Can the Dirac Distribution be Proven Using Integrals?

[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?

 

[arildno]

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

 

[naima]

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

 

[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.

 

[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!

 

[George Jones]

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!

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:

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