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

• galactic
In summary: I finally did it! In summary, when computing the integral \int_ {-\infty}^\infty \mathrm{e}^{ikx}\delta(k^2x^2-1)\,\mathrm{d}x, it is important to note that the delta function must be divided by the absolute value of k^2x. Plugging in x=1/k and x=-1/k into g'(x)=k^2x^2-1 yields the final answer of \frac{1}{2k|k|}[e^i+e^{-i}].
galactic

## Homework Statement

compute the integral:

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

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

galactic said:

## Homework Statement

compute the integral:

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

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.

Dick said:
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).

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

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.

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.

oooohhhhh I gotcha, many, many thanks for the clarification

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

Last edited:
galactic said:

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

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.

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

galactic said:

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

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.

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

finally all good :D ?

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

finally all good :D ?

sorry double posted by accident

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.

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

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|.

Yea!

## 1. What is the Dirac Delta function?

The Dirac Delta function, denoted as δ(x), is a mathematical concept used in the field of calculus and analysis. It is defined as a function that is zero everywhere except at the point x=0, where it is infinite. It is often used to model concentrated point masses or impulses in physics and engineering.

## 2. What is the Dirac Delta Integral?

The Dirac Delta Integral, also known as the Dirac Delta distribution, is an extension of the Dirac Delta function. It is a mathematical tool that allows us to integrate functions containing the Dirac Delta function. It is defined as an integral over an infinitely small interval around the point where the Dirac Delta function is non-zero.

## 3. How do you compute the Dirac Delta Integral?

The Dirac Delta Integral can be computed using the properties of the Dirac Delta function. In the case of the integral given, we can use the property δ(ax) = δ(x)/|a| to simplify the integral. We can also use the fact that the integral of the Dirac Delta function over any interval containing the point of non-zero value is equal to 1. By applying these properties and manipulating the integral, we can solve for the value of the integral.

## 4. What is the significance of the given integral?

This integral is often encountered in physics and engineering, specifically in the study of wave propagation and quantum mechanics. It can be used to model the scattering of a wave by a potential barrier, and the resulting integral can provide information about the transmitted and reflected waves. It is also used in the study of the Schrödinger equation and the behavior of quantum particles.

## 5. Are there any practical applications of the Dirac Delta Integral?

Yes, the Dirac Delta Integral has many practical applications. It is used in signal processing to model impulses in a signal, in quantum mechanics to study the behavior of particles, and in engineering to analyze the response of systems to external forces. It is also used in the study of differential equations and Fourier analysis, among many other fields.

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