Limit of Lorentz distributions

1. Mar 15, 2010

Niles

Hi guys

Can one prove the identity

$$\frac{\epsilon}{x^2 + \epsilon^2} \underset{\epsilon\to 0^+}{\to} \pi \delta(x)$$

or is it just intuitively clear (by looking at a graph)?

2. Mar 15, 2010

Tinyboss

First show that $$\int_{-\infty}^\infty\frac{\epsilon}{x^2+\epsilon^2}dx=\pi$$ for any $$\epsilon>0$$. Then show that for any fixed $$x\ne0$$, $$\lim_{\epsilon\to0^+}\frac{\epsilon}{x^2+\epsilon^2}=0$$.

3. Mar 15, 2010

Niles

4. Mar 15, 2010

Mute

Alternatively, consider

$$\lim_{\epsilon \rightarrow 0^+} \int_{-\infty}^{\infty} dx~\frac{\epsilon}{x^2 + \epsilon^2} f(x)$$
and consider the change of variables $y = x/\epsilon$, and show that the result is $\pi f(0)$. It is in this sense that

$$\frac{\epsilon}{x^2 + \epsilon^2} \underset{\epsilon\to 0^+}{\to} \pi \delta(x).$$