- #1

Pacopag

- 197

- 4

## Homework Statement

I would like to prove that [tex]{d \over {dx}} \sigma(x)=2\delta(x)[/tex],

where [tex]\sigma(x>0)=1[/tex]

[tex]\sigma(x=0)=0[/tex]

[tex]\sigma(x<0)=-1[/tex]

## Homework Equations

## The Attempt at a Solution

I think that I need to show that

[tex]\int_{-\infty}^{\infty}{d \over {dx}} \sigma(x)dx=2=\int_{-\infty}^{\infty}2 \delta(x)dx[/tex].

The integral looks rather harmless, and I would like to write

[tex]\int_{-\infty}^{\infty}{d \over {dx}} \sigma(x)dx=\sigma(x) |^{\infty}_{-\infty}=1-(-1)=2[/tex].

This looks like it works, but it seems to me that there should be some complication when integrating close to zero where we have discontiuous behaviour.