My professor claims: $$\delta(\frac{r^2-a^2}{(r^2+a^2)^3})=4a^5\delta(r-a)$$ where $a>0$ is a constant. I completely don't see this, help?
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I think sth is missing in the equation in the question, see my answer – Physor Sep 13 '20 at 20:45
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Presumably $r>0$. – Qmechanic Sep 13 '20 at 22:39
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@Qmechanic oh yes, I forgot to write, this is in the spherical coordinate system, so r>=0 – fazan Sep 14 '20 at 01:28
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According to my simple knowledge in distributions $$ \delta(f(x)) = \sum_i\frac{1}{|f'(x_i)|}\delta(x-x_i) $$ where {$x_i$ } are simple zeros of $f(x)$. So that means $$ \delta(\frac{r^2-a^2}{(r^2+a^2)^3}) = 4a^5(\delta(x-a)+\delta(x+a)) $$ where $$ f'(x)|_{x=a} = \left(\frac{2 x}{\left(a^2+x^2\right)^3}-\frac{6 x \left(x^2-a^2\right)}{\left(a^2+x^2\right)^4}\right)\Bigg|_{x=a} = \frac{1}{4a^5} $$ and $$ f'(x)|_{x=-a} = \left(\frac{2 x}{\left(a^2+x^2\right)^3}-\frac{6 x \left(x^2-a^2\right)}{\left(a^2+x^2\right)^4}\right)\Bigg|_{x=-a} = -\frac{1}{4a^5} $$
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