# Proving a Dirac delta property

1. Aug 30, 2015

### carlosbgois

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

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

2. Relevant 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.

3. 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?

2. Aug 30, 2015

### blue_leaf77

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.

3. Aug 30, 2015

### vela

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

4. Aug 31, 2015

### HallsofIvy

Staff Emeritus
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.

5. Aug 31, 2015

### 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