# InverseLaplaceTransform involving an ArcTan

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]


• Looks like a bug to me. May be report it to [email protected] Commented Apr 29, 2023 at 20:14
• Just some cursory investigation. InverseLaplaceTransform calls TransformsInverseLaplaceBromwich, which calls TransformsInverseLaplaceBromwichDumpSumResidue, 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. Commented Apr 29, 2023 at 22:41
• Versions 12.0 and 12.1 seem to work ok. Commented Apr 29, 2023 at 23:11
• Reported to Wolfram Support CASE:5025460 Commented May 11, 2023 at 9:15

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:

Integrate[InverseLaplaceTransform[
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]*)


Or:

 LaplaceTransform[InverseLaplaceTransform[
InverseLaplaceTransform[ArcTan[A*Sqrt[s]/Sqrt[2]]/(
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])*)

InverseMellinTransform[InverseLaplaceTransform[
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]]


with

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

Sqrt[2/s]ArcTan[Sqrt[2/s]]


This is identical with Prudnikov et. al. , Integrals an Series, Vol. 5, 2.6.4.16, 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.

    Plot[
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"}]
`

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

• Reported to Wolfram Support, they opened CASE:5025460 Commented May 2, 2023 at 0:03