Sifting property of a Dirac delta inverse Mellin transformation

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Homework Help Overview

The discussion revolves around verifying the sifting property of the inverse Mellin transformation of the Dirac delta function, specifically the expression ##\frac{1}{2\pi i} \int_{-i\infty}^{i\infty} e^{-sa}e^{st} ds##. Participants are exploring the implications of this transformation in relation to the sifting property, which states that ##\int_{-\infty}^{\infty} f(t) \delta(t-a) dt = f(a)##.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to manipulate the integral to demonstrate that it behaves like the Dirac delta function. There are questions about the presence of terms like ##\frac{1}{2\pi}## and how to handle the limits of integration. Some participants suggest substituting the delta function with the Mellin transform and checking the integration results.

Discussion Status

The discussion is ongoing, with participants expressing confusion about the steps needed to verify the sifting property. Some have proposed methods to replace the delta function with the Mellin transform but are struggling to reconcile their results with the expected outcome of obtaining ##f(a)##. There is no explicit consensus yet, but several participants are actively engaging with the problem and offering guidance.

Contextual Notes

Participants note discrepancies between their findings and textbook definitions, particularly regarding the limits of integration and the behavior of the Mellin transform at specific points. There is a shared uncertainty about how to properly apply the sifting property in this context.

  • #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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