My conjecture is $\ln(\delta(x))=-\ln(x)\delta(x)-\gamma$, but I am not sure.
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can have a look at this https://math.stackexchange.com/q/625966/399263, by itself probably undefined since -infinite value almost everywhere, and from this post $\delta\ln(\delta)$ is slightly better but not much. – zwim Oct 24 '22 at 12:24
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How do you define $\ln \delta$? – LL 3.14 Oct 24 '22 at 14:15
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The Dirac Delta is NOT a function. So what is your definition of $\log(\delta(x))$? – Mark Viola Oct 24 '22 at 14:15
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You could use this Wolfram functions point to get $\ln(\delta(x))=-\infty,x\ne0$ and $\ln(\delta(0))=\infty$, but there are multiple definitions of $\delta(x)$ – Тyma Gaidash Oct 25 '22 at 12:16
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This cannot be defined as a distribution. If you consider the logarithmic function $\ln$ as a distribution, the quantity: $\ln \circ g$ can be defined as a distribution only if $g$ is invertible and both $g$ and its inverse are smooth functions. This is not the case for $g=\delta$. See here for why we need $g$ to be smooth. The difficulty stems from the fact that the composition operator is a nonlinear operator and nonlinear operations are hard to handle in the standard theory of distributions.
Saleh
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