# Two different results for the same integral

Calculating the principal value of the integral in version 13 on Windows 10, I obtain

a = Integrate[Exp[I*s]/(1 + s/(s^2 - 1)^2) - Exp[I*s], {s, -Infinity, Infinity},
PrincipalValue -> True]


2 I \[Pi] ((E^(I Root[1 + # - 2 #^2 + #^4& , 1, 0]) Root[ 1 + # - 2 #^2 + #^4& , 1, 0])/ (1 - 4 Root[1 + # - 2 #^2 + #^4& , 1, 0] + 4 Root[1 + # - 2 #^2 + #^4& , 1, 0]^3) + ( E^(I Root[1 + # - 2 #^2 + #^4& , 2, 0]) Root[ 1 + # - 2 #^2 + #^4& , 2, 0])/ ( 1 - 4 Root[1 + # - 2 #^2 + #^4& , 2, 0] + 4 Root[1 + # - 2 #^2 + #^4& , 2, 0]^3) + ( E^(I Root[1 + # - 2 #^2 + #^4& , 4, 0]) Root[ 1 + # - 2 #^2 + #^4& , 4, 0])/( 1 - 4 Root[1 + # - 2 #^2 + #^4& , 4, 0] + 4 Root[1 + # - 2 #^2 + #^4& , 4, 0]^3)

 N[a]


-1.414 + 0.192275 I

That integral can be treated as the value of the Fourier transform at the unit up to a multiplier Sqrt[2*Pi]:

b=FourierTransform[1/(1 + x/(x^2 - 1)^2) - 1, x, s] // ToRadicals;
N[Sqrt[2*Pi]*b /. s -> 1]


-0.998307 - 0.313707 I

I am sure both calculations apply that formula. Three terms in a confirm it in view of

sol = Solve[1 + s/(s^2 - 1)^2 == 0, s] // ToRadicals;N[sol]


{{s -> -1.49022}, {s -> -0.524889}, {s -> 1.00755 - 0.513116 I}, {s -> 1.00755 + 0.513116 I}}

At least one of the results a and b is not true. I cannot numerically verify these because

NIntegrate[Exp[I*s]/(1 + s/(s^2 - 1)^2) - Exp[I*s], {s, -Infinity,sol[[1, 1, 2]],
sol[[2, 1, 2]], Infinity}, WorkingPrecision -> 40, AccuracyGoal -> 4, PrecisionGoal -> 4]


produces -3.62952953475895858897998849318828745270 + 30.33384783506923620920956921799728322119 I and a warning "NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in s near {s} = {-1.49021612009995364811638684235445262003720224040232564447871916489713786748684335358978576}. NIntegrate obtained -3.62952953475895858897998849318828745269524061503385359582374023475829095533142179709562225+<<94>> I and 3.1259552347665095544965989530670535288591983446745504635397922702893866029082199783906738990. for the integral and error estimates."

In a simpler case, both

Integrate[Exp[I*x]/(1 - x^2), {x, -Infinity, Infinity}, PrincipalValue -> True]


and

FullSimplify[Sqrt[2*Pi]*FourierTransform[1/(1 - x^2), x, s] /. s -> 1]


result in \[Pi] Sin[1].

So the question arises: is a or b correct?

sol = Solve[1 + s/(s^2 - 1)^2 == 0, s] //ToRadicals;

-0.998307 - 0.313707 I
confirms b`.