Proving a Dirac delta property

[carlosbgois]

Homework Statement

[/B]
Prove that $$\delta[a(x-x_1)]=\frac{1}{a}\delta(x-x_1)$$

Homework Equations

In my attempt I have used $$\delta(ax)=\frac{1}{a}\delta(x)$$ but I'm not sure I'm allowed to use it in this proof.

The Attempt at a Solution

Some properties of Dirac delta function are proven using a test function. Thence I tried $$I=\int f(x)\delta[a(x-x_1)]=$$

In the following I tried the substitution y = x - x₁, getting $$I=\int f(y+x_1)\delta(ay)=\frac{1}{a}\int f(y+x_1)\delta(y)=\frac{1}{a}f(x_1)=\frac{1}{a}\int f(x)\delta(x-x_1)$$

But I don't know if this proof is correct, as I have used a property similar to the one I'm trying to prove. Is it ok to do this?

 

[blue_leaf77]

Thence I tried $$I=\int f(x)\delta[a(x-x_1)]=$$

Next, do another integration with $\delta[a(x-x_1)]$ replaced by $\frac{1}{a}\delta(x-x_1)$, with the same $f(x)$. Then compare the results with the first integration, taking into account that $f(x)$ is an arbitrary function.

 

[vela]

Homework Statement

[/B]
Prove that $$\delta[a(x-x_1)]=\frac{1}{a}\delta(x-x_1)$$

Homework Equations

In my attempt I have used $$\delta(ax)=\frac{1}{a}\delta(x)$$ but I'm not sure I'm allowed to use it in this proof.

I don't think you're allowed to use that. Try the substitution $u = a(x-x_1)$.

 

[HallsofIvy]

What is your definition of the delta function? Exactly how you would prove this depends upon which definition you are using and I believe there are several.

 

[carlosbgois]

Hallsoflvy: I have used the definition by the integral $$\int f(x)\delta(x)=f(0)$$
vela: with this substitution it worked.

Thank you all for the help