Limit that appears in particle physics calculation

Question

I came across the following statement, which I do not see how to prove it: $$\lim_{\epsilon \to 0} \frac{\epsilon}{x^2+\epsilon^2} =\pi \delta(x)\,.$$ Any ideas how to proceed?

Answer 1

As $\delta(x)$ is not a function you cannot do a simple algebraic manipulation. What you can do is to verify the properties of the delta "function". You can prove that for $x \neq 0$ your limit is $0$. You can also prove that $\int\frac{\epsilon\ dx}{x^2+\epsilon^2}=\arctan(\frac x\epsilon)+c,$ so the integral from $-\infty$ to $\infty$ is $\pi$, or the limit of the integral over any interval that crosses $0$ is $\pi$

Answer 2

The area under the function

$$\int_{-\infty}^\infty\frac{dx}{\pi(1+x^2)}=1$$

is unit.

Then by scaling,

$$\int_{-\infty}^\infty\frac{d\dfrac x\epsilon}{\pi\left(1+\dfrac{x^2}{\epsilon^2}\right)}=\int_{-\infty}^\infty\frac{\epsilon\,dx}{\pi\left(\epsilon^2+x^2\right)}.$$

As $\epsilon$ decreases, the peak gets narrower and narrower.