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Consider the expression

expr = Gamma[1 + s]/Gamma[1 - s] Gamma[-s]^2;

and a slightly simplified version of the same

expr2 = expr//FullSimplify

enter image description here

If we take the InverseMellinTransform of the first shape of the expression expr, we get

InverseMellinTransform[expr, s, x]

Log[1 + x]

On the other hand, if we take the InverseMellinTransform of the simplified form expr2, we get

InverseMellinTransform[expr2, s, x]

Log[1 + 1/x]

Evidently, the two results do not agree. How to make sense of this? I assume, this is not correct behavior? Is there some way to do the transform more carefully in Mathematica, such that both calculations produce the same result consistently?

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    $\begingroup$ at least Mathematica is consistent. It gives back the same result when transforming back :) screen shot !Mathematica graphics may be there is some deep mathematical reason for this. $\endgroup$
    – Nasser
    Apr 28 '20 at 23:44
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expr = Gamma[1 + s]/Gamma[1 - s] Gamma[-s]^2;

imt1 = InverseMellinTransform[expr, s, x, GenerateConditions -> True]

enter image description here

MellinTransform[imt1 // Normal, x, s, GenerateConditions -> True]

enter image description here

Simplifying expr

expr2 = expr // FullSimplify

(* (π Csc[π s])/s *)

imt2 = InverseMellinTransform[expr2, s, x, GenerateConditions -> True]

enter image description here

MellinTransform[imt2 // Normal, x, s, GenerateConditions -> True]

enter image description here

The inverse transforms are for different regions.

InverseMellinTransform[expr, s, x,
  Assumptions -> 0 < Re[s] < 1] // 
   FullSimplify[#, 0 < x < 1] &

(* Log[1 + 1/x] *)
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