How Can I Integrate with Dirac Delta in this Expression?

[Ado]

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 ??

Thanks in advance !

 

[Charles Link]

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).

 

[rude man]

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.

 

[Ado]

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

 

[rude man]

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

Most welcome!