# Dirac Delta Scaling

1. Jan 21, 2010

### donquixote17

Using the defining property of the dirac delta function,
$$\int{dx f(x) \delta(x-c)}$$

show that
$$\delta(ax)=\frac{1}{|a|}\delta(x)$$

I think all I need to do is make the right u substitutions and it will come out right, but I can't think of how to make the substitutions...after a long time working on this problem.

2. Jan 21, 2010

### Dick

You mean delta(ax)=delta(x)/|a|. Not c. Just substitute u=ax. Compare delta(x) with delta(u). Don't forget du=a*dx. And show what you are actually doing this time. The integral of f(x-c)*delta(x)*dx equals f(c). The integral of f(u-c)*delta(u)*du must also equal f(c). What's the relation between the two deltas?

Last edited: Jan 21, 2010
3. Jan 21, 2010

### donquixote17

Thanks Dick, I kind of get the right answer, but it's just not the way I was thinking about it, so starting with the defining property we have
$$\int{dx f(x) \delta(x-c)}$$

I make the u substitution u=ax, and du=a dx
$$\frac{1}{a}\int{du f(u/a) \delta(u/a-c)}$$

So that right there tells me that $$\delta(ax)=\frac{1}{a}\delta(x)$$, but the problem states that it should be $$\delta(ax)=\frac{1}{|a|}\delta(x)$$

So I'm not sure where the absolute value comes in.

4. Jan 21, 2010

### Dick

If a is negative, then the change of variable makes integration limits reverse from -infinity to +infinity to +infinity to -infinity. To turn it back into a proper delta function, you have to turn them back. That introduces an extra sign. See what I'm saying?

5. Jan 21, 2010

### donquixote17

yes that makes sense. Thanks!