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]
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Sign up to join this communityI 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.
InverseLaplaceTransform
callsTransforms`InverseLaplaceBromwich
, which callsTransforms`InverseLaplaceBromwichDump`SumResidue
, which callsSeriesCoefficient[(E^(s t) ArcTan[Sqrt[s]/Sqrt[2]])/(Sqrt[s] (2 + s)), {s, -2, -1}]
. This call returns the offendingLog[s+2]
, which according to the first example in the Possible Issues section in theSeriesCoefficient
docs, it's allowed to do. So if I had to guess, the fix to the bug is thatSumResidue
should fail or try something else if the dummy variable is returned bySeriesCoefficient
. $\endgroup$