Dirac Delta as Gaussian functions

In summary, there is a discussion about expressing the delta Dirac function as a limiting case of the Gaussian function. The first expression provided is correct, but the second one is not. The second expression is not a delta function but an approximation of it, and the use of an integral does not result in a delta function, but a unit step function.
  • #1
rajetk
1
0
I am looking at a problem, part of which deals with expressing delta dirac as a limiting case of gaussian function. I am aware of the standard ways of doing it. In addition, I would also like to know if the following are correct -
[tex]
\delta(x-a) = \lim_{\sigma \rightarrow{0}} \int_{a - \sigma}^{{a + \sigma}} \sqrt{\frac{1}{2\pi \sigma^{2}}}e^{-((x-a)^{2})/(2\sigma^{2})} dx
[/tex]

Or can I say something like the following -
[tex]
\lim_{\sigma \rightarrow{0+}} \int_{a - \sigma}^{{a + \sigma}} \sqrt{\frac{1}{2\pi \sigma^{2}}}e^{-((x-a)^{2})/(2\sigma^{2})} dx
[/tex]

where the second expression is not a delta function but its approximation (since both the a+ and a- regions are considered in the integral)?

Please do excuse me if I am seriously wrong :(. Thanks in advance.
 
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  • #2
The first one is correct, I'm not sure what you're saying about the second one. It looks like the only difference is that sigma is approaching zero from the positive side
 
  • #3
Neither is correct. For one thing, where is the x on the right hand side? It is not the dummy variable of the integration; that you happened to use x is somewhat misleading. You could have called that dummy variable t, u, or anything without changing the value of the integral. It is just a dummy variable.

That missing x is just a tiny nitpick compared to the bigger problem of using an integral. In the limit of sigma going to zero, that integral will become a unit step function, not a Dirac delta distribution.

You just want the Gaussian, non-integrated.
 

1. What is the Dirac Delta function?

The Dirac Delta function, also known as the impulse function, is a mathematical function that is defined as zero everywhere except at a single point, where it is infinite. It is commonly denoted as δ(x) and is used in various areas of mathematics and physics.

2. How is the Dirac Delta function related to Gaussian functions?

The Dirac Delta function can be thought of as a limiting case of a Gaussian function, where the width of the Gaussian approaches zero and the amplitude approaches infinity at the center point. This can be expressed mathematically as the limit of a Gaussian function as the standard deviation approaches zero.

3. What is the significance of using Dirac Delta functions as Gaussian functions?

Using Dirac Delta functions as Gaussian functions allows for a more convenient and compact representation of certain mathematical and physical phenomena. It also helps in simplifying calculations and solving differential equations.

4. How is the Dirac Delta function represented mathematically?

The Dirac Delta function is represented mathematically as an integral of the Gaussian function over an infinitely small interval centered at the origin. In other words, it is the limit of a Gaussian function as the standard deviation approaches zero, multiplied by an infinitely small interval.

5. What are some applications of Dirac Delta functions as Gaussian functions?

Dirac Delta functions as Gaussian functions are used in various areas of science and engineering, such as signal processing, probability theory, and quantum mechanics. They are particularly useful in representing impulses, singularities, and point sources in these fields.

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