Direct Integration
Possible issues in performing the integrations include choice of Assumptions
, branch cuts in the integrands, and how limits are taken. Addressing the first of these gives five solutions.
il[f_, s_, t_] := Module[{r}, 1/(2 π I) Integrate[f Exp[s t], {s, r - I ∞, r + I ∞},
Assumptions -> r > 2 && t > 0]]
Grid[ParallelMap[{InverseLaplaceTransform[#, s, t], il[#, s, t]} &,
Extract[ex, {{1}, {2}, {5}, {6}, {10}}], Method -> "FinestGrained"], Frame -> True]

The last result does not agree with the corresponding result from InverseLaplaceTransform
, perhaps due to the location of the branch cuts of ArcTan[2/s]
.
Indefinite Integrals
Although Integrate
returns unevaluated for all definite integrals, il
, except those in the preceding table, Integrate
returns indefinite integrals for many of the expressions in ex
. For instance,
int = Grid[ParallelMap[{InverseLaplaceTransform[#, s, t],
Integrate[# Exp[s t]/(2 π I), s, Assumptions -> t > 0]} &,
Extract[ex, {{1}, {4}, {5}, {6}, {14}, {15}}], Frame -> True];
returns results in which s
appears only in terms of the form, ExpIntegralEi[s t + c[t]]
, where c[t]
is a polynomial in t
. For instance, ex[[4]]
integrates to
(* -((I E^(I t) ExpIntegralEi[(-I + s) t])/(4 π))
- (I E^(-I t) ExpIntegralEi[(I + s) t])/(4 π) *)
For st
a large positive or negative imaginary number, c[t]
can be ignored, and
ExpIntegralEi[s t] - ExpIntegralEi[- s t] == 2 Pi I
in this limit, as can be seen from Eq (7) the Wolfram MathWorld description of this function or simply by plotting the left side of this last expression.
Plot[Evaluate[ReIm[ExpIntegralEi[I/x] - ExpIntegralEi[-I/x]]], {x, 0, 0.2}]

Hence, int
can be simplified in this limit to
(int /. ExpIntegralEi[_] -> 2 Pi I) // Expand // ExpToTrig // FullSimplify

Inverse Laplace Transforms obtained by integration have, of course, been obtained in the first table for elements {1, 5, 6}
of ex
, and those transforms agree with those in the second table. Incidentally, it may not be obvious that the two results for ex[[15]]
in the second table are equal, They are.
FullSimplify[1 - (E^(-t/2) (Sqrt[3] Cos[(Sqrt[3] t)/2] + Sin[(Sqrt[3] t)/2]))/Sqrt[3]]
- (E^(-(-1)^(1/3) t) (-1 - I Sqrt[3] + 2 (1 + (-1)^(1/3)) E^((-1)^(1/3) t)
- 2 E^(I Sqrt[3] t)))/(2 (1 + (-1)^(1/3)))
(* 0 *)
Thus, we now have results for {1, 2, 4, 5, 6, 10, 14, 15}
of ex
, although the result for ex[[10]]
does not agree with the InverseLaplaceTransform
result, as explained above.
At first glance, the indefinite integral of ex[[13]]
should be evaluated in a similar fashion.
(* -((I E^(1 + s (1 + t)) ExpIntegralEi[-s])/(2 π (1 + t)))
+ (I E ExpIntegralEi[s t])/(2 π (1 + t)) *)
The second term in this expression reduces in the limit of s t
a large imaginary number to -(E/(1 + t))
, the result returned InverseLaplaceTransform
. However, the first term does not vanish in this limit but instead oscillates at an every faster rate. Probably, this is so, because ex[[13]]
has a branch cut along the entire positive real axis, so the path integral in the question cannot be carried out legitimately.
Indefinite integrals also can be computed for ex[[3]]
in terms of Erf
and Erfi
functions, and for ex[[9]]
(and also ex[[2]]
) in terms of HypergeometricPFQ
functions. I have not attempted to obtain limits of these expressions for s t
a large imaginary number. Integrate
again returns unevaluated for elements {7, 11, 12, 16}
of ex
.
Integrate[]
, correct? $\endgroup$