Dirac Delta Integral: Compute \int_ {-\infty}^\infty e^{ikx}δ(k^2x^2-1)dx

[galactic]

Homework Statement

compute the integral:

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

Homework Equations

none that I have

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

 

[Dick]

Homework Statement

compute the integral:

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

Homework Equations

none that I have

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

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.

 

[SammyS]

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.

That's very cool (in my estimation).

Direct link to the section Dick mentioned: LINK .

 

[galactic]

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

 

[Dick]

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

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

 

[galactic]

thank you!

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

 

[Dick]

thank you!

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

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?

 

[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"

 

[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}]

 

[Dick]

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"

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.

 

[Dick]

\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?

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.

 

[galactic]

oooohhhhh I gotcha, many, many thanks for the clarification

 

[galactic]

with your correction I got

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

 

[Dick]

with your correction I got

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

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.

 

[galactic]

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

 

[Dick]

with your correction I got

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

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

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.

 

[galactic]

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

finally all good :D ?

 

[galactic]

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

finally all good :D ?

sorry double posted by accident

 

[Dick]

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

finally all good :D ?

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

 

[galactic]

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

 

[Dick]

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

No problem. Here's hoping you got cos(1)/|k|.

 

[galactic]

Yea!