Dirac delta definition with $i \epsilon$.

Question

I am asked to prove that: $$\delta(x) = \frac{1}{\pi} \Im \bigg(\frac{1}{x-i\epsilon}\bigg)$$ defines a Dirac delta function.

For one, $\Im\big(\frac{1}{x-i \epsilon}\big) = \frac{\epsilon}{x^2+\epsilon^2}$ and integrating over $x$ gives $\arctan(\frac{x}{\epsilon})$ so, for all $\epsilon >0$ we come up with $1$.

I'm left with showing $$\lim\limits_{\epsilon\to 0}\int\frac{\epsilon}{x^2 + \epsilon^2} f(x) dx = f(0)$$

I tried a Taylor expansion of $\frac{\epsilon}{x^2 + \epsilon^2}$ around $\epsilon \approx 0$ but that didn't simply the problem for me much. Can someone show me the proof?

Answer 1

The change of variables $x=\epsilon t$ shows that $$\int_{\mathbb{R}}\frac\epsilon{x^2+\epsilon^2}f(x)dx=\int_{\mathbb{R}}\frac1{t^2+1}f(\epsilon t)dt$$ Now $\lim\limits_{\epsilon\to0}f(\epsilon t)=f(0)$ and by the dominated convergence theorem $$\lim\limits_{\epsilon\to0}\int_{\mathbb{R}}\frac1{t^2+1}f(\epsilon t)dt= \int_{\mathbb{R}}\frac1{t^2+1}f(0)dt=\pi f(0)$$

Answer 2

Hint: For any $t\gt0$, what is $$ \lim_{\epsilon\to0}\frac1\pi\int_{-t}^t\frac{\epsilon}{x^2+\epsilon^2}\,\mathrm{d}x $$ Next, find a $t\gt0$ so that for $|x|\le t$, $|f(x)-f(0)|$ is small.