(adsbygoogle = window.adsbygoogle || []).push({}); [SOLVED] Proofs for Dirac delta function/distribution

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

Prove that

[tex]

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

[/tex]

2. Relevant equations

[tex]\delta(x)[/tex] is defined as

[tex]

\delta(x)=\left\{\stackrel{0 for x \neq 0}{\infty for x=0}

[/tex]

It has the properties:

[tex]

\int^{\infty}_{-\infty}\delta(x)dx=1

[/tex]

[tex]

\int^{\infty}_{-\infty}f(x)\delta(x-a)dx=f(a)

[/tex]

Two expressions [tex]D_{1}[/tex] and [tex]D_{2}[/tex] involving [tex]\delta(x)[/tex] are considered equivalent if

[tex]

\int^{\infty}_{-\infty}f(x)D_{1}dx = \int^{\infty}_{-\infty}f(x)D_{2}dx

[/tex]

3. The attempt at a solution

Starting from

[tex]

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

[/tex]

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

[tex]

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

[/tex]

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

[tex]

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

[/tex]

Evaluate left side.

[tex]

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

[/tex]

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

[tex]

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

[/tex]

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

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# Homework Help: Proofs for Dirac delta function/distribution

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