In response to the question here: https://community.wolfram.com/groups/-/m/t/3295143
On way is:
In[]:=
f[s_]:=ArcCoth[Sqrt[s]]/Sqrt[s];​​InverseLaplaceTransform[FullSimplify[InverseMellinTransform[f[s]//TrigToExp,s,w,GenerateConditions->False]/.w->Exp[t],Assumptions->t>0],t,x]
Out[]=
HeavisideTheta[1-x]
2
x
In[]:=
Simplify[%,Assumptions->0<x<1]
Out[]=
1
2
x
Another way:
In[]:=
Simplify[InverseMellinTransform[Simplify[InverseMellinTransform[f[s]//TrigToExp,s,Exp[t],GenerateConditions->False],Assumptions->t>0],t,Exp[-x],GenerateConditions->False],Assumptions->x>0]
Out[]=
HeavisideTheta[1-x]
2
x
Update: I managed to find the Inverse Stieltjes transform is similar like Hilbert transform:
In[]:=
InverseStieltjesTransform=1/π^2*Integrate[1/(s-x)*f[s],{x,-Infinity,Infinity},PrincipalValue->True]==Simplify[-1/Pi*InverseFourierTransform[-ISign[z]FourierTransform[f[s],s,z,FourierParameters->{1,-1}],z,x,FourierParameters->{1,-1}],Assumptions->x>0]==Simplify[1/Pi*InverseFourierTransform[I*(2*HeavisideTheta[z]-1)*FourierTransform[f[s],s,z,FourierParameters->{1,-1}],z,x,FourierParameters->{1,-1}],Assumptions->x>0]
Stieltjes transform of:
Let:
In[]:=
$Version
Out[]=
14.1.0 for Microsoft Windows (64-bit) (July 16, 2024)
In[]:=
func={Sin[x],Cos[x],Exp[Ix],1/(2Sqrt[x]),Piecewise[{{1/(2*Sqrt[x]),0<x<1}},0]};(*Functionexamples.*)
In[]:=
A=Integrate[1/(s-x)*(#),{x,-Infinity,Infinity},PrincipalValue->True]&/@func(*StieltjesTransform*)
Out[]=

-πCos[s] if s∈
,
πSin[s] if s∈
,
-
s

π if s∈
,
-
π
2
-s
if Re[s]<0&&Im[s]0
,
ArcTanh
1
s

s
Re[s]≥1||Re[s]≤0||Im[s]>0||Im[s]<0
ArcTanh[
s
]
s
True

In[]:=
A1=Integrate[1/(s-x)*(#),{x,-Infinity,Infinity},PrincipalValue->True,GenerateConditions->False]&/@func(*StieltjesTransform*)
Out[]=

-πCos[s]
Arg[s]≥π||Arg[s]≤0
-π(Cos[s]+2Sin[s])
True
,
πSin[s]
Arg[s]≥π||Arg[s]≤0
π(-2Cos[s]+Sin[s])
True
,
-
s

π
Arg[s]≥0

s

π
True
,
1
4
-2π
1
s
-
-2Log-
1
s
+2Log
1
s
+Log-
1
s
+Log[s]
s
,
2ArcCoth[
s
]+Log-
1
s
-Log
1
s

2
s

In[]:=
B=1/π^2*Integrate[1/(s-x)*(#),{s,-Infinity,Infinity},PrincipalValue->True]&/@A(*InverseStieltjesTransform*)​​Simplify[B,Assumptions->x>0](*OK,can'tcalculate*)
In[]:=
B1=1/π^2*Integrate[1/(s-x)*(#)//Expand,{s,-Infinity,Infinity},PrincipalValue->True,Assumptions->x>0]&/@A1[[1;;4]](*OK!InverseStieltjesTransform*)
Out[]=
Sin[x],Cos[x],
x

