# Starting with the definition of the Dirac delta function,

1. Mar 4, 2009

### skrtic

1. The problem statement, all variables and given/known data

Starting with the definition of the Dirac delta function, show that $\delta( \sqrt{x})$

um... i have looked in my book and looked online for a problem like this and i really have no clue where to start. the only time i have used the dirac delta function is in an integral with another function and never with it in this form, only like delta(x-a).

looking for guidance.

2. Mar 4, 2009

### Staff: Mentor

Show that $\delta( \sqrt{x})$ is what?

3. Mar 4, 2009

equal to 0

4. Mar 4, 2009

5. Mar 4, 2009

### skrtic

from what i get out of the text. it is what replaces an inner product that vanishes if x doesn't equal x'.

I also know that it's integral is unity.

6. Mar 4, 2009

### Avodyne

$$\int_{-\infty}^{+\infty}dx\,f(x)\delta(x)=f(0)$$

Now, if only the argument of the delta function was $\sqrt{z}$ instead of $x$ ... hmmm, how could we make this happen?

7. Mar 4, 2009

### skrtic

let x= $\sqrt{z}$ then dx=(1/2)z^(-1/2)dz

and we get $\int_{-\infty}^{\infty}dz z^{-1/2} f(z^{-1/2})\delta(z^{-1/2})$

8. Mar 4, 2009

### skrtic

is this a valid assumption

treat $\delta(\sqrt{x})$ as $\delta(\sqrt{x}-0)$

9. Mar 4, 2009

### Hurkyl

Staff Emeritus
Aside: I know the hand-wavy argument is clear, but does anyone know of a source that actually defines the result of composing a distribution with a function of some sort?

10. Mar 5, 2009

### skrtic

so assuming i made the substitution right and that was what i was supposed to do, i still dont see how i proved that it equals zero.

is that supposed to say if i plug in zero for x i get zero?

and even if i treat $\delta(\sqrt{x})$ as $\delta(\sqrt{x}-0)$
i still don't have a function to plug zero into.

i am still quite confused about what i am trying to do.

11. Mar 5, 2009

### George Jones

Staff Emeritus
Section 7.4.d Composition of $\delta$ with a function, from Mathematics for Physics and Physicists by Walter Appel.

The idea, as usual, is to to use the distribution obtained by integrating a locally integrable function against test functions to motivate a more general definition.

1. Use locally integrable $g$ and integration to generate a distribution $G$.

2. Use locally integrable $g \circ f$ and integration to generate a distribution denoted by $G \circ f$.

3. If $x$ is the integration variable in 2., make the substitution $y = f(x)$.

4. Relate $G \circ f$ to $G$.

5. Use 4. to motivate the definition of $T \circ f$ in tems of an arbitrary distribution $T$ and differentiable and bijective function $f$.

If I get time tomorrow, I might type in the details.

Last edited: Mar 5, 2009