How can I prove $\lim_{\epsilon \to 0} \space \text{Im}\frac{1}{x+i \epsilon}=-\pi\delta(x)$?

Question

I want to show that:

$$\lim_{\epsilon \to 0} \space \text{Im}\frac{1}{x+i \epsilon}=-\pi\delta(x)$$

This is my attempt:

1. I assumed that $\text{Im}$ stands for the imaginary part. Therefore, $$\frac{1}{x+i \epsilon} \cdot \frac{x-i \epsilon}{x-i\epsilon}=\frac{x-i\epsilon}{x^2+\epsilon^2}=\frac{x}{x^2+\epsilon^2}-i \left (\frac{ \epsilon}{x^2+\epsilon^2} \right ) \implies\Im(z)=-\frac{1}{\frac{x^2}{\epsilon}+\epsilon}$$
2. Taking the limit: $$\lim_{\epsilon \to 0} -\frac{1}{\frac{x^2}{\epsilon}+\epsilon}=0$$

But this would mean that $$\implies0=-\pi \delta(x) $$

However, this equality can't be true because:

$$\delta(x)=\begin{cases}\infty, \space \space\space x=0 \\ 0, \space \space \space x \not= 0 \end{cases}$$

which would mean that at $x=0$ the equality $0=-\pi \delta(x)$ is not true. What am doing wrong?

Answer 1

In THIS ANSWER and THIS ONE, I provide primers on the Dirac Delta.

For any smooth test function $\phi$, we have for any $\nu >0$, there exists a $\delta>0$ such that $|\phi(x)-\phi(0)|<\nu$ whenever $|x|<\delta$. Then, we can write

$$\begin{align} \lim_{\epsilon \to 0}\int_{-\infty}^\infty \phi(x)\text{Im}\left(\frac{1}{x+i\epsilon}\right)\,dx&=-\lim_{\epsilon \to 0}\int_{-\infty}^\infty \phi(x)\left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx\\\\ &=-\lim_{\epsilon \to 0}\int_{|x|\ge \delta} \phi(x)\left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx-\lim_{\epsilon \to 0}\int_{-\delta}^\delta \phi(x)\left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx\\\\ &-\lim_{\epsilon \to 0}\int_{-\delta}^\delta \phi(x)\left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx\\\\ &=-\lim_{\epsilon \to 0}\left(\int_{-\delta}^\delta (\phi(x)-\phi(0))\left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx-\phi(0)\int_{-\delta}^\delta \left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx\right)\\\\ &=-\pi \phi(0)-\lim_{\epsilon \to 0}\int_{-\delta}^\delta (\phi(x)-\phi(0))\left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx\\\\ \end{align}$$

Note that we have for any $\nu>0$

$$\left|\int_{-\delta}^\delta (\phi(x)-\phi(0))\left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx\right|\le 2\nu \arctan(\delta/\epsilon)\le \pi \nu $$

Therefore, for any smooth tests function $\phi$, we find

$$\lim_{\epsilon \to 0}\int_{-\infty}^\infty \phi(x)\text{Im}\left(\frac{1}{x+i\epsilon}\right)\,dx=-\pi \phi(0)$$

This is equivalent to the statement that

$$\lim_{\epsilon \to 0}\text{Im}\left(\frac{1}{x+i\epsilon}\right)\sim -\pi \delta(x)$$

in the sense of a regularization of the Dirac Delta.