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$$ \int_x \delta(x)~\ln(\delta(d))~dx = 0 ? $$

Where $\delta(x)$ denotes the Dirac-Delta function, $ln(\cdot)$ is a logarithm, and $dx$ is simply the diferential of $x$ for the integral.

I'm working on an applied probability problem whose behavior can be neatly explained if I manage to prove the above equality. To be honest, it appears that the integral is equal to zero, but I can't really be sure.

Somehow it makes sense to me: $\delta$ is zero everywhere but at $x=0$, hence the first $\delta$ in the integral should become equal to $1$, and the other should lead to a logarithm of 1, which is zero (i.e. $\int \ln(\delta) = \log(1) = 0$); which in the end should lead to $1\cdot 0 = 0$.

However, I completely fail to come up with a formal explanation. I'm probably missing some important step or property when working it out on a paper. Or worst case, the integral does not even evaluate to 0. How should I tackle the problem?

P.S.: Feel free to change the tags of my question, I might have gotten them wrong.

andresgongora
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    What is $d$? in your notation? Did you mean $\ln(\delta(x))$? – quarague Feb 15 '19 at 11:47
  • And in every cases it depends on your definitions. With $f(\delta(.)) \overset{def}= \lim_{n \to \infty} f(n e^{-\pi n^2 .^2})$ then $\int_{-\infty}^\infty \delta(x) \ln(\delta(x))dx\overset{def}= \lim_{n \to \infty}\int_{-\infty}^\infty n e^{-\pi n^2 x^2} \ln(n e^{-\pi n^2 x^2}) dx$ $=\lim_{n \to \infty}\int_{-\infty}^\infty e^{-\pi y^2} (\ln( n)-\pi y^2) dy= \infty$ – reuns Feb 15 '19 at 12:02
  • @quarague Indeed. I just changed the question to be clearer – andresgongora Feb 15 '19 at 12:17
  • @reuns I want to adhere as strictly as possible to the definition of $\delta$. That is, $\delta(x)=0$ everywhere but for $x=0$, and $\int \delta(x)=1$ – andresgongora Feb 15 '19 at 12:19
  • You didn't define what you need. Do you see why the definition I proposed makes sense to $\int_a^b f(\delta(x))g(x)dx$ as a (possibly divergent) limit for any $f,g$ ? Then the point is to find for which $f,g$ the result doesn't depend on the chosen definition. – reuns Feb 15 '19 at 12:50

1 Answers1

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Use per partes, the definition of Dirac delta function and the fundamental theorem of calculus as follows:

$$ u = \ln\delta(x)\quad \Rightarrow\quad du = \frac{d}{dx}\ln\delta(x) \\ dv = \delta(x)dx\quad \Rightarrow\quad v = \int\delta(x)dx = 1 \\ \int\delta(x)\ln\delta(x)dx = uv - \int vdu = \ln\delta(x) - \int\frac{d}{dx}\ln\delta(x)dx = \ln\delta(x) - \ln\delta(x) = 0 $$

Nejc Kejzar
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    I couldn't see anything wrong with this derivation, and the integration by parts is relatively straightforward to follow, yet it seems to contradict more complicated answers for similar questions that imply the entropy of the Dirac delta function is $-\infty$. For example, see questions here and here. Can anyone comment of the disagreement in results? – Ross Allen Jan 28 '21 at 01:45
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    In response to my previous comment, the only problem I could see is that the derivative of the delta function is undefined so the term $du$ is undefined – Ross Allen Jan 28 '21 at 02:17