Bug introduced in 11.1 or earlier and fixed in 11.3

On Mathematica the Mellin transform of $x^p$ is evaluated as $\delta(p+s)$, while I think it should be $2\pi\,\delta(p+s)$:

In:= MellinTransform[x^p, x, s, GenerateConditions -> True]  
Out:= DiracDelta[p + s]  

edited posting after Daniel Lichtblau's comment

I initially did not understand this result, but this 2004 paper has explained to me how to arrive at the Dirac delta function, however, with an additional factor of $2\pi$. I checked that this is not a matter of a different definition of the Mellin transform. (I summarized the calculation in this Mathoverflow posting.)

Missing factor $2\pi$ is fixed in Mathematica 11.3.0:

 In:= MellinTransform[x^p, x, s, GenerateConditions -> True]  
 Out:= 2π DiracDelta[i(p + s)]  

consequence: before 11.3 Integrate[MellinTransform[1, x, s], {s, -Infinity, Infinity}] returned 1, now it returns $2\pi\int_{-\infty}^\infty\delta(is)ds$
Q: is this v. 11.3 change in the implementation of MellinTransform documented somewhere?

  • 3
    $\begingroup$ See last example in documentation under Scope Elementary Functions. It should be noted that this is a generalization of the integral definition, not unlike the case for FourierTransform. $\endgroup$ – Daniel Lichtblau Jul 9 '17 at 15:27
  • $\begingroup$ thank you, Daniel, for the feedback, I understand things a bit better now and have edited my posting accordingly --- my problem has been reduced to a missing factor $2\pi$... $\endgroup$ – Carlo Beenakker Jul 9 '17 at 19:06
  • 2
    $\begingroup$ What specific definition is used is not particularly important so long as the MellinTransform and InverseMellinTransform are inverses of each other. Both x^p == InverseMellinTransform[ MellinTransform[x^p, x, s], s, x] and DiracDelta[p + s] == MellinTransform[ InverseMellinTransform[DiracDelta[p + s], s, x], x, s] evaluate to True $\endgroup$ – Bob Hanlon Jul 9 '17 at 22:41
  • $\begingroup$ @BobHanlon --- but if we assume that the factor of $2\pi$ is absorbed in the definition of DiracDelta, then Integrate[MellinTransform[1, x, s], {s, -Infinity, Infinity}] should return $2\pi$, while instead it returns 1. $\endgroup$ – Carlo Beenakker Jul 10 '17 at 6:18
  • $\begingroup$ The integral of DiracDelta should be one. $\endgroup$ – Bob Hanlon Jul 10 '17 at 14:48

The Mellin transforms for $x^j$ reported by Mathematica 11.2 didn't make sense to me, so on 11/28/2017 I submitted the following question on Math StackExchange.

Questions on Mellin Transform of $x^j$ and Interpretation of Distributions with Complex Arguments

I ended up deriving the answer to my own question and on 12/7/2017 I submitted a problem report to Wolfram technical support where I attached a Mathematica notebook illustrating the problem and the correct solution (CASE:3980660).

I received an email from Wolfram technical support on 12/13/2017 indicating my analysis was accepted as correct and a report was being filed with the developers. The correct solution was subsequently implemented in Mathematica 11.3.

Note that not only was the $2\,\pi$ prefix missing, but $i$ was also missing in the $\delta$ function parameter.

I subsequently posted the correct solution in answers to related questions on both Math StackExchange and MathOverflow StackExchange.

Delta function with imaginary argument

Dirac Delta function with a complex argument


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.