Dirac delta integral

1. Jan 27, 2013

galactic

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

compute the integral:

$$\int_ {-\infty}^\infty \mathrm{e}^{ikx}\delta(k^2x^2-1)\,\mathrm{d}x$$

2. Relevant equations

none that I have

3. The attempt at a solution

I don't actually have any work by hand done for this because this is more complex than any dirac delta integral I have done...I assume that we must break down the x^2 in the delta function because there would be 2 roots, and thus, 2 answers? I'm not totally sure how to start this problem and can't find anything on the internet like it.

I know the delta function (from finding the roots of x) occurs at $$x=-1/k$$ and $$x=1/k$$. But not sure where to go from there

2. Jan 27, 2013

Dick

Here's something on the internet that's relevant. http://en.wikipedia.org/wiki/Dirac_delta_function Skip down to the "Composition with a function" section.

3. Jan 27, 2013

SammyS

Staff Emeritus
That's very cool (in my estimation).

4. Jan 27, 2013

galactic

I ended up getting cos(k) as the final answer. Is that right?

5. Jan 27, 2013

Dick

How did you get that? It's not what I get.

6. Jan 27, 2013

galactic

thank you!

my last step before the final answer was $$\frac{1}{2}(e^{ik}+e^{-ik}) = cos(k)$$

7. Jan 27, 2013

Dick

You're welcome. But the cool reference I pointed you to says the delta function is the same as the sum of δ(1/k) and δ(-1/k) each divided by |g'| evaluated at those points. What happened to all of that stuff?

8. Jan 28, 2013

galactic

i started out with:

$$\delta[(x+1)(x-1)] = \frac{1}{2}[\delta(x+1)+\delta(x-1)]$$
$$\frac{1}{2} \int_{-\infty}^\infty e^{ikx}\delta(x+1)\,dx + \frac{1}{2} \int_{-\infty}^\infty e^{ikx} \delta(x-1)\,dx$$

from there i boiled it down to cos(k)

did I mess up in the beginning?

I just read your last post..sorry i was in the middle of typing this while you posted last. So i found this in the composition section function however as you point out it looks like I started off wrong and need the include the "k"

9. Jan 28, 2013

galactic

$$\frac{1}{2} \int_{-\infty}^\infty e^{ikx}\delta(x+\frac{1}{k})\,dx + \frac{1}{2} \int_{-\infty}^\infty e^{ikx} \delta(x-\frac{1}{k})\,dx$$

that should be how I solve it using what you said, correct?

and the final solution i get is $$\frac{1}{2}[e^{i}+e^{-i}]$$

10. Jan 28, 2013

Dick

Yes, if you had δ(x^2-1) that would be fine. But you have δ(k^2*x^2-1). Makes a difference. You had the right general idea originally that the delta function should involve x=1/k and x=(-1/k). You just need to correct that a little.

11. Jan 28, 2013

Dick

You're getting closer. Look at the |g'(x)| factor you are supposed to divide those delta functions by. g(x)=k^2*x^2-1.

12. Jan 28, 2013

galactic

oooohhhhh I gotcha, many, many thanks for the clarification

13. Jan 28, 2013

galactic

$$\frac{1}{2xk^2}[e^i+e^{-i}]$$

Last edited: Jan 28, 2013
14. Jan 28, 2013

Dick

Closer yet. But if g(x)=k^2*x^2-1, g'(x)=2*k^2*x. Putting x=1/k I get |g'(1/k)|=|2k|, not 2k^2.

15. Jan 28, 2013

galactic

Oh i see how you did that plugging in 1/k for x, I didnt think to do that

16. Jan 28, 2013

Dick

Right. There shouldn't be any x in the final answer. You should put x=1/k,-1/k into g'(x) and the absolute value is pretty important.

17. Jan 28, 2013

galactic

$$\frac{1}{2k}[e^i+e^{-i}]$$

finally all good :D ?

18. Jan 28, 2013

galactic

$$\frac{1}{2k}[e^i+e^{-i}]$$

finally all good :D ?

sorry double posted by accident

19. Jan 28, 2013

Dick

Absolute value. Absolute value. The bars around |g'| mean something. And you can simplify e^i+e^(-i) just like you did before.

20. Jan 28, 2013

galactic

many thanks for the help and putting up with my careless errors at midnight :D