# Question on commutativity of Mellin convolution evaluations

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 For the function f[x] defined in [1] below, evaluation [5] below is inconsistent with evaluations [2] to [4] below. Note the change in sign preceding y in evaluation [5] below. In[1]:= f[x_] := DiracDelta'[x - 1] In[2]:= MellinConvolve[f[x], h[x], x, y, GenerateConditions -> True] Out[2]= h[y] + y Derivative[1][h][y] In[3]:= MellinConvolve[h[x], f[x], x, y, GenerateConditions -> True] Out[3]= h[y] + y Derivative[1][h][y] In[4]:= Integrate[f[x] h[y/x] 1/x, {x, 0, \[Infinity]}, GenerateConditions -> True] Out[4]= h[y] + y Derivative[1][h][y] In[5]:= Assuming[y > 0, FullSimplify[ Integrate[h[x] f[y/x] 1/x, {x, 0, \[Infinity]}, GenerateConditions -> True]]] Out[5]= h[y] - y Derivative[1][h][y] Question 1: Does evaluation [5] above represent an error in the implementation of the Mathematica Integrate function?For the function g[x] defined in [6] below, evaluations [7] and [8] below are inconsistent with evaluations [9] and [10] below. Note Mellin convolution is claimed to be commutative. In[6]:= g[x_] := x DiracDelta'[x - 1] In[7]:= MellinConvolve[g[x], h[x], x, y, GenerateConditions -> True] Out[7]= y Derivative[1][h][y] In[8]:= Integrate[g[x] h[y/x] 1/x, {x, 0, \[Infinity]}, GenerateConditions -> True] Out[8]= y Derivative[1][h][y] In[9]:= MellinConvolve[h[x], g[x], x, y, GenerateConditions -> True] Out[9]= 2 h[y] - y Derivative[1][h][y] In[10]:= Assuming[y > 0, FullSimplify[ Integrate[h[x] g[y/x] 1/x, {x, 0, \[Infinity]}, GenerateConditions -> True]]] Out[10]= 2 h[y] - y Derivative[1][h][y] Question 2: Do the discrepancies between evaluations [7] to [10] above represent errors in the Mathematica MellinConvolve and Integrate functions or is Mellin convolution not always commutative with respect to evaluations involving distributions?