,
1
2
x

Hard example to invert :
In[]:=
1/π^2*Integrate[1/(s-x)*(-(π/(2Sqrt[-s]))),{s,-Infinity,Infinity},PrincipalValue->True,Assumptions->x>0](*InverseStieltjesTransform.OkWorksfine*)
Out[]=
1
2
x
In[]:=
1/π^2*Integrate[1/(s-x)*(1/4(-2IπSqrt[1/s]-(-2Log[-(1/s)]+2(Log[1/s]+Log[-(1/Sqrt[s])])+Log[s])/Sqrt[s])),{s,-Infinity,Infinity},PrincipalValue->True,Assumptions->x>0](*InverseStieltjesTransform.OkWorksfine*)
Out[]=
1
2
x
1/π^2*Integrate[1/(s-x)*(Piecewise[{{ArcTanh[1/Sqrt[s]]/Sqrt[s],Re[s]>=1||Re[s]<=0||Im[s]>0||Im[s]<0}},ArcTanh[Sqrt[s]]/Sqrt[s]]),{s,-Infinity,Infinity},PrincipalValue->True,Assumptions->x>0](*InverseStieltjesTransform.IAbortedtolongtimetocompute!*)
1/π^2*Integrate[1/(s-x)*((2ArcCoth[Sqrt[s]]+Log[-(1/s)]-Log[1/s])/(2Sqrt[s])),{s,-Infinity,Infinity},PrincipalValue->True,Assumptions->x>0](*InverseStieltjesTransform.IAbortedtolongtimetocompute!*)
Simplify[-1/Pi*InverseFourierTransform[-ISign[z]*FourierTransform[Piecewise[{{ArcTanh[1/Sqrt[s]]/Sqrt[s],Re[s]>=1||Re[s]<=0||Im[s]>0||Im[s]<0}},ArcTanh[Sqrt[s]]/Sqrt[s]],s,z,FourierParameters->{1,-1}],z,x,FourierParameters->{1,-1}],Assumptions->x>0](*InverseStieltjesTransform.IAbortedtolongtimetocompute!*)
Workaround for one hard example, I use a Trick:
L=Assuming[x∈Reals,1/π^2*Integrate[1/(s-x)*(D[Piecewise[{{ArcTanh[Q/Sqrt[s]]/Sqrt[s],Re[s]>=1||Re[s]<=0||Im[s]>0||Im[s]<0}},ArcTanh[Q*Sqrt[s]]/Sqrt[s]],Q]),{s,-Infinity,Infinity},PrincipalValue->True]]​​​​L1=FullSimplify[L,Assumptions->{0<Q<1,x∈Reals}]​​Integrate[L1,{Q,0,1},Assumptions->x∈Reals]​​(*Toolongtimetocompute!!!*)
We compute numerically:
In[]:=
R[x_]:=NIntegrate[Piecewise[{{Log[-(((-1+Q^2)*x)/(-1+x))]/(-1+Q^2*x)-Log[-(((-1+Q^2)*x)/(Q^2*(-1+x)))]/(Q^2-x),x>1},{((-I)*Pi+Log[((-1+Q^2)*x)/(-1+x)])/(-1+Q^2*x)-Log[-(((-1+Q^2)*x)/(Q^2*(-1+x)))]/(Q^2-x),x<=0}},Log[((-1+Q^2)*x)/(-1+x)]/(-1+Q^2*x)-Log[((-1+Q^2)*x)/(Q^2*(-1+x))]/(Q^2-x)]/Pi^2,{Q,0,1}];​​​​Plot[{R[x],Piecewise[{{1/(2*Sqrt[x]),0<x<1}},0]},{x,0,2},PlotStyle->{Red,{Dashed,Black}},PlotLabels->{"numeric","analytically"}]
Out[]=
Computing Inverse Stieltjes transform numerically:
In[]:=
N[Piecewise[{{1/(2*Sqrt[x]),0<x<1}},0]/.x->1/2,20]
Out[]=
0.70710678118654752440
In[]:=
g[s_]:=(Piecewise[{{ArcTanh[1/Sqrt[s]]/Sqrt[s],Re[s]>=1||Re[s]<=0||Im[s]>0||Im[s]<0}},ArcTanh[Sqrt[s]]/Sqrt[s]])
In[]:=
1/π^2*NIntegrate[1/(s-x)*g[s]/.x->1/2,{s,-Infinity,1/2,Infinity},Method->"PrincipalValue",WorkingPrecision->20]
Out[]=
0.70710678118654752494
Another example:
In[]:=
X=LaplaceTransform[MomentGeneratingFunction[NormalDistribution[μ,σ],t],t,s]
Out[]=
-
-
2
(s-μ)
2
2
σ

π
2
-
2
σ
+σErfi
-s+μ
2
σ

2
σ
In[]:=
X1=X/.μ->1/2/.σ->1/3
Out[]=
-9
-
9
2
2
-
1
2
+s

π
2

3
+
1
3
Erfi
3
1
2
-s
2

In[]:=
1/π^2*NIntegrate[1/(s-x)*X1/.x->1,{s,-Infinity,1,Infinity},Method->"PrincipalValue",WorkingPrecision->20]
Out[]=
0.38855278699767518268+0.71823717421850285373
and here’ s the puzzle, I don’ t know why I have to take the real part
In[]:=
Re[%]
Out[]=
0.38855278699767518268
In[]:=
N[PDF[NormalDistribution[μ,σ],x]/.μ->1/2/.σ->1/3/.x->1,20]
Update: If we don’t use the principal-value integral can be written explicitly:

Visualizing density given by its Stieltjes transformation

In response to the question here: https://mathematica.stackexchange.com/q/286914/27878
Using Numeric Inverse Stieltjes Transform from here: https://community.wolfram.com/groups/-/m/t/3302428
Method One
Method Two
Two Methods on one Plot
But that plot doesn’t look like @JimB answer!.
Update: From comments @Yaroslav Bulatov Harder example, to invert:
Let try:
Using:
Then:
Inverse Stieltjes Transform - method 1
Inverse Stieltjes Transform - method 2
Inverse Stieltjes Transform - method 3
Inverse Stieltjes Transform - method 4
Inverse Stieltjes Transform - method 5
Inverse Stieltjes Transform - method 6
Inverse Stieltjes Transform - method 7
Computing with Numerics:
Could say that the solution is distribution nature that’s why numeric’s methods fail.

CITE THIS NOTEBOOK

Inverse Stieltjes transform implementation​
by Mariusz Iwaniuk​
Wolfram Community, STAFF PICKS, October 18, 2024
​https://community.wolfram.com/groups/-/m/t/3302428