSimplify;
g = Table[{t, InverseLaplaceTransform[f, s, t]}, {t, 0.01, 10.,0.1}];
model = a*Exp[-b*t] + c;
fit = NonlinearModelFit[g, model, {a, b, c}, t];
Show[ListPlot[g], Plot[fit[t], {t, 0, 10}, PlotStyle -> Red],
Frame -> True, GridLines -> Automatic] ListPlot[fit["FitResiduals"],DataRange -> {0, 10}, Filling -> Axis] • so it's useful in a sense that the solution is clearly not exponential, I'm trying to figure out what the correct functional form is May 9 at 9:09
• The solution for closely related problem is $\sqrt{\frac{\pi}{4 t}}$ so this may also be some kind of power law May 9 at 9:11
• @Yaroslav Bulatov A huge improvement is obtained with the model = a - Exp[b/Sqrt[t]]. Good luck!
– rmw
May 9 at 15:33
• interesting! If we extrapolate, this will underestimate the true value, while power-law fit overestimates, so the solution is neither pure exponential nor power law May 9 at 16:06
• @Yaroslav Bulatov Please consider that the solution only applies to the numerically calculated range, i.e. in this case for t in the interval [0.01,10]. For this range, the solution is only an approximation and by no means a general analytical solution.
– rmw
May 9 at 19:57

Defining

approx[n_] := approx[n] =
InverseLaplaceTransform[Normal[Series[expr, {s, ∞, n}]],s, t]


I find that the results are computed quite quickly.

This seems to give consistent values for small values of t

For example

Plot[Evaluate[Table[approx[i], {i, 16, 32}]], {t, 0, 5},
PlotRange -> All] I've no idea if the results are mathematically valid or helpful to you.

• Interesting ! I wonder if this representation is connected to the series in the theorem, writing expression as infinite partial fraction representation May 8 at 7:33

Hope I didn't misunderstand the question:

If only numerical solution is asked for try

invLaplace[ t_?NumericQ] :=
InverseLaplaceTransform[1/(s/2 + Sqrt[s]/(Sqrt ArcTan[Sqrt/Sqrt[s]])),s, t]

Plot[invLaplace[t], {t, 0.001, 10}] • That looks like correct graph. This is another example of the integro-differential equation you provided Nested DSolve solution for earlier, the technique to get Laplace Transform of solution is documented here, I'm was looking to understand how fast the solution decays to zero May 14 at 14:04
• But the original integro-differential equation leads to a more complicated x-dependent inverse Laplace-Transformation I think May 14 at 15:26
• They are slightly different (due to different values of $a,b$) but you can see that asymptotically they both have $1/\sqrt{t}$ decay May 14 at 17:09