Dirac Delta as Gaussian functions

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The discussion focuses on expressing the Dirac delta function as a limiting case of Gaussian functions. The first proposed expression, involving the limit as sigma approaches zero, is deemed correct, while the second expression is criticized for not accurately representing the delta function. The main issue highlighted is the misuse of the integration variable, which should not affect the integral's value. Additionally, it is pointed out that the integral approaches a unit step function rather than a Dirac delta distribution as sigma goes to zero. The consensus emphasizes the need for a non-integrated Gaussian to accurately represent the delta function.
rajetk
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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 -
<br /> \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<br />

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

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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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
 
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.
 

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