InverseLaplaceTransform[ArcTan[s], s, t]

Questions
,
1

Why

InverseLaplaceTransform[ArcTan[1/s], s, t]

Out[]= Sin[t]/t

works and

InverseLaplaceTransform[ArcTan[s], s, t]

does not work?

2

Good question. I’d like to know why does not work? Anyone know the answer?

3

This is a limitation in InverseLaplaceTransform since it evaluates for In[2] below, and
ArcTan[s] is equivalent to In[1] for s > 0.

In[1]:= expr = Pi/2 - ArcTan[1/s];

In[2]:= InverseLaplaceTransform[expr, s, t] // InputForm

Out[2]//InputForm=
(Pi*DiracDelta[t])/2 - Sin[t]/t

In[3]:= FullSimplify[expr, s > 0]

Out[3]= ArcTan[s]

Thank you for reporting this issue with InverseLaplaceTransform.

4

By the way, these examples are where Mathematica has issue:

InverseLaplaceTransform[1/(s^\[Alpha] - a), s, t] == t^(\[Alpha] - 1)*MittagLefflerE[\[Alpha], \[Alpha], a*t^\[Alpha]]
InverseLaplaceTransform[s^\[Alpha]/(s*(s^\[Alpha] + a)), s, t] == MittagLefflerE[\[Alpha], -a*t^\[Alpha]]
InverseLaplaceTransform[a/(s*(s^\[Alpha] + a)), s, t] == 1 - MittagLefflerE[\[Alpha], -a*t^\[Alpha]]
InverseLaplaceTransform[s^(\[Alpha] - \[Beta])/(s^\[Alpha] - a), s, t] == t^(\[Beta] - 1)*MittagLefflerE[\[Alpha], \[Beta], a*t^\[Alpha]]

and should also solve:

LaplaceTransform[t*f[t], t, s] == -D[LaplaceTransform[f[t], t, s], s]
LaplaceTransform[t*f'[t], t, s] == -LaplaceTransform[f[t], t, s] - s*D[LaplaceTransform[f[t], t, s], s]
 ...
LaplaceTransform[t^2*f[t], t, s] == D[LaplaceTransform[f[t], t, s], {s, 2}]
LaplaceTransform[t^2*f'[t], t, s] == 2*D[LaplaceTransform[f[t], t, s], s] + s*D[LaplaceTransform[f[t], t, s], {s, 2}]
 ...