What is the interpretation of $\delta(x)\ln\delta(x)$, where $\delta(x)$ is the Dirac's delta function?

Question

What's the result of the following integral?

$$\int f(x)\delta(x)\ln\delta(x)\mathrm{d}x$$

where $f(x)$ is a smooth function (continuous derivatives of as high order as needed).

Answer 1

The simplest approximation to $\delta$ (located at $0$ on the real line) is the sequence $$\phi_n(x) = \begin{cases} n/2 ,\quad & |x|\le 1/n \\ 0 \quad & |x|>1/n\end{cases}$$ It is reasonable to interpret $\phi_n \ln \phi_n $ as $0$ when $\phi_n=0$, since linear decay beat logarithm. Hence,
$$ \int_{\mathbb R} f (x) \phi_n(x)\ln \phi_n(x)\,dx = \frac{n}{2} \ln\frac{n}{2} \int_{-1/n}^{1/n} f(x)\,dx $$ Since $f(x)-f(0)=O(x)$, it follows that $$ \frac{n}{2}\int_{-1/n}^{1/n} f(x)\,dx = f(0) + \frac{n}{2}\int_{-1/n}^{1/n} (f(x)-f(0))\,dx = f(0)+O(1/n) $$ Thus, $$ \int_{\mathbb R} f (x) \phi_n(x)\ln \phi_n(x)\,dx = f(0) \ln\frac{n}{2} + O\left(\frac{\ln n}{n}\right)$$

As $n\to \infty$, the limit is infinite if $f(0)\ne 0$, and zero otherwise. This is not a continuous functional on the space of smooth functions; in other words, not a distribution.