# Dirac distribution

1. Jun 3, 2006

### naima

bonjour from france,

I thought that sum of dirac(x - xi)/g'(xi), where the xi verify g(xi) = 0, was a definition for dirac(g(x)). It was proposed, as an exercise, to prove the equality of the 2 terms.
can one help me
thanks

ps : can i write this in latex?

2. Jun 3, 2006

### arildno

Yes, you may write this in LATEX. Please do so.

3. Jun 3, 2006

### naima

thanks for your help but it was only the second question.

4. Jun 3, 2006

### StatusX

The delta function is defined by its action under an integral. Try integrating both sides and verify that they behave the same in an integral expression.

5. Jun 4, 2006

### naima

Yes, you are true but if the equality I try to prove is not a definition, I do not know what means the first term. In this case I cannot integrate it with a function to compare!

6. Jun 4, 2006

### George Jones

Staff Emeritus
I think StatusX means that what you want to show is true if and only if is true "under an integral", i.e., if

$$\int^{\infty}_{-\infty} f(x) \delta \left( g \left( x \right) \right) dx = \int^{\infty}_{-\infty} f(x) \sum_{i} \frac{\delta \left( x - x_{i} \right)}{\left| g' \left( x_{i} \right) \right|} dx,$$

where $g \left( x_{i} \right) = 0$ for each $x_i$, and $f$ is an arbitrary test function.

Note that I've inserted an absolute value and a summation.

Hint:

$$dx = \frac{1}{\frac{dg}{dx}} dg.$$

Last edited: Jun 4, 2006