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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?

cmk
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hal
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  • What $\delta (x)$ representing here ? – Rishi Sep 30 '19 at 18:51
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    @Rishi It's got to be Dirac's $\delta$-function. (Which is not a function, but a distribution, so the mathematically sound proof would need to come from the theory of distributions: https://en.wikipedia.org/wiki/Distribution_(mathematics) ) –  Sep 30 '19 at 18:53
  • @hal It would be helpful if you added a bit more context here. Where is this statement appearing (in particular, what's immediately before it)? What mathematical background do you have (this can be helpful in terms of where to pitch an explanation; for example, do you know much about the Dirac function)? – Kevin P. Costello Sep 30 '19 at 19:12

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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 $\pi$

Ross Millikan
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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.