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Proving the scaling property of the Delta function

  • Thread starter caesius
  • Start date
  • #1
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Homework Statement



Prove that [tex]\delta(at)=\frac{1}{abs(a)}\delta(t)[/tex]

Hint: Show that [tex]\int\phi(t)\delta(at)dt=\frac{1}{abs(a)}\phi(0)[/tex]

(the limits of integration are from -inf to +inf btw, I couldn't find how to put them in..)

Homework Equations





The Attempt at a Solution



Ok. I understand that the integral is only defined for at = 0, i.e., t = 0. But if I follow this logical then I evaluate the integral to be phi(0), not phi(0)/a. Where does the a come from?

And I'm not entirely sure how proving that relationship helps with the proof of the scaling function...
 

Answers and Replies

  • #2
CompuChip
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Homework Helper
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That trick only works when the argument to the delta is the bare integration variable.
So try substituting u = a t, so you get something like
[tex]\int_{-\infty}^{\infty} f(u) \delta(u) \, du = f(0)[/tex]
(and click the formula to see how I did the boundaries)
 
  • #3
24
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Ok I've substituted x = at, now I get

[tex]
\int_{-\infty}^{\infty}\phi(\frac{x}{a})\delta(x)d\frac{x}{a} = \frac{1}{a}\int_{-\infty}^{\infty}\phi(\frac{x}{a})\delta(x)dx = \frac{1}{a}\phi(\frac{0}{a}) = \frac{1}{a}\phi(0)
[/tex]

I don't quite see how I'm allowed to take the 1/a out of the d/dx operater and put it at the front though. And why should it be the abs(a) not a?

And I still can't see how this helps prove anything.

Cheers, Benjamin
 
  • #4
CompuChip
Science Advisor
Homework Helper
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You are allowed to take the a outside, because it is just a constant: du = a dt, so dt = (1/a) du, and you can put constants outside the integration as in
[tex] \int \tfrac12 x^2 \, dx = \tfrac12 \int x^2 \, dx[/tex]

The absolute value of a comes from transforming the integration boundaries. If a < 0, then -infinity goes to +infinity in the transformed integral, and vice versa. However, in the final formula you simply want -infinity to be the lower boundary, and +infinity as the upper boundary. So if a < 0 you have to swap them, at the cost of a minus sign, whereas for a > 0 you don't. Do you have to multiply the whole thing by sgn(a), and then you can shorten sgn(a) / a to 1/abs(a).
 

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