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

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Homework Help Overview

The discussion revolves around the computation of the integral involving the Dirac delta function: \(\int_ {-\infty}^\infty e^{ikx}\delta(k^2x^2-1)\,\mathrm{d}x\). Participants explore the implications of the delta function's roots and the complexity of the integral, particularly in relation to the presence of \(k\) in the expression.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to identify the roots of the delta function, noting that there are two roots at \(x=-1/k\) and \(x=1/k\). There is uncertainty about how to proceed from this point, with some participants expressing confusion regarding the application of the delta function and its composition with other functions.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and corrections. Some have proposed methods for evaluating the integral using the properties of the delta function, while others are questioning the accuracy of their approaches and the necessary adjustments related to the factor of \(k\). There is a recognition of the importance of absolute values in the context of the derivative of the function within the delta function.

Contextual Notes

Participants are navigating the complexities of the Dirac delta function and its application in integrals, with specific attention to the implications of the variable \(k\) and the need for careful evaluation of derivatives. There is a shared acknowledgment of the challenges posed by the problem, particularly in relation to the roots and the evaluation process.

galactic
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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
 
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galactic said:

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

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

my last step before the final answer was \frac{1}{2}(e^{ik}+e^{-ik}) = cos(k)
 
galactic said:
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?
 
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"
 
\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
galactic said:
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.
 
  • #11
galactic said:
\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.
 
  • #12
oooohhhhh I gotcha, many, many thanks for the clarification
 
  • #13
with your correction I got

\frac{1}{2xk^2}[e^i+e^{-i}]
 
Last edited:
  • #14
galactic said:
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.
 
  • #15
Oh i see how you did that plugging in 1/k for x, I didnt think to do that
 
  • #16
galactic said:
with your correction I got

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

galactic said:
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.
 
  • #17
\frac{1}{2k}[e^i+e^{-i}]

finally all good :D ?
 
  • #18
\frac{1}{2k}[e^i+e^{-i}]

finally all good :D ?

sorry double posted by accident
 
  • #19
galactic said:
\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.
 
  • #20
many thanks for the help and putting up with my careless errors at midnight :D
 
  • #21
galactic said:
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|.
 
  • #22
Yea!
 

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