# A question about Dirac Delta Function

1. Sep 9, 2014

### sinaphysics

For proving this equation:

$$\delta (g(x)) = \sum _{ a,\\ g(a)=0,\\ { g }^{ ' }(a)\neq 0 }^{ }{ \frac { \delta (x-a) }{ \left| { g }^{ ' }(a) \right| } }$$

We suppose that
$$g(x)\approx g(a) + (x-a)g^{'}(a)$$

Why for Taylor Expansion we just keep two first case and neglect others? Are those expressions so small? if yes how we can explain it?

2. Sep 9, 2014

### PaulDirac

OK. Let me explain it to you. We start from decomposing the integral

$\int_{+\infty}^{-\infty} f(x)\ \delta(g(x))\,dx = \sum_{a} \int_{a + ε}^{a - ε} f(x)\ \delta((x - a) g^{'}(a))$

into a sum of integrals over small intervals containing the zeros of g(x). In these intervals, since x is supposed to be very near to the a we can approximate g(x) as g(a) + (x - a)$g^{'}(a)$ (note also that it is just the definition of derivative of g(x) when x goes toward a). Now we have proved it if we employ the equation $\delta(αx) = 1/α\ \delta(x)$ on the right-hand side of the integral.

Last edited: Sep 9, 2014