Derivative of Dirac Delta function

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The discussion focuses on evaluating the derivative of the Dirac delta function, specifically δ'(g(x)), in a distributional sense. Participants clarify that the integral of δ'(g(x)) with a test function φ(x) can be approached using integration by parts after a change of variables. It is noted that if g(x) is not injective, the evaluation will yield a sum of terms corresponding to each zero of g. Concerns are raised about the well-defined nature of the distribution when both g and its derivative g' are zero at any point. Overall, the conversation emphasizes the complexities involved in handling derivatives of the Dirac delta function in various contexts.
LedPhoton
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Hello I'm trying to figure out how to evaluate(in the distribution sense)
\delta'(g(x)). Where \delta(x) is the dirac delta function. Please notice that what I want to evaluate is not \frac{d}{dx}(\delta(g(x))) but the derivative of the delta function calculated in g(x).
If anyone could post a proof, an idea to find the proof or a link it would be greatly appreciated!
 
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\delta' is a linear operator on functions. Are you referring to its value when you pair it with g(x), as in
$$ \langle \delta',\, g\rangle = \int_{\mathbb{R}} \delta'(x)g(x)\, dx \, ?$$
I'll assume you are. In that case, you use the integral notation above and then symbolically do integration by parts. Don't worry if it is not a well-defined operation because the answer you get is literally the definition of what you want.

Look here under distributional derivatives for more info:
http://en.wikipedia.org/wiki/Dirac_delta_function
 
No I'm sorry if I wasn't clear. I understand the value of
\int \delta'(x)g(x)dx
I'm asking the value of
\int \delta'(g(x))\phi(x)dx
Where \phi(x) is the test function. Here it is not immediately obvious to me how to integrate by parts. I thought about this(but I am unsure of whether it is correct):
Assume that g(x) is an invertible function with as many derivatives as necessary(to keep things simple for now), so we substitute y = g(x) and get
\int \delta'(y)\frac{\phi(g^{-1}(y))}{g'(g^{-1}(y))} dy Now I could integrate by parts and get
-\int \delta(y)\frac{d}{dy}(\frac{\phi(g^{-1}(y))}{g'(g^{-1}(y))}) dy
Do you think my reasoning is correct up to here?
 
The dirac delta is just a normal distribution who's standard deviation approaches 0. Take the derivative of the normal dist. then take the limit as stdev =>0. I'm not sure if that's a valid way to do the problem, but its what I would try.
 
Ok thanks, I'll try that
 
Your calculation looks right to me, and what Aero said makes sense too. As far as proof goes, I can't remember exactly how general the rules are for changing variables like that. A book like Friedlander would probably have it...

Check out the very last post here for a similar problem:
https://www.physicsforums.com/showthread.php?t=201774&page=2
There is no proof, but there is a citation.

If g is not injective, then in the end, when you evaluate against a test function, you should get a sum of terms, one for each zero of g. If g' and g are simultaneously 0 at any point, then I don't think the distribution is well-defined.
 

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