Proofs for Dirac delta function/distribution

[Wishe Deom]

[SOLVED] Proofs for Dirac delta function/distribution

Homework Statement

Prove that
$$\delta(cx)=\frac{1}{|c|}\delta(x)$$

Homework Equations

$$\delta(x)$$ is defined as
$$\delta(x)=\left\{\stackrel{0 for x \neq 0}{\infty for x=0}$$

It has the properties:
$$\int^{\infty}_{-\infty}\delta(x)dx=1$$
$$\int^{\infty}_{-\infty}f(x)\delta(x-a)dx=f(a)$$

Two expressions $$D_{1}$$ and $$D_{2}$$ involving $$\delta(x)$$ are considered equivalent if
$$\int^{\infty}_{-\infty}f(x)D_{1}dx = \int^{\infty}_{-\infty}f(x)D_{2}dx$$

The Attempt at a Solution

Starting from

$$\int^{\infty}_{-\infty}f(x)\delta(cx)dx = \int^{\infty}_{-\infty}f(x)\frac{1}{|c|}\delta(x)dx$$

I make a substitution u=cx, and du=cdx.

$$\frac{1}{c}\int^{\infty}_{-\infty}f(u/c)\delta(u)du = \frac{1}{|c|}\int^{\infty}_{-\infty}f(x)\delta(x)dx$$

Substitute f(u/c)=g(u), and on right side evaluate integral.

$$\frac{1}{c}\int^{\infty}_{-\infty}g(u)\delta(u)du = \frac{1}{|c|}f(0)$$

Evaluate left side.

$$\frac{1}{c}g(0)= \frac{1}{|c|}f(0)$$

By x=u/c, g(0)=f(0), so


$$\frac{1}{c}f(0)= \frac{1}{|c|}f(0)$$


So, I got this far, but I don't understand where the absolute value on the right side comes from.

 

[Oerg]

see, when c is negative, your limits on the integral after the substitution change signs, therefore, for negative c, you should have a modulus sign.

 

[Dick]

If c is negative, when you do the substitution u=cx the u-limits become +infinity to -infinity, instead of -infinity to +infinity for x.

 

[Wishe Deom]

Ok, thank you both. In retrospect, that was kind of obvious :P