This produces result that doesn't make sense, it has both s and t variables. Any idea for a work-around?

InverseLaplaceTransform[(\[Pi] - 2 ArcTan[Sqrt[s]/Sqrt[2]])/(
 2 Sqrt[2] Sqrt[s]), s, t]

enter image description here

  • 4
    $\begingroup$ Looks like a bug to me. May be report it to [email protected] $\endgroup$
    – Nasser
    Apr 29, 2023 at 20:14
  • 1
    $\begingroup$ Just some cursory investigation. InverseLaplaceTransform calls Transforms`InverseLaplaceBromwich, which calls Transforms`InverseLaplaceBromwichDump`SumResidue, which calls SeriesCoefficient[(E^(s t) ArcTan[Sqrt[s]/Sqrt[2]])/(Sqrt[s] (2 + s)), {s, -2, -1}]. This call returns the offending Log[s+2], which according to the first example in the Possible Issues section in the SeriesCoefficient docs, it's allowed to do. So if I had to guess, the fix to the bug is that SumResidue should fail or try something else if the dummy variable is returned by SeriesCoefficient. $\endgroup$
    – Greg Hurst
    Apr 29, 2023 at 22:41
  • $\begingroup$ Versions 12.0 and 12.1 seem to work ok. $\endgroup$
    – Bill Watts
    Apr 29, 2023 at 23:11
  • $\begingroup$ Reported to Wolfram Support CASE:5025460 $\endgroup$ May 11, 2023 at 9:15

2 Answers 2


I will only deal with the second part, I mean: $\frac{\tan ^{-1}\left(\frac{\sqrt{s}}{\sqrt{2}}\right)}{\sqrt{2} \sqrt{s}}$

f = (\[Pi] - 2 ArcTan[Sqrt[s]/Sqrt[2]])/(2 Sqrt[2] Sqrt[s]) // Expand

(*\[Pi]/(2 Sqrt[2] Sqrt[s]) - ArcTan[Sqrt[s]/Sqrt[2]]/(Sqrt[2] Sqrt[s])*) 

$$\mathcal{L}_s^{-1}\left[\frac{\tan ^{-1}\left(\frac{\sqrt{s}}{\sqrt{2}}\right)}{\sqrt{2} \sqrt{s}}\right](t)=\frac{\sqrt{\frac{\pi }{2}} \text{erfc}\left(\sqrt{2} \sqrt{t}\right)}{2 \sqrt{t}}$$ Workaround:

D[ArcTan[A*Sqrt[s]/Sqrt[2]]/(Sqrt[2] Sqrt[s]), A] // Factor, s, t], {A, 0, 1}]

(*ConditionalExpression[(Sqrt[\[Pi]/2] Erfc[Sqrt[2] Sqrt[t]])/(2 Sqrt[t]), Re[t] >= 0]*)


  Sqrt[2] Sqrt[s]), A, q], s, t] // Expand, q, A] /. 
 A -> 1 // FullSimplify

 (*(Sqrt[\[Pi]/2] Erfc[Sqrt[2] Sqrt[t]])/(2 Sqrt[t])*)

 MellinTransform[ArcTan[A*Sqrt[s]/Sqrt[2]]/(Sqrt[2] Sqrt[s]), A, 
  q], s, t] // ExpandAll, q, A, Assumptions -> -1 < Re[q] < 0] /. 
 A -> 1 // FullSimplify

 (*(Sqrt[\[Pi]/2] Erfc[Sqrt[2] Sqrt[t]])/(2 Sqrt[t])*)

The logs cancel in the complex plane with a cut along the negative real line s <-2.

But the result is doubious for this reason:

Vs 6 yields

InverseLaplaceTransform[(pi/2 - ArcTan[Sqrt[s/2]]) / Sqrt[s/2]],s,t]

Sqrt[pi/(2 t)]Erf[ Sqrt[2 t]]


LaplaceTransform[Sqrt[pi/(2 t)]Erf[ Sqrt[2 t]]]


This is identical with Prudnikov et. al. , Integrals an Series, Vol. 5,, p. 103, but with a condition (p>0, -Re a^2) whatever that means.

The result from vs 13 yields in vs6 yiedls for the direct Laplace transform

LaplaceTransform[Sqrt[pi/t]] +Sqrt[2]/t (1/2-Exp[-2t]), t, s]

pi/Sqrt[s] + Log[2/s+1]/Sqrt[2]

All results together show, that there is something wrong. The production of conditons for s >-2 in t has been abondoned somehow in the internal procedure. Its the condition, that the path has to be to pass on the right of the largest singularity.

 Evaluate[{(-Sqrt[2] (1/2 - E^(-2 t)/2) + (E^(-2 t) Log[(2 + 4)/8])/(
    2 Sqrt[2]) + (E^(-2 t) (Log[8] - Log[2 + 4]))/(2 Sqrt[2]))/(2 t),
   Sqrt[\[Pi]/(2 t)] Erf[Sqrt[2 t]],   ReIm[Sqrt[ 2] (Sqrt[\[Pi]]/(
       2 Sqrt[t]) - (-Sqrt[2] (1/2 - E^(-2 t)/2) + 
    (  E^(-2 t) Log[(2 - 4)/8])/(2 Sqrt[2]) + ( E^(-2 t) (Log[8] - Log[2 - 4]))/(2 Sqrt[2]))/( 2 t))]}], {t, -4, 4}, 
 PlotLegends -> {"13", "6", "Re 13 s=-4", "Im 13 s=-4"}]

inverseLaplace arctan

Of course, its not so difficult, to evaluate the integrals with s/2 -> u^2 directly.

  • $\begingroup$ Reported to Wolfram Support, they opened CASE:5025460 $\endgroup$ May 2, 2023 at 0:03

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