How to prove this Dirac Delta Function property?

Question

How to prove the equation below, using Dirac Delta function properties? $$ \delta(x^2-m^2)=\frac{1}{2|w|}(\delta(x-w)+\delta(x+w)) $$ where $$ w^2=|x|^2+m^2 $$ I tried to show it using $$ \delta(f(x))=\sum\frac{\delta(x-x_i)}{|f'(x_i)|} $$ but could not success.

Answer 1

I think I got it. As Emilio mentioned I will start with the definition. $$ \delta(f(x))=\sum_{i}\frac{\delta(x-x_i)}{|f'(x_i)|} $$ $$ \delta(x^2-m^2)=\frac{\delta(x-m)+\delta(x+m)}{2|m|} $$ with the transformation $w^2=x^2+m^2$ $$ \delta(x^2-m^2)=\frac{1}{2|m|}(\delta(x-\sqrt {w^2-x^2})+\delta(x+\sqrt{w^2-x^2})) $$ Let me derive the first and second term, using similar definition $$ \delta(f(x))=\frac{\delta(x-x_0)}{|f'(x_0)|} $$ $$ \delta(x-\sqrt {w^2-x^2})+\delta(x+\sqrt{w^2-x^2})= \frac{\delta(x-w)}{\frac{2|w|}{2|m|}}+\frac{\delta(x+w)}{\frac{2|w|}{2|m|}} $$ and eventually come to the end $$ \delta(x^2-m^2)=\frac{1}{2|w|}(\delta(x-w)+\delta(x+m)) $$

Answer 2

The starting point is the definition: $$ \delta(f(x))=\sum_i\dfrac{\delta(x-x_i)}{|f'(x_i)|} $$ where $x_i$ are all the roots of $f(x)$ ( the formula in OP is not correct)

For $f(x)=x^2-y^2$ the roots are $x_1=-y$ and $x_2=y$ and $f'(x)=2x$, so we have:

$$ \delta(x^2-y^2)=\dfrac{\delta(x-y)}{2|+y|}+\dfrac{\delta(x+y)}{2|-y|}=\dfrac{\delta(x-y)+\delta(x+y)}{2|y|} $$