Delta function of a function with multiple zeros

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SUMMARY

The discussion centers on the computation of delta functions for functions with multiple zeros, specifically addressing the integral of the form \int_{-\infty}^{\infty}dx g(x)\delta(x^2). The user seeks clarification on how to handle integrals involving \delta(f(x)) when f(x) has multiple zeros. The established formula \delta(f(x)) = \sum_{i=1}^{N}\frac{\delta(x-x_{i})}{|f'(x_{i})|} applies only to simple zeros, making it inadequate for this scenario. The user proposes a transformation for \delta(x^2-a^2) but notes that as a \rightarrow 0, the result diverges.

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maverick280857
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Hi everyone,

I was wondering how to deal with delta functions of functions that have double zeros.

For instance, how does one compute an integral of the form

[tex]\int_{-\infty}^{\infty}dx g(x)\delta(x^2)[/tex]

where g(x) is a well behaved continuous everywhere function?

In general how does one find

[tex]\int_{-\infty}^{\infty}dx g(x)\delta(f(x))[/tex]

where f(x) has a finite number of multiple zeros along with some simple zeros. I know that

[tex]\delta(f(x)) = \sum_{i=1}^{N}\frac{\delta(x-x_{i})}{|f'(x_{i})|}[/tex]

where [itex]x_{i}[/itex]'s (for i = 1 to N) are simple zeros of [itex]f(x)[/itex] and it is known that [itex]f(x)[/itex] has no zeros of multiplicitiy > 1.

but this is of course not valid here. Using this, however I could write

[tex]\delta(x^2-a^2) = \frac{1}{2|a|}\left(\delta(x-a) + \delta(x+a)\right)[/tex]

But the limit of this as [itex]a \rightarrow 0[/itex] tends to infinity.

Any ideas?

Thanks in advance.
Cheers.
 
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Maybe this link helps you:

http://ocw.mit.edu/NR/rdonlyres/Physics/8-07Fall-2005/40686CBE-369E-4373-95AD-BAEA1F2B8A37/0/deltafun.pdf
 
Last edited by a moderator:
Thank you haushofer.
 

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