# How Can I Integrate with Dirac Delta in this Expression?

In summary, when integrating a function with a Dirac Delta term, the order of integration does not matter and the substitution u = vk is useful when integrating with respect to the variable in the delta function.

## Homework Statement

I need to integrate this expression :

P(k, w) = A * δ(w-k*v) * f(k, w)

A is constant and δ, Dirac Delta.

## Homework Equations

[/B]
There is double integration :

I = ∫0 dk ∫0 P(k,w) dw
= A ∫∫0 δ(w-k*v) * f(k, w) dw dk

## The Attempt at a Solution

[/B]
I'm confused with Delta Dirac for calculating this integral.

Let us proceed the w-variable integration firstly. Since the δ-term impose w = k*v, we need to calculate :
0 f(k=w/v, w) dw
But after that, there is no longer any k-dependence...

Of course I'm wrong.. can you explain me how to proceed ??

I believe you could do the integration in either order, ## dk ## or ## d \omega ## first. The delta function should get removed with the first integration, and both methods should give the same answer. When doing the ## dk ## integration first, I believe the factor ## v ## will affect the delta function, and the substitution ## u=vk ## would be useful. ## \\ ## Suggestion: Try something such as ## \int\limits_{0}^{+\infty} \int\limits_{0}^{+\infty} \exp(-\omega^2) \, \delta(\omega-vk) \, d \omega \, dk ##. When you do the ## dk ## integration first, nothing happens to the ## \omega ## in the exponential, since any ## k ## term would get converted, but ## \omega ## terms are unaffected. And yes, to get agreement with the answer where you do the ## d \omega ## first, you need the substitution ## u=vk ## when doing ## dk ## first. With this substitution, ## dk=\frac{du}{v} ##, and the delta function with the ## du ## integration integrates to unity. ## \\ ## The process when the ## d \omega ## integration is performed first is a little more straightforward. ## \\ ## (Note: Here we let ## f(k, \omega)=\exp(-\omega^2) ## with no ## k ## dependence).

Last edited:
fresh_42

## Homework Statement

I need to integrate this expression :
P(k, w) = A * δ(w-k*v) * f(k, w)
A is constant and δ, Dirac Delta.

## Homework Equations

There is double integration :
I = ∫0 dk ∫0 P(k,w) dw
= A ∫∫0 δ(w-k*v) * f(k, w) dw dk

## The Attempt at a Solution

I'm confused with Delta Dirac for calculating this integral.
Let us proceed the w-variable integration firstly. Since the δ-term impose w = k*v, we need to calculate :
0 f(k=w/v, w) dw
You're forcing w to be a function of k but this is not right. k and w are both variables. Go
I = A ∫ ∫ f(k,w) δ(w - kv) dw dk
Sampling characteristic of the delta function leads to
I = A f(k, w=kv)dk
provided (lower limit of integration) < kv < (upper limit of integration) over w.

You are right rude man, thank's for your help :)

You are right rude man, thank's for your help :)
Most welcome!

## 1. What is the Dirac delta function and why is it important in integration?

The Dirac delta function, denoted by δ(x), is a mathematical function that is defined to be zero everywhere except at the origin, where it is infinitely tall and narrow. It is important in integration because it allows us to model point sources or impulses in a continuous system, and thus helps us solve problems in physics, engineering, and other fields.

## 2. How is the Dirac delta function used in integration?

The Dirac delta function is used in integration as a tool for solving integrals involving discontinuous functions or functions with singularities. It is commonly used to represent a point charge or point mass in physics, and allows us to simplify complex integrals by replacing them with simpler ones involving the delta function.

## 3. What are the properties of the Dirac delta function?

The Dirac delta function has several important properties, including the sifting property, scaling property, and sampling property. The sifting property states that the integral of the delta function over any interval that includes the origin is equal to 1. The scaling property allows us to manipulate the delta function by multiplying it with a constant. The sampling property states that the delta function can be used to extract the value of a function at a specific point.

## 4. Can the Dirac delta function be integrated like a regular function?

No, the Dirac delta function cannot be integrated like a regular function because it is not a true function in the traditional sense. It is a generalized function or distribution that is defined in terms of its properties and not by a specific formula. However, we can still use it in integration by treating it as a limit of a sequence of functions that approximate it.

## 5. How does the Dirac delta function relate to the Kronecker delta function?

The Dirac delta function and the Kronecker delta function are closely related, but they are used in different contexts. The Dirac delta function is defined in the context of continuous systems, while the Kronecker delta function is defined in the context of discrete systems. Both functions have similar properties, but they are defined differently and cannot be directly interchanged in integrals or sums.

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