"With hands", what is the reason why $\pi$ appears in this result :
$\lim_{\epsilon\to 0}\frac{\epsilon}{\epsilon^2+x^2}=\pi\delta(x)$
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7Because $$\int_{-\infty}^\infty \frac{dx}{1+x^2}=\pi.$$ – Giuseppe Negro May 19 '20 at 14:42
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1Or this one: What's the deal with this $\frac1\pi$? – Martin R May 19 '20 at 14:42
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Thank you @Martin R. The link is very technical and does not give a justification "with hands" : it just show the mathemaical explanation : so $\pi$ appears from nowhere, from mathematics. The answer from Dldier_ explains the principle. Thanks – Mathieu Krisztian May 19 '20 at 14:49
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Er...no it doesn't, it is a very vague nod to the first link above ( and the second comment does a better job too). The answer you like below has the $\pi$ appearing from nowhere. Just sayin' – Paul May 19 '20 at 15:57
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Because the left hand side is the derivative of $\text{arctan}$ (by rescaling), so it has to do with trigonometry and $\pi$.
Alapan Das
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Didier
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Suppose you know that $$ \frac{\epsilon}{\epsilon^2+x^2}\to C\delta, $$ and you want to compute $C$. Taking integrals termwise, $$ \int_{-\infty}^\infty \frac{\epsilon}{\epsilon^2+x^2}\, dx\to C, $$ and the change of variable $y=\epsilon x$ shows that the left-hand side equals $$ \int_{-\infty}^\infty \frac{dy}{1+y^2}=\pi.$$
Giuseppe Negro
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