Moving a function under an integral

[NanakiXIII]

I've got an integral of the form

$$\int d^3x e^{-i \vec{k} \cdot \vec{x}} \int d^3k e^{i \vec{k} \cdot \vec{x}} f(\vec{k})$$

and I'm wondering whether or not the following is a valid approach. I want to move the factor

$$e^{-i \vec{k} \cdot \vec{x}}$$

under the k-integral, so I relabel my k so that I end up with

$$\int d^3x \int d^3k e^{i (\vec{k}-\vec{k}') \cdot \vec{x}} f(\vec{k}).$$

One now just needs to remember that the primed k is actually the same as the old k, but we don't want to integrate over it. Now the x-integral turns the exponential factor into a delta-function, so that we get (apart from some factors of $2\pi$)

$$\int d^3k \delta(\vec{k}-\vec{k}') f(\vec{k}),$$

which just yields

$$f(\vec{k}') = f(\vec{k}).$$

Now, maybe it's just me, but what I just did sounds too easy to be true, but I'm not sure what might be flawed about my reasoning. Can anyone tell me whether there is anything wrong with it?

 

[tiny-tim]

Hi NanakiXIII!

One now just needs to remember that the primed k is actually the same as the old k, but we don't want to integrate over it. Now the x-integral turns the exponential factor into a delta-function, so that we get (apart from some factors of 2π)

Your very first line didn't make sense …

how you could you possibly start with the same k both outside and inside an integral?

(where does this come from?)

 

[NanakiXIII]

Hi NanakiXIII!


Your very first line didn't make sense …

how you could you possibly start with the same k both outside and inside an integral?

(where does this come from?)

I'm not sure what you mean. Why couldn't there be a function of the same variable k outside the integral, as well as a function of k inside? Perhaps I didn't write it down clearly. My expression is

$$\int d^3x \left( e^{-i \vec{k} \cdot \vec{x}} \int d^3k \left( e^{i \vec{k} \cdot \vec{x}} f(\vec{k}) \right) \right)$$

This integral arose in a QFT problem I was doing. Basically there's a Fourier expansion for a function (the integral over k) and I'm transforming that back (but with a slightly different transformation).

 

[tiny-tim]

perhaps I should have been more specific …

how you could you possibly start with the same k both outside an integral and as the dummy variable of integration inside it?

 

[NanakiXIII]

perhaps I should have been more specific …

how you could you possibly start with the same k both outside an integral and as the dummy variable of integration inside it?

I still don't understand. Perhaps you're tripping over some mathematical sloppiness I'm not aware of.

 

[tiny-tim]

I don't understand what makes you think that they were the same k in the first place …

One now just needs to remember that the primed k is actually the same as the old k, but we don't want to integrate over it. Now the x-integral turns the exponential factor into a delta-function, so that we get (apart from some factors of 2π)

 

[NanakiXIII]

I don't understand what makes you think that they were the same k in the first place …

Oh. I see what you mean now. I had forgotten what I was integrating over (i.e. that it's not an indefinite integral). I see why you reacted so confused. Thanks for helping me see that. It was just a sloppiness, but one I forgot to keep track of.

This makes things even easier. At least now I'm convinced it's also correct. You didn't see any error in the rest, right?

 

[tiny-tim]

No, seems fine!

 

[NanakiXIII]

Thanks for the help!