Message Boards Message Boards

Question on commutativity of Mellin convolution evaluations

GROUPS:

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?

POSTED BY: Steven Clark
Answer
2 months ago

Group Abstract Group Abstract