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One property of the Dirac Delta Distribution is $x \delta'(x) = -\delta(x)$ because of $\int x \delta'(x) f(x) dx = -\int \delta(x) (xf(x))' dx = -\int \delta(x) (f(x)+xf'(x)) dx = -\int \delta(x) f(x) dx = -f(0)$. This property is in the same vein as $x \delta(x) =0$.

I will now say that $x \delta'(x) = -\delta(x) \implies \delta'(x) = -\delta(x)/x$. My question is now, when is this a sensible statment? I would argue that we should assume $f(0) = 0$ to avoid divergence. This brings us to $\int \frac{\delta(x)}{x} f(x) dx = \lim_{x \rightarrow 0 } \frac{f(x)}{x} = f'(0) $, where the last equal arises from the limit definition of the derivative of $x=0$ for $f(x)=0$ or as well by L'Hospital, making it consistent.

Any ideas if this is ever useful or if this can be made more generally? This can't be a true equality, I would be also thankful for additional information for the distribution of $1/x$.

Thank you!

theta_phi
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  • Are you familiar with the construction of the Laplace transform via the field of convolution fractions? That might provide a helpful interpretation here. – user3716267 May 11 '24 at 21:25
  • I'm not yet familiar with convolution fractions but with Laplace transformation. Can you provide some insight how they are related and how this may help me? Maybe some further reading that you found well written... – theta_phi May 11 '24 at 22:03
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    You are correct. The distribution $\delta(x)/x$ is not defined. See THIS ANSWER – Mark Viola May 11 '24 at 23:25
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    Thanks for the answer, yes it helped. I also thought about the Taylor approximation which led me to posting this question. I feared that the 1/x problem was too big as I knew that 1/x leads to the Cauchy principal value, but I was hoping for a neat trick or something else. But sometimes a wrong manipulation of a formula is simply a wrong manipulation, leading to void statments and more contradictions then exciting math. – theta_phi May 11 '24 at 23:38
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    @theta_phi You're welcome. My pleasure. Sometimes we are tempted to carry out operations that naively appear elementary , but are not justifiable rigorously. – Mark Viola May 11 '24 at 23:43

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