Proving the Delta Function Property: \delta(ax) = {\delta(x) \over {|a|}}

[Pacopag]

Homework Statement

I would like to prove that $$\delta(ax)={\delta(x) \over {|a|}}$$.
My problem is that I don't know how the absolute value brackets arise.

Homework Equations

$$\int_{-\infty}^{\infty} \delta(x)dx = 1$$

The Attempt at a Solution

I start from $$\int_{-\infty}^{\infty} \delta (ax) dx$$, and make the substitution $$u=ax$$, so that $$du=adx$$. Then I arrive at the answer, but without the absolute value brackets.
I think that it is true that $$\delta(ax)=\delta(-ax)=\delta(|a|x)$$. I realize that if a<0, then we would reverse the direction of integration upon the substitution for u, so we would get the negative of the expected result. So do we just manually insert the absolute value brackets to make sure we get the correct sign, or is there a way to get the absolute value brackets to come out in the math?

 

[jhicks]

I think that it is true that $$\delta(ax)=\delta(-ax)=\delta(|a|x)$$... or is there a way to get the absolute value brackets to come out in the math?

That's not math? . If you aren't convinced yourself, you can apply the definition of the delta function to see that this is indeed true.

 

[eok20]

Everything you said is correct and the absolute value comes out if you consider the two cases a>=0 and a < 0. If a < 0 then, as you said, you get that $$\int_{-\infty}^{\infty} \delta (ax) dx = \int_{\infty}^{-\infty} \frac{\delta (u)}{a} du = -\int_{-\infty}^{\infty} \frac{\delta (u)}{a} = \int_{-\infty}^{\infty} \frac{\delta (u)}{|a|}$$ since -a = |a|.

 

[Pacopag]

Great! Thank you both.