How Do Delta Functions Behave Under Scaling Transformations?

[kq6up]

Homework Statement

Delta functions said to live under the integral signs, and two expressions ($D_1(x)$ and $D_2(x)$) involving delta functions are said to be equal if:

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

(a) Show that:

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

Where $c$ is a real constant. (Be sure to check the case where $c$ is negative.)

Homework Equations

Posted above.

The Attempt at a Solution

Let $u=cx$ $\therefore$ $\frac{1}{c}du=dx$

This yields:

$\frac{1}{c}\int_{-\infty}^{\infty}{f(u/c)\delta (u)du}=\frac{1}{|c|}\int_{-\infty}^{\infty}{f(x)\delta (x)dx}$. This works for a test case where $c > 0$, but obviously fails when $c < 0$. I am not sure where this absolute value came from.

Hints please.

Thanks,
Chris

 

[TSny]

Don't forget to think about the limits of integration when making a change of integration variable.

 

[kq6up]

Don't forget to think about the limits of integration when making a change of integration variable.

Isn't still negative infinity to infinity. Oh wait, not of the sign changes, so the sign flips. I think I got it now.

Thanks,
Chris

 

[ChrisVer]

you can choose to try for $f(x)=1$

 

[kq6up]

I used f(x)=x+2, and it works if I consider that using $-c$ would flip the signs of the infinities on the integral thus flipping the sign back to positive.

Chris