Integration using delta function and step function

In summary, the conversation discusses the evaluation of a specific integral using the Dirac delta function and the Heaviside step function. The correct answer is supposed to be 1/omega, but it is unclear if there was a mistake made in the steps leading up to that answer. The use of a change of variable is suggested as a potential way to simplify the integral.
  • #1
spaghetti3451
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I would like to evaluate the following integral:

##\displaystyle{\int_{-\infty}^{\infty} dp^{0}\ \delta(p^{2}-m^{2})\ \theta(p^{0})}##

##\displaystyle{= \int_{-\infty}^{\infty} dp^{0}\ \delta[(p^{0})^{2}-\omega^{2}]\ \theta(p^{0})}##

##\displaystyle{= \int_{-\infty}^{\infty} dp^{0}\ \delta[(p^{0}+\omega)(p^{0}-\omega)]\ \theta(p^{0})}##

##\displaystyle{= \frac{1}{2\omega}\int_{-\infty}^{\infty} dp^{0}\ \delta[(p^{0}+\omega)+(p^{0}-\omega)]\ \theta(p^{0})}##

##\displaystyle{= \frac{1}{2\omega}\int_{-\infty}^{\infty} dp^{0}\ \delta(2p^{0})\theta(p^{0})}##

Have I made a mistake somewhere?

The answer is supposed to be ##\frac{1}{\omega}## but I'm not sure how to proceed next.
 
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  • #2
I don't follow some of your steps so far. But if you want to be able to get rid of the integral, try a change of variable, making the variable over which we are integrating ##u=2p^0##.

The integral can then be evaluated quite simply. However I then get the answer ##1/4\omega##, which is why I am wondering about your earlier steps.
 
  • #3
Would you recommend me to post this on the Relativity forum?

This question is better suited for that forum.
 
  • #4
[tex]
\delta(x^2 - \alpha^2) = \frac{1}{2|\alpha|} \left[\delta(x-\alpha) + \delta(x+\alpha) \right] \neq \frac{1}{2|\alpha|} \left[\delta( (x-\alpha) + (x+\alpha)) \right]
[/tex] which is what you seem to have done.
Are you sure the answer is supposed to be ##1/\omega## and not ##1/2\omega## though?
 
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1. What is the purpose of using delta function and step function in integration?

The delta function is used to represent an infinitesimally narrow pulse, while the step function is used to represent a sudden change in a function's value. These functions are useful in integration because they allow us to model and solve problems involving sudden changes or impulses in a system.

2. How do you use the delta function in integration?

The delta function is used in integration by multiplying it with another function, known as the integrand. This results in a new function that is zero everywhere except at the point where the delta function is located. The value of the new function at this point is equal to the value of the integrand at that point.

3. What is the relationship between the delta function and the step function?

The delta function and the step function are closely related. The delta function is the derivative of the step function, and the step function is the integral of the delta function. This means that they are inverse operations of each other.

4. Can you use the delta function and step function to solve differential equations?

Yes, the delta function and step function can be used to solve differential equations that involve impulses or sudden changes. By representing these changes using these functions, we can solve the differential equations using integration techniques.

5. Are there any real-world applications of integration using delta function and step function?

Yes, these functions have many real-world applications, particularly in physics and engineering. They are used to model and solve problems involving impulse forces, sudden changes in voltage or current, and other similar phenomena. They are also used in signal processing and control systems.

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