Calculate integral with dirac delta

Question

I want to show that $$\lim_{\epsilon \rightarrow 0} \left( \int_{\epsilon}^{2\epsilon} \frac{f(x)}{x} dx \right) = \log(2) \delta_0$$ with $\delta_0: f \rightarrow f(0)$. My thought was to integrate by parts so that $$\lim_{\epsilon \rightarrow 0} \left( \int_{\epsilon}^{2\epsilon} \frac{f(x)}{x} dx \right) = \lim_{\epsilon \rightarrow 0} \left( f(2\epsilon)\log(2\epsilon)-f(\epsilon)\log(\epsilon)-\int_{\epsilon}^{2\epsilon} f'(x) \log(x) dx \right)$$ Would that integral then become 0? I know that $\log(2\epsilon)-\log(\epsilon) = \log(2)$, but I'm unsure how the dirac delta might come into it.

Answer 1

Follow the hint given in the comments, let $x=\epsilon u$ to get:

$$\lim_{\epsilon\to 0} \int_1^2 \frac{f(\epsilon u)}{u} du$$

Since $f\in C_c^{\infty}(\Bbb{R})$ (or $S$. I'm assuming $f$ is well behaved because you're discussing distributions), we may use dominated converge theorem (I'll leave it to you to fill out the details. Why can we use this?) to get:

$$\int_1^2 \lim_{\epsilon \to 0} \frac{f(\epsilon u)}{u} du$$

Since $f$ is continuous we may move the limit inside to get:

$$\int_1^2 \frac{f(0)}{u} du=f(0)(\log(2)-\log(1))=\log(2)f(0)=\log(2)\delta_0(f)$$