Proofs for Dirac delta function/distribution

[Wishe Deom]

[SOLVED] Proofs for Dirac delta function/distribution

Homework Statement

Prove that
<br /> \delta(cx)=\frac{1}{|c|}\delta(x)<br />

Homework Equations

\delta(x) is defined as
<br /> \delta(x)=\left\{\stackrel{0 for x \neq 0}{\infty for x=0}<br />

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

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

The Attempt at a Solution

Starting from

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

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

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

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

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

Evaluate left side.

<br /> \frac{1}{c}g(0)= \frac{1}{|c|}f(0)<br />

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


<br /> \frac{1}{c}f(0)= \frac{1}{|c|}f(0)<br />


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