Sifting property of a Dirac delta inverse Mellin transformation

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SUMMARY

The discussion focuses on verifying the sifting property of the inverse Mellin transformation of the Dirac delta function, specifically the integral ##\frac{1}{2\pi i} \int_{-i\infty}^{i\infty} e^{-sa}e^{st} ds##. Participants analyze the relationship between this integral and the property ##\int_{-\infty}^{\infty} f(t) \delta(t-a) dt = f(a)##, expressing confusion over the presence of the ##\frac{1}{2\pi}## factor and the implications of integrating with respect to ##t##. The conversation emphasizes the need to demonstrate that the Mellin transform behaves like the delta function, ultimately leading to the conclusion that further clarification on the integration process is necessary.

PREREQUISITES
  • Understanding of inverse Mellin transformations
  • Familiarity with the Dirac delta function and its properties
  • Knowledge of complex integration techniques
  • Ability to manipulate integrals involving exponential functions
NEXT STEPS
  • Study the properties of the Dirac delta function in the context of Fourier and Mellin transforms
  • Learn about complex analysis techniques relevant to evaluating integrals
  • Explore the derivation of the sifting property for various integral transforms
  • Investigate the implications of the ##\frac{1}{2\pi}## factor in Mellin transformations
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Mathematicians, physicists, and engineers working with signal processing, integral transforms, and complex analysis who need to understand the sifting property of the Dirac delta function in the context of Mellin transformations.

  • #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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