Sifting property of a Dirac delta inverse Mellin transformation

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The discussion revolves around verifying the sifting property of the inverse Mellin transformation of the Dirac delta function, specifically the integral expression involving the exponential function. Participants express confusion about how to manipulate the integral to demonstrate that it behaves like the delta function, particularly in relation to the sifting property, which states that integrating a function multiplied by the delta function yields the value of the function at the delta's argument. There is a debate over the correct approach to substitute the delta function with the Mellin transform and how to handle the resulting integrals, especially concerning the appearance of terms like 2π and potential divisions by zero. Ultimately, the complexity of the integral and the participants' uncertainty about the steps needed to prove the sifting property highlight the challenges in this mathematical verification. The conversation emphasizes the need for clarity in handling the integral transformations and their implications.
  • #31
EpselonZero said:
Since I have to verify that, I don't think I can use that right away. I see what you meant now.
However, if this is the thing I have to show it makes no sense to use it.
Exactly what are you supposed to verify?
Please state it once more.
 
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  • #32
EpselonZero said:
Since I have to verify that, I don't think I can use that right away. I see what you meant now.
However, if this is the thing I have to show it makes no sense to use it.
I wouldn't say that I'm using the result I'm trying to show.
It's true that the two integrals follow from each other by a change of variable, but I guess that's the point of the exercise. I agree it's not a very good exercise.

I would also question that the integral you quote as the inverse Mellin transform of the delta function is actually that. This integral is clearly just another way of writing the integral representation of the delta function.
So to me it seems this integral is the delta function and not its Mellin transform.
In short the point of the exercise seems to be to show that the integral in question is the delta function and therefore it has the sifting property.
Unless, of course, you're supposed to verify the sifting property of the delta function, but that's a completely different problem and not what you stated in your original post.
 

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