# Mellin transform of $x^p$ seems to miss a factor of $2\pi$

Bug introduced in 11.1 or earlier and fixed in 11.3

On Mathematica 11.1.1.0 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?

• 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. – Daniel Lichtblau Jul 9 '17 at 15:27
• 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 $2\pi$... – Carlo Beenakker Jul 9 '17 at 19:06
• 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 – Bob Hanlon Jul 9 '17 at 22:41
• @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. – Carlo Beenakker Jul 10 '17 at 6:18
• The integral of DiracDelta should be one. – 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